Questão 4 (Física)

**Considering gas dynamics and pistons** I’m thinking about the piston at a turning point where it’s momentarily at rest, but the gas might still be dynamic. In ideal gas thermodynamics, I might use the adiabatic equation for uniform equilibrium states, but the process isn't necessarily quasi-static, which means the PV^γ relation generally wouldn’t apply. I wonder, at maximum displacement if the piston is transient, how does that affect gas oscillations? There’s definitely a lot to consider here!**Evaluating gas behavior** I'm looking at the equation and thinking about how rearranging it leads to the condition where the sudden load creates irreversible expansion. I wonder if they expect energy conservation principles when applying the adiabatic law. It doesn't seem to fit with the entropy generation involved. I need to consider the pressure and internal energy relationships in a gas confined within a piston, especially at maximum height. The dynamics of movement seem complex since acceleration isn't necessarily zero despite the piston being stationary.**Characterizing gas behavior** I'm considering how pressure at the turning point might be different from mg/A and how this relates to the gas's thermodynamic state. For an ideal gas with no internal friction, the dynamic behavior can be complex, but textbooks often simplify it. Using the first law, I can derive a differential equation for internal energy. However, I'm recognizing that a nonstatic situation may lead to discrepancies between boundary and bulk pressures. The sudden removal of sand likely causes an initial net upward force, leading to adiabatic gas expansion.**Finding maximum displacement** I'm looking to find the maximum displacement by integrating energy using the PV law. I realize I need to formulate an equation that relates hmax, m, M, and hi, although it might not have a closed form. This requires careful consideration of how these variables interact, and I need to ensure that the energy calculations align with the proper physical principles. It seems like there’s quite a bit to work through to get a clear expression!**Considering gas dynamics** I'm exploring the relationship between sudden sand removal and gas behavior. It seems the exam might expect me to use dynamic equations rather than quasi-static ones. Part b might be designed to test my understanding of energy conservation for gases. They could want the pressure law regarding adiabatic processes, similar to part a. If I consider an instantaneous change versus a gradual one, the endpoint could differ. However, if entropy is conserved, then the process isn't reversible, so I might need to model it using the first law together with mechanical motion for uniform gas pressure.**Considering gas processes** It sounds like the prompt is asking us to treat gas using quasi-equilibrium principles. The first part specifies a quasistatic process, while the second part refers to instant removal, which might suggest a distinction. If part b involves adiabatic conditions that aren't quasi-static, I'll need to apply energy conservation and perhaps maximum force. I need to identify a valid answer mathematically based on the information provided.**Analyzing energy balance** I’m considering energy balance and the equation U=3/2 nRT. But it seems there’s some misunderstanding at the highest point, as pressure from force balance contradicts. I think the desired equation might involve gas internal energy combined with mechanical energy. To find U_f, they could be using maximum velocity at zero, under specific conditions. Maybe if they treat "maximum height" as equilibrium without friction — but that seems off. I could derive h_max using first-law integration with gas work, but the process path affects the work integral.

For part a, I set up the adiabatic relation with γ=5/3 for the monoatomic gas, using initial pressure from the weight of sand plus piston and final pressure from just the piston, then solve for the final height: h_f = h_i((m+M)/m)^{3/5}. For part b, I'm now considering what happens when the sand is removed suddenly instead of quasi-statically, thinking through how the piston's mass affects the work done by the gas as it expands. Since the process isn't quasi-static, I treat it via energy conservation instead: using that the piston has zero velocity at maximum height, I relate the gas's internal energy change to the piston's gravitational potential energy change. I compute U_i and U_max in terms of pressure, volume, and height using the adiabatic relation, then set U_i − U_max equal to mg(h_max − h_i), yielding a transcendental equation linking h_max to m, M, and h_i. I also note there's an alternative view where the gas doesn't follow a quasi-static adiabatic path at all during this sudden expansion.

For part a, I'm using adiabatic condition for a monoatomic gas with γ=5/3: initial pressure is (m+M)g/A and final is mg/A, so applying PV^γ=const gives h_f = h_i((m+M)/m)^(3/5). For part b, I'm considering sudden removal of the mass — the piston shoots up and the gas expands, but now I need to think about whether this expansion is quasi-static or not since the piston accelerates. Treating the gas as still following PV^γ=const internally while energy conservation governs the piston's rise, I equate the work done by the gas (via the adiabatic energy expression) to mg times the height gained, since the piston has zero velocity at maximum height. This gives a transcendental relation: 3(m+M)h_i[1-(h_i/h_max)^(2/3)] = 2m(h_max - h_i), which should be the equation the problem wants.

For the monoatomic gas with γ=5/3, I set up the initial pressure from piston plus weight and the final pressure from just the piston's weight, then use the adiabatic relation to get the quasi-static final height in terms of mass ratios. For the sudden removal case, I'm now thinking through the energy balance as the piston moves against the constant force of its own weight with vacuum above it. Applying energy conservation, the work done by the expanding gas equals the loss in internal energy, and at the piston's maximum height the kinetic energy is zero, giving me mg(h_max - h_i) = (3/2)nR(T_i - T_max). The tricky part is that at the turning point the piston isn't in pressure equilibrium—its velocity is zero but acceleration isn't, so I can't just assume gas pressure equals mg/A there, meaning I need another relation connecting T_max and h_max rather than relying on quasi-static assumptions. I'm modeling the gas as slowly oscillating with a reversible adiabatic relation, so the internal energy scales as U(h) ∝ h^{-2/3} given the initial state. Setting up energy conservation between gravitational potential energy and this internal energy expression should give me an equation for h_max in terms of h_i. One root is trivially h=h_i, so the physically relevant solution is the other root above h_i, since removing sand lowers the equilibrium pressure and lets the piston rise higher before settling at some intermediate h_f.

The problem is from the 2027 ITA Physics second phase. The problem involves a vertical cylinder with an ideal monoatomic gas and a thermally insulating piston. The piston has negligible mass and slides without friction. Sand grains of total mass M rest on the piston. The system is initially in thermal and mechanical equilibrium at height h_i above the base. The problem has two parts. First, find the final height h_f when sand is removed quasi-statically. Second, find the maximum height h_max when all sand is removed instantly. The walls and piston are adiabatic and frictionless. The small-angle approximation hint may not be needed. For the quasi-static removal, the process is adiabatic and reversible. The initial pressure balances the weight of the piston plus sand, and the final pressure balances only the piston's weight. Using the adiabatic relation P_i V_i^{5/3} = P_f V_f^{5/3}, the final height becomes h_f = h_i [(m+M)/m]^{3/5}. The instantaneous removal of sand makes the process irreversible, so the adiabatic relation P_i V_i^{5/3} = P_f V_f^{5/3} cannot be applied. Energy conservation must be used instead to find the maximum height. The instantaneous removal causes oscillations and non-uniform pressure, so the process is irreversible. At maximum height the piston momentarily stops, meaning its velocity is zero. The energy conservation approach will focus on the gas and piston system, since the sand is no longer present after removal. The system is the gas plus the piston. The initial state has zero velocity and internal energy U_i. The only external force is gravity. The gas does work against this force, and the adiabatic condition means this work equals the decrease in internal energy. The piston's gravitational potential energy must also be included in the energy balance. The total energy of the gas and piston is conserved. The adiabatic condition ensures that work done by the gas equals its decrease in internal energy, which is converted into the piston's mechanical energy. Whether pressure remains uniform throughout the expansion is still uncertain. Energy conservation applies even to irreversible processes, provided the work is tracked correctly. In this idealized problem, however, the internal energy of the gas depends on temperature, which remains unknown at maximum height. The energy conservation equation is U_i - U_f = m g (h_max - h_i). Since the gas internal energy equals n C_v T, this becomes n C_v (T_i - T_f) = m g (h_max - h_i). At maximum height the piston accelerates and pressure is not uniform, so T_f and V_f are not related by equilibrium. The final temperature cannot be determined without knowing the process. The problem likely asks for an equation rather than a numerical solution. The small-angle approximation hint may indicate small oscillations, but it is unclear how it relates to the maximum height. Perhaps the piston oscillates after the sand is removed. The equilibrium shifts from (m+M)g to mg. Small oscillation theory would linearize this, but removing all sand may produce large-amplitude motion. The maximum height could then be found using adiabatic oscillation mechanics. The oscillation can be treated as a slow, adiabatic process. If the piston's motion is slow enough, pressure remains uniform and the gas stays in equilibrium. The adiabatic relation P V^{5/3} = const then applies. This allows the maximum height to be found using energy conservation. The internal energy of the monoatomic gas is U = P V / (γ − 1). The total mechanical energy E = U + m g h + (1/2) m v² is constant. The initial state has zero velocity and height h_i. The initial internal energy is U_i = (m+M) g h_i / (γ − 1). At maximum height the adiabatic relation gives U_max = U_i (h_i / h_max)^{γ−1}. Energy conservation then yields U_i − U_max = m g (h_max − h_i). The energy conservation equation is now fully derived. Substituting γ − 1 = 2/3 and U_i = (3/2)(m+M)g h_i gives the final relation: (3/2)(m+M)h_i [1 − (h_i / h_max)^{2/3}] = m(h_max − h_i). This equation relates the maximum height to the masses and initial height. The derivation is internally consistent because the adiabatic work done by the gas equals the change in the piston's mechanical energy. The dynamic maximum height exceeds the quasi-static final height. The oscillation amplitude is determined by the energy equation, with the piston reaching its highest point at the turning point. The derived equation depends only on the masses and heights, with the gas constant, cross-sectional area, and gravity cancelling out. This confirms it is independent of the specific gas properties. The energy conservation equation was verified by deriving it directly from the first law of thermodynamics. For a small displacement, dU = −P dV, and the piston’s mechanical work is P A dh − mg dh = m v dv. Combining these gives −dU − mg dh = d(KE), confirming that U + m g h + KE is constant. The problem explicitly states the system is frictionless. Without friction, the piston oscillates indefinitely without damping, so the maximum height is reached at the first turning point. The reversibility of the thermodynamic process is now being reconsidered. Even without mechanical friction, rapid piston motion can create shock waves and non-uniform pressure, which would violate the adiabatic relation. However, in idealized exam problems the assumption is that the process remains slow enough to be treated as adiabatically reversible. The fact that the final equation depends only on the masses and heights confirms this conservative model. The final answer will be presented as a) h_i ((m+M)/m)^{3/5} and b) the energy conservation equation. A small mass limit was checked: when M is much smaller than m, the equilibrium height shifts by approximately (3/5)(M/m)h_i. The small mass limit is being checked against the derived equation. For a harmonic oscillator released from rest at an offset position, the amplitude equals the equilibrium shift, giving h_max ≈ h_i [1 + (6/5)(M/m)]. The energy conservation equation is now being expanded for small δ = M/m to verify this scaling.

We need answer physics problem in Portuguese. Need solve. Let's restate. Problem: Vertical cylinder, cross-sectional area A, insulating walls (adiabatic), contains n moles ideal monatomic gas. Closed at top by piston negligible thickness mass m, also thermally insulating, can slide vertically without friction. Initially, a container of negligible mass containing amount of sand total mass M at rest on piston. System in thermal and mechanical equilibrium, piston height h_i relative to base. g local, universal gas constant R, vacuum above piston. Find: a) Sand removed slowly quasi-static. Determine final height h_f of piston when no more sand, as function of m, M, h_i. b) New situation entire sand removed instantaneously. Determine equation relating maximum height h_max attained by piston and m, M, h_i. Need solve. Need be careful thermodynamics. Cylinder adiabatic? Walls insulating, piston insulating. Gas monatomic. Sand removal. Part a: Quasi-static adiabatic expansion? Initially external pressure on gas = (M+m)g/A? Plus atmospheric? Vacuum above piston, so force on piston from gas + weight? Let's analyze equilibrium. Cylinder vertical, base? Piston at height h, gas below piston. Above piston vacuum, and sand on piston? Actually sand in recipient on piston. Piston mass m, sand mass M. External downward force due to gravity on piston+sand? Also maybe gas pressure upward. Equilibrium: P_gas A = (m+M)g. Yes since vacuum above, no atmospheric pressure. Gas pressure P_i = (m+M)g/A. Volume initial V_i = A h_i. Temperature T_i = P_i V_i/(nR) = (M+m)g h_i/(nR). Gas internal energy U = (3/2)nRT = (3/2)P V for monatomic ideal. In part a, sand removed slowly quasi-static, adiabatic (insulating). Process is reversible adiabatic? "quase estático" means quasi-static, but adiabatic expansion. Since sand removed slowly, gas expands doing work, pressure decreases. Process is adiabatic and quasi-static => isentropic (reversible adiabatic). For ideal monatomic gas, P V^γ = constant with γ=5/3. Or T V^{γ-1}= constant. Need final pressure when no sand: external pressure from piston only = m g / A. Gas pressure final P_f = m g / A (mechanical equilibrium at final; quasi-static so P_gas = external pressure throughout? Yes if frictionless and quasi-static, gas pressure equals external pressure m(t)g/A plus maybe? Wait external pressure includes piston weight and current sand mass; removing slowly ensures equilibrium: P_gas = (m+M_current)g/A. At end M_current=0, so P_f = m g/A. Then adiabatic relation P_i V_i^γ = P_f V_f^γ, with P_i=(M+m)g/A, V_i=A h_i, P_f=m g/A, V_f=A h_f. Then (M+m) h_i^γ = m h_f^γ? Since P proportional (M+m), A cancels. h_f = h_i [ (M+m)/m ]^{1/γ} = h_i [(M+m)/m]^{3/5}. Wait check: P_i V_i^γ = P_f V_f^γ => (M+m) A? P_i = (M+m)g/A; P_f = m g/A. V_i^γ=(A h_i)^γ; V_f^γ=(A h_f)^γ. Then (M+m)g/A * A^γ h_i^γ = m g/A * A^γ h_f^γ => (M+m) h_i^γ = m h_f^γ => h_f = h_i ((M+m)/m)^{1/γ}. γ=5/3 => exponent 3/5. So h_f = h_i ((M+m)/m)^{3/5}. That likely part a. But need be careful: maybe process not adiabatic? Walls adiabatic, piston insulating. Sand removal slow quasi-static => reversible adiabatic. Yes. Part b: Entire sand removed instantaneously. Process? Gas expands rapidly; piston moves upward. No heat exchange (adiabatic walls/piston). But process is irreversible (sudden removal), not quasi-static; gas pressure not uniform? Piston may accelerate. Need determine maximum height h_max. This is tricky. Need apply energy conservation? Gas internal energy converts to kinetic/potential energy of piston, then max height when velocity zero. But gas also expands non-equilibrium, pressure? Need use first law for entire process from initial equilibrium to final max height state? At maximum height, piston momentarily at rest, but gas state? Is gas in equilibrium? Not necessarily? Usually problem assumes after oscillations, maybe gas reaches equilibrium at max? Hmm. We need derive equation relating h_max and m,M,h_i. Need consider adiabatic free expansion? The sand removed instantaneously, external pressure abruptly becomes m g/A (only piston). The gas expands against constant external pressure? Wait piston mass only, vacuum above, no sand. But while piston moves, gas pressure may vary; external load constant m g/A (plus inertia of piston). If piston moves slowly? But instant removal causes sudden decrease in external pressure from (m+M)g/A to m g/A. Gas expands; piston may accelerate. The process is adiabatic but irreversible. Need model. This is likely an ITA exam problem? Maybe involves applying conservation of energy to piston+gas system? Since adiabatic and no external work? Let's think. Initial state: gas at P_i=(M+m)g/A, V_i=A h_i, T_i=P_i V_i/(nR). Piston at rest. Sand removed instantaneously (mass M removed). Then external force on piston downward = mg (plus maybe atmospheric? no vacuum). Gas pressure upwards. Initially gas pressure P_i = (M+m)g/A > mg/A, so net upward force = (P_i A - mg) = M g, piston accelerates upward. Gas expands, pressure drops. Piston moves until gas pressure maybe less than mg/A? It may overshoot due to kinetic energy. The maximum height when piston velocity zero; gas maybe not equilibrium. Since adiabatic and no work against external other than lifting piston? Wait gas does work on piston: W = ∫ P_gas A dx = increase in gravitational potential energy of piston + kinetic energy. At maximum height, kinetic zero, so gas work equals m g (h_max - h_i) plus? Maybe also work done by gas against external? No other. But if gas pressure nonuniform, internal energy decreases by work; energy conservation for gas+piston system: ΔU_gas + ΔU_piston(grav) = Q + W_ext? Since Q=0, no external work except gas does work on piston? The piston is part of system? Let's choose gas + piston as system. Then gravitational potential energy changes. There is no external work (cylinder rigid, vacuum; no force from outside except gravity). The container? The gas does work on piston, but internal to system. The wall base? The gas may do work on bottom? Rigid base no displacement, no work. So for system gas+piston: ΔU_gas + ΔE_grav = 0 (if no kinetic at max). Initially piston at rest, final at rest. Thus final gas internal energy U_f = U_i - m g (h_max - h_i). That's an energy equation. Need also determine final gas state? At maximum height, is gas in equilibrium? If process is rapid, maybe after oscillations? The problem asks "equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." Maybe use energy conservation and perhaps assume gas undergoes an adiabatic process with constant external pressure? Hmm. Alternative: For instantaneous removal, gas expands irreversibly; as piston moves, the external pressure is just m g/A (constant) if piston mass dominates? Actually gas pressure on piston equals external pressure plus m a/A (Newton). At maximum height, acceleration maybe? At max height velocity zero but acceleration downward? The gas pressure may be less than mg/A, piston acceleration downward, so h_max determined by energy. Need derive final state relation. Maybe treat gas expansion against constant external pressure? Because instant removal: piston feels constant external force mg downward (weight) and gas force upward. As long as piston moves upward, gas pressure must exceed mg/A? Wait if gas pressure less than mg/A, piston would accelerate downward. For piston to reach h_max > h_i, initially gas pressure > mg/A; during motion, pressure may fall below mg/A near end? At maximum height, velocity zero; acceleration may be downward if P_f < mg/A. So final gas pressure not necessarily mg/A. But perhaps if piston is mass m, at max height instantaneous rest, not equilibrium. Then energy equation plus ideal gas relation? Need find U_f at max? Need know final volume V_f = A h_max, but final temperature? The gas may not be in equilibrium; but we might assume final state at max height is equilibrium? Not necessarily; but for equation maybe use energy conservation with final internal energy corresponding to some effective pressure? Need more. Maybe problem expects using first law for the gas only: Q=0, ΔU = -W_gas_by_piston? Work done by gas equals ∫ P_gas A dx. For irreversible sudden expansion, maybe can approximate as expansion against constant external pressure? Wait external pressure is just piston weight, constant mg/A, but gas pressure during acceleration not equal external. However work done by gas is ∫ P_gas A dx = kinetic + potential of piston = ΔK + mg(h_max-h_i). At max K=0, W = mg(h_max-h_i). Thus ΔU = -mg(h_max-h_i). That's same energy. But without relation for final U_f? We also have initial U_i = (3/2)nRT_i = (3/2)(M+m)g h_i? Wait P_i V_i = (M+m)g A h_i / A? Let's compute: U_i = (3/2) P_i V_i = (3/2)(M+m)g h_i? Since P_i A = (M+m)g, V_i=A h_i => P_i V_i = (M+m)g h_i. Yes U_i = (3/2)(M+m)g h_i. Final U_f = (3/2) P_f V_f if equilibrium? But P_f? At max, not necessarily equilibrium. However if we assume final gas state has pressure P_f = ? Maybe because process is adiabatic irreversible? Wait no heat, but gas pressure not uniform; but after piston stops at max, gas may still have kinetic energy (internal macroscopic) and pressure waves; but in ideal quasi-static limit? Hmm. Maybe there is a known result for sudden removal of mass in adiabatic piston: The gas expands doing work against constant external pressure equal to final external pressure? Let's examine. If sand removed instantaneously, external pressure is mg/A constant throughout? Actually the piston may have mass; but the external force (mg) constant. The gas expands against constant external pressure? If piston massless? No, mass m significant. For a piston of mass m, gas pressure minus external pressure = m a/A. Work done by gas = work done on piston = ΔK + mg Δh. So W = mg Δh + K. At max K=0, so W=mgΔh. But if the expansion is fast, pressure inside may not be uniform; but work done by gas at piston equals ∫ P_piston A dx, where P_piston gas pressure at piston. Not necessarily average gas pressure. Could use energy conservation plus assumption that final state is equilibrium with same entropy? No, irreversible so entropy increases. But for maximum height, maybe use energy and final gas internal energy expression? Need additional relation from dynamics? Maybe use equation of motion and adiabatic relation? Let's explore. Process: gas expands; we can model as adiabatic irreversible expansion of ideal gas in piston. The gas pressure at piston P(t) = (m/A) a + mg/A? Actually Newton: m a = P_gas A - mg (upward positive; if a upward, P_gas A - mg = m a). So P_gas = (mg + m a)/A. Not constant. At initial, a = M g / m? Wait P_i A = (M+m)g, so net = M g; a = M g / m upward. Large acceleration. Gas expands rapidly; pressure drops. To find h_max, maybe integrate equation of motion with adiabatic relation? But irreversible: gas state maybe not uniform; however maybe consider gas undergoes adiabatic expansion with P V^γ = constant during motion? Is that valid for the gas as a whole when piston moves? If expansion is not quasi-static, P not uniform; but for idealized "adiabatic piston" maybe gas pressure at piston follows adiabatic relation P V^γ? Actually in a rapid adiabatic expansion, if no internal friction? An ideal gas has no viscosity; pressure waves reflect. Not simple. But many textbook problems treat gas as always uniform and apply P V^γ = const even for non-quasistatic? That would be incorrect but maybe intended? Let's see. If assume gas always uniform and undergoes adiabatic expansion P V^γ = constant (isentropic) even though rapid, then final at max height? But at max height, gas would have pressure P_f = P_i (V_i/V_f)^γ = (M+m)g/A (h_i/h_max)^γ. At max height, piston velocity zero but acceleration? If we require mechanical equilibrium? Not necessarily. If we set P_f = mg/A (external), then same as part a? But instantaneous removal would give same final if equilibrium? No, if final equilibrium after oscillations, gas pressure = mg/A; that would be same as quasi-static if adiabatic reversible? Wait irreversible process could have same final pressure but different temperature/volume? For adiabatic expansion against constant external pressure, final equilibrium pressure equals external pressure, and work done by gas equals P_ext ΔV. First law: ΔU = -P_ext ΔV. For ideal gas U = (3/2)nRT = (3/2)PV. So (3/2)(P_f V_f - P_i V_i) = -P_f (V_f - V_i) (if P_ext=P_f). This yields P_f V_f difference? Let's solve: (3/2)(P_f V_f - P_i V_i) = -P_f(V_f - V_i) => (3/2)P_f V_f - (3/2)P_i V_i = -P_f V_f + P_f V_i => (5/2)P_f V_f = (3/2)P_i V_i + P_f V_i => etc. This would give final volume maybe different from adiabatic reversible? Let's solve with P_f=mg/A, P_i=(M+m)g/A, V_i=A h_i, V_f=A h_f: (3/2)(P_f A h_f - P_i A h_i)= -P_f A (h_f - h_i) Divide A: (3/2)(P_f h_f - P_i h_i)= -P_f(h_f - h_i) => (3/2)P_f h_f - (3/2)P_i h_i = -P_f h_f + P_f h_i => (5/2)P_f h_f = (3/2)P_i h_i + P_f h_i => h_f = h_i [ (3/2)P_i + P_f ] / (5/2 P_f) = h_i [ (3 P_i + 2 P_f)/(5 P_f) ] = h_i [ (3(M+m)+2m)/(5m) ] = h_i (3M+5m)/(5m)? Wait P_i=(M+m)g/A, P_f=mg/A: (3*(M+m)+2*m)/(5m)= (3M+5m)/(5m). That differs from quasi-static γ=3/5: ((M+m)/m)^{3/5}. Is this maybe part b? But if sand removed instantaneously, external pressure suddenly drops to mg/A; gas expands irreversibly against constant external pressure mg/A. Final equilibrium after oscillations? But question asks maximum height, not final equilibrium. The piston may overshoot equilibrium due to kinetic energy, so h_max > h_f (equilibrium). The final equilibrium (after damping) would be same as quasi-static? Actually if after oscillations gas reaches equilibrium at P=mg/A, but process irreversible; final state might satisfy energy equation with constant external pressure? Wait if final equilibrium at P=mg/A, the gas work done against constant external pressure mg/A equals P_ext ΔV = mg(h_f-h_i). But the piston's kinetic energy converted to? At final equilibrium, piston at rest; gas internal energy change equals -P_ext ΔV if all work done against constant external pressure (piston weight). This would give h_f as above, which differs from adiabatic reversible? But adiabatic irreversible expansion against constant external pressure final equilibrium indeed satisfies ΔU = -P_ext ΔV, and final pressure = P_ext. That is standard for "adiabatic expansion against constant external pressure" (irreversible) final equilibrium. But part b asks max height, not final equilibrium. For irreversible adiabatic expansion against constant external pressure with a piston of mass? If piston has mass, after sudden removal, it will oscillate and eventually damp to equilibrium with P=mg/A. But initially no damping? Heat? Gas viscosity? The problem may ignore damping and ask maximum height before returning? Need model. Let's parse wording: "Considere uma nova situação em que toda a quantidade de areia é completamente retirada instantaneamente. Determine a equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." They likely expect energy conservation for the gas + piston, and perhaps because removal instantaneous, the gas expands adiabatically but irreversibly against the constant pressure of the piston (mg/A). At maximum height, kinetic energy zero; the gas has done work equal to m g (h_max - h_i). But to determine final internal energy at max, maybe assume final state is equilibrium? Hmm. Actually if removal instantaneous, the external pressure on piston is just mg/A (constant). The gas expands, pushing piston upward. At maximum height, piston stops; but gas may still have not equilibrated (pressure waves). However in an idealized quasi-static? No. But for an adiabatic process with a moving piston of mass m, maybe the gas and piston exchange energy; at max height, all energy from gas goes into gravitational potential. The final gas internal energy can be related to final volume and pressure? But if piston at max, its acceleration? The gas pressure at piston must equal mg/A + m a/A; at max velocity zero, acceleration may be? Let's derive: Equation of motion m dv/dt = P A - mg. Multiply by v: d(1/2 m v^2)/dt = P A v - mg v = P dV/dt - mg dh/dt. Integrate from initial to max: 0 = ∫ P dV - mg (h_max - h_i). So ∫ P dV = mg Δh. But ∫ P dV is work done by gas, depends on actual path. If we approximate gas pressure P during rapid expansion as given by adiabatic relation P V^γ = const? Then ∫ P dV = (P_i V_i - P_f V_f)/(γ-1) for γ=5/3? Wait for adiabatic ideal gas, relation P V^γ = const gives work = (P_i V_i - P_f V_f)/(γ-1). Then energy equation: (P_i V_i - P_f V_f)/(γ-1) = mg Δh. Also P_f? At max, if same relation, P_f = P_i (V_i/V_f)^γ = P_i (h_i/h_max)^γ. Then solve for h_max. This would give a different result. Let's test. Using reversible adiabatic relation for a rapid process maybe if piston mass? But in reversible adiabatic, gas pressure equals external pressure throughout? Not here because external pressure mg/A is less than initial; but if gas expands as if isolated with no dissipation? Actually reversible adiabatic expansion requires external pressure matched to gas pressure, which would need sand removed gradually (part a). If instant removal, the gas expands freely? No, piston mass, external pressure smaller, irreversible. The gas pressure will not be uniform; P V^γ not valid. However many physics contest solutions treat sudden removal as "adiabatic free expansion against constant pressure"? Let's search memory: This problem might be from ITA (Instituto Tecnológico de Aeronáutica) 2027 2ª fase? Could be about "pistão adiabático com areia removida" maybe answer involves energy conservation and polytropic exponent? Let's reason. Maybe there is an exact result using conservation of energy and momentum? Let's derive with gas as ideal monatomic. Consider initial state: gas pressure P_i = (M+m)g/A, volume V_i. Piston mass m, sand mass M. Sand removed instantaneously: initially the piston (with gas) experiences net upward force M g. The sand removed no longer interacts. The system (gas + piston) has no external work except gravity on piston; adiabatic. The gas will expand; its internal energy is converted into gravitational potential energy and kinetic energy of piston. At maximum height, kinetic zero, so: ΔU_gas = - m g (h_max - h_i). (1) This is universal, no need details. But to find h_max, need know final internal energy U_f. The final gas state may be non-equilibrium, but maybe if piston is at maximum height, it has zero velocity and maybe also zero acceleration? Not necessarily. But if we assume that at maximum height, the gas has expanded to a new equilibrium state? Actually "altura máxima atingida pelo pistão" in an oscillatory motion; after some time, gas may have internal dissipative processes (viscosity, turbulence) that eventually equilibrate. The maximum height is first turnaround point, not equilibrium. But if no damping, oscillations continue; gas pressure at turnaround maybe not uniform. The problem likely expects using energy conservation plus treating the process as an adiabatic reversible expansion? Wait part a says quasi-static, part b instantaneous. Could compare: in a an isentropic expansion; in b instantaneous, perhaps use energy conservation with the gas expanding adiabatically but irreversibly against constant pressure mg/A until equilibrium? But they ask h_max not h_f. Let's identify from known problem. I recall a classic problem: A piston of mass m contains gas, with additional mass M on top. When M is suddenly removed, the piston rises to a maximum height. Find h_max. It uses energy conservation and the fact that the gas expands adiabatically; often assume the gas follows P V^γ = constant during the expansion, because the piston moves fast? Actually if M removed suddenly, the gas expands "adiabatically but not quasi-statically"; in many solutions, they use the adiabatic relation P V^γ = const because the gas does work only on piston and no heat; but the process is reversible? Wait if the piston moves rapidly, the gas may not be in equilibrium, but for ideal gas, the relation P V^γ = constant holds for any adiabatic process of a fixed mass of gas if the process is reversible; but not for irreversible. However some textbook solutions erroneously use it. Let's examine. Another approach: Use conservation of energy and the fact that at maximum height, the piston velocity is zero, but the gas may still be in a state with same entropy? Actually sudden removal: external pressure suddenly mg/A; gas expands irreversibly, entropy increases. But perhaps for a monatomic gas and a frictionless piston, the gas expansion can be described by a polytropic process with exponent? The equation of motion can be integrated if assume gas pressure uniform and adiabatic? Let's solve with Newton + adiabatic relation without assuming quasi-static? Let's see. If assume gas pressure uniform and related to volume by P V^γ = const (maybe valid because no heat and reversible? But not reversible due to finite pressure difference. However if gas has no internal friction and piston mass? Actually the process is reversible if the gas passes through equilibrium states; even if external pressure is lower, if piston mass, the gas pressure may differ, but if we consider the gas alone, it expands against a piston whose external pressure is mg/A; the gas pressure at the piston is not equal to mg/A except at equilibrium. But if no friction and gas inviscid, the gas may not be in equilibrium. So no. Let's search memory: In an adiabatic process with a piston of mass m, when external load changes suddenly, the maximum displacement can be found by energy conservation and "adiabatic relation" for the gas? Example: A mass M on a piston of mass m in a cylinder with gas; M removed; find max height. There is a well-known result using energy conservation and adiabatic condition: P V^γ = const. Let's test. Assume P V^γ = const throughout. Initial P_i V_i^γ = P_f V_f^γ. Final pressure at max? Not necessarily mg/A. The work done by gas = (P_i V_i - P_f V_f)/(γ-1) = m g (h_max - h_i). Also P_f = P_i (V_i/V_f)^γ = (M+m)g/A (h_i/h_max)^γ. Then: (P_i A h_i - P_f A h_max)/(γ-1) = m g (h_max - h_i) Divide A g? Let P_i=(M+m)g/A, P_f=(M+m)g/A (h_i/h_max)^γ. Then: g/(γ-1)[(M+m) h_i - (M+m) h_i^γ h_max^{1-γ}? Wait P_f h_max = (M+m)g/A h_i^γ h_max^{1-γ}? Let's compute: P_f = (M+m)g/A (h_i/h_max)^γ. P_f h_max = (M+m)g/A h_i^γ h_max^{1-γ}. Multiply A: A P_f h_max = (M+m)g h_i^γ h_max^{1-γ}. So work = [ (M+m)g h_i - (M+m)g h_i^γ h_max^{1-γ} ]/(γ-1). Set = m g (h_max - h_i). Cancel g: [(M+m) h_i - (M+m) h_i^γ h_max^{1-γ}]/(γ-1) = m (h_max - h_i). For γ=5/3, maybe can solve? Let's test. Let x = h_max/h_i. Then: [(M+m) - (M+m) x^{1-γ}]/(γ-1) = m (x - 1)? Wait h_i^γ h_max^{1-γ} = h_i (h_i/h_max)^{γ-1} = h_i x^{1-γ}. Yes. Then left = (M+m) h_i (1 - x^{1-γ})/(γ-1). Divide h_i: (M+m)(1 - x^{1-γ})/(γ-1) = m (x-1). For γ=5/3, 1-γ = -2/3, γ-1=2/3: (M+m)(1 - x^{-2/3})/(2/3) = m(x-1) => (3/2)(M+m)(1 - x^{-2/3}) = m(x-1). This is one equation for x. Does this match any known? Let's test with M=0 (no sand? then x=1). If M large, x maybe? For M>>m, left ~ (3/2)M(1 - x^{-2/3}) = m(x-1). For large M, x maybe large? Actually gas expands a lot; h_max roughly? Let's approximate: 1 - x^{-2/3} ~ 1, left ~1.5M, so m(x-1) enormous? That can't be? Wait if M large, initial pressure high, sudden removal gives large upward force, piston attains high speed, but gas expands and pressure drops; max height maybe? Let's solve for M=100m: (3/2)*101m*(1 - x^{-2/3}) = m(x-1) => 151.5(1 - x^{-2/3})=x-1. For x large, LHS ~151.5, RHS large if x>152? Actually LHS max 151.5, so x-1 ~151.5 => x≈152.5. So h_max ~152 h_i. But energy: initial gas internal energy ~ (3/2)(M+m)g h_i ~151.5 m g h_i; if all converted to potential, h_max-h_i ~ U/(mg)=151.5 h_i, so h_max ~152.5 h_i. That matches. So this relation plausible. But is the assumption P V^γ valid for sudden removal? Let's check with energy approach if final pressure not mg/A. At max, piston velocity zero; if using adiabatic relation, final pressure is not mg/A generally. Does that make sense? The piston at max may have acceleration downward; gas pressure can be greater or less than mg/A? At maximum height, velocity zero, but acceleration could be downward if P_f A < mg. If P_f = P_i (h_i/h_max)^γ, for large M, P_f very small, so acceleration downward large; it's a turnaround point after overshoot, so gas pressure indeed less than external? Wait at max, if gas pressure were greater than mg/A, piston would accelerate upward, so it would continue up; so at max, net force must be downward or zero. So P_f ≤ mg/A (if no other forces). Thus final pressure at max is less than external if overshoot. That's okay. The adiabatic relation gives P_f small. Good. But is there another possible relation from energy conservation with "adiabatic expansion against constant external pressure" and final pressure mg/A? That would give h_f equilibrium after damping, not max. For max, final pressure not mg/A; so relation must include dynamics. The energy conservation alone insufficient; need a process relation. The adiabatic relation P V^γ = const might be the only way. But is it valid for irreversible? Many physics problems use the "adiabatic law" for any adiabatic process of an ideal gas regardless of reversibility? Actually no, P V^γ = constant holds for a quasistatic adiabatic process. For irreversible adiabatic expansion against constant pressure, the relation is different: P_i V_i^γ? No. Example: free expansion into vacuum: P V^γ not constant. So cannot blindly apply. Maybe the correct part b uses energy conservation plus the fact that the process is adiabatic and the gas expands irreversibly against constant external pressure mg/A until maximum? Wait at maximum, the gas may have zero kinetic energy? Hmm. Let's consider another angle: The problem statement: "a) Em um dado instante, os grãos de areia são retirados lentamente em um processo quase estático. Determine a altura h_f do pistão no estado final, quando não há mais areia..." So part a is final equilibrium state after slow removal. Part b: "toda a quantidade de areia é completamente retirada instantaneamente. Determine a equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." They likely expect using conservation of energy and mechanical energy; maybe because removal is instantaneous, the gas has no time to exchange heat, but the piston gains kinetic energy; at maximum height, kinetic energy zero. The gas's internal energy change equals potential energy. But to compute internal energy change, they may treat the gas expansion as adiabatic but with "the gas pressure equal to external pressure plus inertial term"? Let's derive from first law for gas and equation of motion for piston. Let's set up differential equations: Gas: dU = -P dV (adiabatic, no heat; assuming uniform state even if irreversible? Actually first law for gas: dU = δQ - δW = -P dV, where P is pressure at moving boundary. If gas state nonuniform, dU ≠ -P dV? For a closed system, dU = δQ - δW, and δW = P_boundary dV. Yes, even if nonuniform, work done by gas is P_at_piston dV. And dU = n Cv dT if equilibrium? If nonuniform, total internal energy may not be simple. But if we assume homogeneous, then dU = n Cv dT = (3/2)nR dT = (3/2)d(PV). Then (3/2)d(PV) = -P dV, where P is pressure at piston? For homogeneous, P same. Then (3/2)(P dV + V dP) = -P dV => (5/2)P dV + (3/2)V dP = 0 => dP/P + (5/3)dV/V=0 => P V^{5/3}=const. This derivation assumes homogeneous and reversible? Actually it uses P same inside; if P_at_piston = P, then first law alone gives adiabatic relation for any adiabatic process of homogeneous ideal gas? Let's check: dU = (3/2)d(PV). Work δW = P dV. For a quasi-static process with uniform P, yes. But for non-quasi-static, if gas is uniform and P is thermodynamic pressure, work is P dV? In a rapid expansion, gas pressure may not be uniform, but if we model as uniform with a single P, then relation holds. Many contest problems ignore non-uniformity for entire gas. So likely intended. But maybe part b uses energy conservation and relation P V^γ = const? Let's solve and see if the equation is simple. Equation: (P_i V_i - P_f V_f)/(γ-1) = mg(h_max - h_i), with P_i=(M+m)g/A, P_f=P_i(V_i/V_f)^γ. Let V_i=A h_i, V_f=A h_max. Then: (P_i A h_i - P_f A h_max)/(γ-1) = mg(h_max - h_i) => [(M+m)g h_i - (M+m)g h_i^γ h_max^{1-γ}]/(γ-1) = mg(h_max - h_i) Cancel g, divide m? Let μ = M/m maybe. Then: (M+m)/m * [h_i - h_i^γ h_max^{1-γ}]/(γ-1) = h_max - h_i. Let x = h_max/h_i, μ=M/m: (1+μ) [1 - x^{1-γ}]/(γ-1) = x - 1. For γ=5/3: (1+μ) [1 - x^{-2/3}]/(2/3) = x - 1 => (3/2)(1+μ)(1 - x^{-2/3}) = x - 1. This is an equation relating x and M/m. Maybe answer: (3/2)(1+M/m)(1 - (h_i/h_max)^{2/3}) = h_max/h_i - 1? Could be. But is there alternative using conservation of energy and final pressure? Let's test with small M: M=0, x=1. Expand around μ small: left ~ (3/2)(1+μ)(1 - (1 - (2/3)ε)? if x=1+ε, x^{-2/3}≈1 - (2/3)ε, so left≈(3/2)(1+μ)*(2/3)ε≈(1+μ)ε. Right ε. So μ=0 gives identity; for small μ, ε≈? (1+μ)ε ≈ ε => ε≈0? Actually need second order. Let's solve small μ: (3/2)(1+μ)(1 - x^{-2/3}) = x - 1. Let x=1+ε small. LHS ≈ (3/2)(1+μ)( (2/3)ε + (1/9)ε^2?) because x^{-2/3}= (1+ε)^(-2/3)=1 - (2/3)ε + (5/9)ε^2? Let's compute: (1+ε)^n≈1+nε+n(n-1)/2 ε^2; n=-2/3: 1 - (2/3)ε + (5/9)ε^2. So 1 - x^{-2/3}≈(2/3)ε - (5/9)ε^2. Then LHS≈(3/2)(1+μ)[(2/3)ε - (5/9)ε^2] = (1+μ)[ε - (5/6)ε^2]. RHS=ε. So ε[(1+μ)-1] - (5/6)(1+μ)ε^2 ≈0 => μ ε ≈ (5/6)(1+μ)ε^2 => ε≈ (6/5) μ/(1+μ). So x≈1+1.2 μ for small μ. If μ=0.1, x≈1.109. Seems okay. But actual physical? Let's test energy for small mass: If M small, sudden removal of small extra load causes small oscillation. Max height slightly above initial. Good. But wait if M small, the gas expansion is not negligible? Actually initial pressure (M+m)/m >1; work available (3/2)M g h_i; potential increase mg(h_max-h_i) ≈ (3/2)M g h_i /? Wait from energy naive: h_max-h_i ≈ (3/2)(M/m) h_i? Actually if all internal energy change to potential? But gas also does work? For small M, ΔU ≈ (3/2)(P_i V_i - P_f V_f) ≈ (3/2)(M+m)g h_i (1 - x^{-2/3}) ≈ (3/2)(M+m)g h_i*(2/3)(x-1) ≈ (M+m)g h_i (x-1). Set = mg(h_max-h_i)=mg h_i (x-1). Then (M+m) ≈ m, so okay. Thus M small gives x-1 maybe not determined by energy alone; actually change in internal energy arises from pressure difference? Hmm. But maybe the correct part b answer could be simply from energy conservation with constant external pressure and final gas equilibrium? Let's compare. If using irreversible adiabatic expansion against constant external pressure mg/A, the work done by gas is P_ext ΔV = mg(h_max-h_i) (if final pressure at max equals external? But not necessarily). For maximum, perhaps the gas pressure at turnaround equals external pressure? No, Newton says if P_f = mg/A, acceleration zero; then piston would be in equilibrium, not necessarily max? Actually at max, acceleration could be zero if also velocity zero; then mechanical equilibrium. But if process overshoots, it passes through equilibrium at h_f where P=mg/A, continues due to kinetic energy, reaches max where P<mg/A. So final pressure not mg/A. Could the problem be asking for "equação" not solved for h_max, because it's implicit? They say "Determine a equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." So answer may be an equation involving h_max, m, M, h_i. In part a explicit formula. So part b expected an equation (not solved). We can provide implicit equation from energy + adiabatic relation. Let's investigate if there is known result for this exact ITA problem. Maybe from "ITA 2027 Física" could be future? The problem statement likely from an exam; answer might be: a) h_f = h_i [(M+m)/m]^{3/5} b) \frac{3}{2}(M+m)\left(1 - \left(\frac{h_i}{h_{max}}\right)^{2/3}\right) = m\left(\frac{h_{max}}{h_i}-1\right)? Let's derive in terms of M,m,h_i maybe. Actually if using energy + adiabatic relation, the equation becomes: \frac{3}{2}(M+m)\left[1 - \left(\frac{h_i}{h_{max}}\right)^{2/3}\right] = m\left(\frac{h_{max}}{h_i}-1\right)? Let's check dimensions: left has mass, right mass. Yes. But let's verify derivation: U_i = (3/2)(M+m)g h_i? Wait dimensions: (M+m)g h_i = energy (Force*length). Yes U_i = (3/2)(P_i V_i) = (3/2)( (M+m)g/A * A h_i) = (3/2)(M+m)g h_i. Correct. U_f = (3/2) P_f V_f = (3/2) [ (M+m)g/A (h_i/h_max)^{5/3} ] A h_max = (3/2)(M+m)g h_i (h_i/h_max)^{2/3}? Let's compute: P_f = P_i (h_i/h_max)^{5/3}; V_f=A h_max; P_f V_f = P_i A h_max (h_i/h_max)^{5/3} = (M+m)g h_max (h_i/h_max)^{5/3} = (M+m)g h_i^{5/3} h_max^{-2/3}? Wait h_max * h_max^{-5/3}=h_max^{-2/3}; times h_i^{5/3}. So P_f V_f = (M+m)g h_i^{5/3} h_max^{-2/3} = (M+m)g h_i (h_i/h_max)^{2/3}. Yes. So U_f = (3/2)(M+m)g h_i (h_i/h_max)^{2/3}. Then ΔU = U_f - U_i = (3/2)(M+m)g h_i [(h_i/h_max)^{2/3} - 1] = - (3/2)(M+m)g h_i [1 - (h_i/h_max)^{2/3}]. Work done by gas = -ΔU = (3/2)(M+m)g h_i [1 - (h_i/h_max)^{2/3}] = mg(h_max - h_i) = mg h_i (h_max/h_i -1). So equation: (3/2)(M+m) [1 - (h_i/h_max)^{2/3}] = m (h_max/h_i -1). Yes. This is a relation involving (h_i/h_max)^{2/3}. Let's denote y = h_max/h_i >1. Then (1/y)^{2/3}. Equation: (3/2)(M+m)[1 - y^{-2/3}] = m(y-1). Or equivalently: 3(M+m)(1 - y^{-2/3}) = 2m(y-1). This could be b. But is there another possibility: Maybe part b uses conservation of energy and the fact that the expansion is adiabatic but the final internal energy is given by something else because the process is a free expansion? But the piston mass m means not free expansion; however the gas expands against constant external pressure mg/A. For an adiabatic irreversible expansion of an ideal gas against constant external pressure, the final equilibrium temperature can be found by ΔU = -P_ext ΔV, and final pressure = P_ext. If the problem asked final equilibrium height h_f after instantaneous removal, then h_f would be h_i [ (3(M+m)+2m)/(5m) ]? Let's derive for final equilibrium after oscillations (if damping) and compare with part a. But they specifically ask maximum height, not final equilibrium. The maximum height occurs before damping; energy at maximum doesn't require final pressure = mg/A. So our equation is appropriate. But we need consider work done by gas when external pressure is only mg/A; however during the rapid expansion, the gas pressure at piston may not be uniform, but energy conservation still holds: gas internal energy change equals work done by gas plus heat (0). The work done by gas equals ∫ P_piston A dx. This equals increase in potential energy of piston plus kinetic energy. At maximum, kinetic zero, so W = mg(h_max-h_i). Thus ΔU = -mgΔh. This is exact regardless of reversibility. So if we can determine U_f at maximum, we can solve. Our assumption of P V^γ constant gives U_f. But is that assumption justified? At maximum, the gas may not be in equilibrium, but if we treat the gas's internal energy as total kinetic energy of molecules, it depends only on temperature; if the gas is not in equilibrium, temperature may not be defined. However ideal gas internal energy is sum of kinetic energies; even if nonuniform, total internal energy may not be (3/2)PV if pressure not uniform. But many contest problems assume equilibrium. The phrase "gás ideal monoatômico" and "processo adiabático" likely allows using adiabatic law for the gas during the expansion, as a model. Let's see. Maybe they expect an equation from energy conservation only, but with the gas pressure at max equal to mg/A? Let's test if possible: At maximum height, perhaps the piston has zero velocity but may be in equilibrium? No, if it were equilibrium, it would stay; maximum is first extremum, not equilibrium. But maybe due to "instantaneous removal" the gas expands so quickly that it is essentially a free expansion? Wait if the sand is removed instantaneously, the piston (mass m) initially is not in equilibrium; the gas pressure is much higher, so the piston accelerates. The gas expands, but because the piston has mass, the gas pressure might remain uniform? Actually the speed of sound in gas finite; pressure nonuniform. Hmm. Let's search memory: There is a well-known problem "A vertical cylinder contains n moles of an ideal gas at temperature T. The piston has mass m and cross-section A. A block of mass M is placed on piston. The block is removed suddenly. Find maximum height reached by piston. Assume adiabatic process." The solution often uses energy conservation and the fact that the process is adiabatic and "the gas expands according to PV^γ = const" because it's adiabatic. Yes, because in many textbooks, an adiabatic process of ideal gas is PV^γ=const even if not quasi-static? Actually they may state "the expansion is adiabatic and the gas obeys PV^γ=const" as approximation. The problem likely designed to distinguish quasi-static (part a) from rapid (part b); part b uses energy conservation and adiabatic law. Let's derive part a with adiabatic law; part b also with adiabatic law but energy. Another possible part b: Using conservation of energy, but the final gas pressure equals external pressure m g/A (ignoring kinetic energy) would give same as irreversible constant pressure final, not max. But they ask max, so energy with kinetic. Maybe part b answer is simply: \[ \frac{3}{2}(M+m) g h_i \left[1 - \left(\frac{h_i}{h_{max}}\right)^{2/3}\right] = m g (h_{max}-h_i) \] or equivalently \[ \frac{3}{2}(M+m)\left[1-\left(\frac{h_i}{h_{max}}\right)^{2/3}\right] = m \left(\frac{h_{max}}{h_i}-1\right). \] This is an equation (not explicit). They may accept. But let's ensure initial internal energy: The gas pressure initially is due to piston + sand: P_i = (m+M)g/A. Volume V_i=A h_i. So U_i = (3/2)P_i V_i = 3/2 (m+M)g h_i. Yes. At max, gas volume V_f=A h_max; if process adiabatic reversible/ideal, P_f V_f^{5/3} = P_i V_i^{5/3}. Thus U_f = (3/2)P_f V_f = (3/2)P_i V_i^{5/3} V_f^{-2/3} = (3/2)(m+M)g h_i (h_i/h_max)^{2/3}. Then ΔU = U_f - U_i = (3/2)(m+M)g h_i[(h_i/h_max)^{2/3}-1] = -mg(h_max-h_i). So equation as above. Let's solve for y maybe to provide explicit? They ask "determine a equação que relaciona", so implicit is fine. But maybe they expect "equação" not "solução"; can leave implicit. Need also consider if "gás ideal monoatômico" γ=5/3. Yes. Let's check part a derivation with pressure at final: When all sand removed, final equilibrium pressure P_f = mg/A. Initial P_i = (m+M)g/A. For adiabatic reversible: P_i V_i^{5/3}=P_f V_f^{5/3}. Then (M+m) h_i^{5/3}= m h_f^{5/3} => h_f = h_i ((M+m)/m)^{3/5}. Yes. But wait part a: "os grãos de areia são retirados lentamente em um processo quase estático. Determine a altura h_f do pistão no estado final, quando não há mais areia, em função de m, M e h_i." If sand is removed grain by grain, the process is quasi-static. At each intermediate state, external pressure = (m+M_current)g/A; gas pressure equals that. The final state: no sand, external pressure = mg/A. The gas has undergone adiabatic expansion; since quasi-static, adiabatic, P V^γ = const. Final volume V_f=A h_f. So yes. But maybe there is a subtlety: The walls are adiabatic; piston also adiabatic; process quasi-static means gas temperature changes. Yes. Let's consider if the piston has mass m and sand is removed; initially equilibrium: pressure of gas = (M+m)g/A. At final without sand, pressure = mg/A. For adiabatic reversible expansion, temperature decreases, gas does work. The formula h_f = h_i ((M+m)/m)^{3/5}. This is standard. Now part b: "Considere uma nova situação em que toda a quantidade de areia é completamente retirada instantaneamente. Determine a equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." We should maybe derive with energy and adiabatic relation. Let's formulate clearly in Portuguese. Need present as candidate solving: Show development item by item. Let's write: a) Inicialmente: P_i A = (M+m)g => P_i = (M+m)g/A; V_i = A h_i. Final: P_f A = mg => P_f = mg/A; V_f = A h_f. Processo adiabático quase estático de gás ideal monoatômico: \(P V^\gamma = \text{cte}\) com \(\gamma=5/3\). \[ P_i V_i^{5/3}=P_f V_f^{5/3} \] \[ \frac{(M+m)g}{A}(A h_i)^{5/3} = \frac{mg}{A}(A h_f)^{5/3} \] \[ (M+m) h_i^{5/3} = m h_f^{5/3} \] \[ h_f = h_i\left(\frac{M+m}{m}\right)^{3/5}. \] b) Para a remoção instantânea, a expansão é adiabática, mas não quase estática. Como o pistão adquire energia cinética e potencial, usamos conservação de energia. No estado de altura máxima, a velocidade do pistão é nula. O trabalho realizado pelo gás é igual ao aumento da energia potencial gravitacional do pistão: \[ W_{\text{gás}} = m g (h_{\max}-h_i). \] Admitindo que o gás ideal ainda obedeça à relação adiabática entre os estados inicial e de altura máxima (ou tratando o gás como um sistema que se expande adiabaticamente segundo \(PV^{5/3}=\text{cte}\)): \[ P_i V_i^{5/3}=P_{\max} V_{\max}^{5/3} \] onde \(P_i=(M+m)g/A\), \(V_i=A h_i\), \(V_{\max}=A h_{\max}\). O trabalho do gás é \[ W_{\text{gás}} = \frac{P_i V_i - P_{\max}V_{\max}}{\gamma-1}, \quad \gamma=5/3. \] Assim, \[ \frac{3}{2}(P_i V_i - P_{\max}V_{\max}) = m g (h_{\max}-h_i). \] Mas \(P_{\max}=P_i (V_i/V_{\max})^{5/3}=P_i (h_i/h_{\max})^{5/3}\), logo \[ P_{\max}V_{\max}=P_i V_i \left(\frac{h_i}{h_{\max}}\right)^{2/3} = (M+m)g h_i \left(\frac{h_i}{h_{\max}}\right)^{2/3}. \] Então \[ \frac{3}{2}\left[(M+m)g h_i - (M+m)g h_i \left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m g (h_{\max}-h_i). \] Cancelando \(g\): \[ \boxed{\frac{3}{2}(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m\left(\frac{h_{\max}}{h_i}-1\right)} \] ou \[ \boxed{\frac{3}{2}(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m\left(\frac{h_{\max}-h_i}{h_i}\right)} \] Wait second has h_i? Let's check: m g(h_max - h_i) = m g h_i (h_max/h_i -1). So after cancel g: \[ \frac{3}{2}(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m\left(\frac{h_{\max}}{h_i}-1\right). \] Yes. If multiply both sides by h_i? No. The equation has dimensions mass. Good. But is there a possibility to express as \[ \frac{3}{2}(M+m)\left(1 - \sqrt[3]{\frac{h_i^2}{h_{\max}^2}}\right) = m\left(\frac{h_{\max}}{h_i}-1\right) \] equivalent. Use \( (h_i/h_max)^{2/3}\). Need ensure final answer line with a) and b). The problem says "Determine a equação que relaciona ..." so for b answer equation. Should we solve for h_max? It's implicit. They might expect "equação" not explicit; but final line should include. Let's see if the equation could be written differently using energy conservation directly: At maximum, \(U_i - U_f = mg(h_max-h_i)\). \[ \frac32 (M+m)g h_i - \frac32 (M+m)g h_i\left(\frac{h_i}{h_{\max}}\right)^{2/3} = mg(h_max-h_i) \] which is same. It might be simpler. Let's consider if part b should use "equação que relaciona a altura máxima ... e m, M e h_i" maybe they expect: \[ h_{\max} = h_i\left[1+\frac{3(M+m)}{2m}\left(1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right)\right] \] No, not explicit. The implicit equation is fine. Let's double-check b using another method: If the gas expands against constant external pressure mg/A only (piston weight), then energy conservation gives: \[ \Delta U = -mg(h_max-h_i) \] and if we assume the gas pressure at max is not mg/A, but we can use the adiabatic relation. Yes. However there is an issue: The problem statement says "recipiente de massa desprezível contendo uma quantidade de grãos de areia, que totaliza uma massa M em repouso sobre o pistão." So the sand is in a recipient (bucket) on piston. When removed instantaneously, the bucket also removed? "toda a quantidade de areia é completamente retirada instantaneamente" means sand removed; bucket maybe negligible. So external mass M removed. Good. Let's think if the initial gas pressure includes piston + sand + bucket? Recipiente de massa desprezível, so no. What about the piston's mass m? Yes. Potential issue: The "cilindro vertical, de seção transversal de área A e com paredes isolantes (adiabáticas), contém n mols de um gás ideal monoatômico. O cilindro é fechado na parte superior por um pistão de espessura desprezível de massa m, também isolante térmico, que pode deslizar verticalmente sem atrito." So cylinder closed at top by piston; above piston vacuum. The gas below. Good. Part a: "Em um dado instante, os grãos de areia são retirados lentamente em um processo quase estático. Determine a altura h_f do pistão no estado final, quando não há mais areia, em função de m, M e h_i." We did. Part b: "Considere uma nova situação em que toda a quantidade de areia é completamente retirada instantaneamente. Determine a equação que relaciona a altura máxima h_max atingida pelo pistão e m, M e h_i." We did. Let's maybe derive equation more elegantly: No item b, como a remoção é instantânea, o processo não é quase estático. Porém, como as paredes são adiabáticas, \(Q=0\). Pela conservação da energia para o sistema gás + pistão: \[ \Delta U + \Delta E_{pg} + \Delta K = 0. \] No ponto de altura máxima, \(K=0\). Então: \[ U_f - U_i + mg(h_{\max}-h_i)=0 \Rightarrow U_i - U_f = mg(h_{\max}-h_i). \] Usando a lei adiabática para o gás entre os estados inicial e de altura máxima (ou considerando que a expansão adiabática do gás ideal fornece \(P_i V_i^{5/3}=P_f V_f^{5/3}\)): \[ U_i = \frac{3}{2}(M+m)g h_i,\quad U_f = \frac{3}{2}(M+m)g h_i\left(\frac{h_i}{h_{\max}}\right)^{2/3}. \] Substituindo: \[ \frac{3}{2}(M+m)g h_i\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = mg(h_{\max}-h_i). \] Cancelando \(mg\): \[ \boxed{\frac{3}{2}(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m\left(\frac{h_{\max}}{h_i}-1\right)}. \] Good. Let's think if there is an alternative equation using the pressure at max. Maybe they might want: \[ P_{\max} = \frac{mg}{A} - \frac{m}{A}\ddot{h}? \] No. Let's verify dimensions of b equation: LHS mass, RHS mass. Good. Let's maybe solve for h_max explicitly? For final answer maybe we can leave implicit. But user says "Termine com uma linha neste formato: Resposta final: <o resultado de cada item, de forma curta>" So final line should include a) and b). We can include the equation as answer. It might be acceptable: "b) \frac32(M+m)[1-(h_i/h_{max})^{2/3}]=m(h_{max}/h_i-1)". But maybe they want an equation equaling zero? We'll provide. Let's also consider if part b should include n? No, n cancels. R cancels? R not needed because energy expressed via P,V. Good. Let's ensure no missing factor: Initial gas pressure P_i = (M+m)g/A. Final after sand removed instantaneously, the external pressure is mg/A. But if the piston is at max height, the gas pressure P_max = P_i (h_i/h_max)^γ. Then work done by gas: W = integral P dV. For adiabatic relation P V^γ=const, W = (P_i V_i - P_max V_max)/(γ-1) = [P_i V_i - P_i V_i (V_i/V_max)^{γ-1}]/(γ-1) = P_i V_i [1 - (V_i/V_max)^{γ-1}]/(γ-1). γ-1=2/3, so W = (3/2)P_i V_i [1 - (h_i/h_max)^{2/3}]. P_i V_i=(M+m)g h_i. Thus W=(3/2)(M+m)g h_i [1 - ...]. Equate to mg(h_max-h_i). Yes. But if the process were irreversible, the actual work might differ; but the problem likely assumes the same adiabatic relation. Let's see if there is any clue: Part a "processo quase estático" uses adiabatic law; part b "instantaneamente" maybe they want compare: For quasi-static, process reversible; for instantaneous, irreversible, but they still say "determine a equação" maybe using energy conservation with same adiabatic law? In some exams, they expect using "primeira lei" and "equação de estado" but not necessarily P V^γ = const. Let's try to derive without P V^γ for part b using the fact that at maximum height, the gas has done work equal to \(m g(h_max-h_i)\). But can we get another relation from the fact that when sand is removed instantaneously, the gas expands freely? Wait if the piston has mass, it's not free expansion; but maybe the gas pressure at max equals the external pressure? Let's suppose at maximum height, the piston is momentarily at rest, but there is no damping, so it will reverse; the gas pressure could be less than external. However if we assume the gas reaches mechanical equilibrium at max (net force zero), then P_f=mg/A. Then energy equation: \[ \frac32(P_f h_f - P_i h_i) = -mg(h_f-h_i) \] with P_f=mg/A, P_i=(M+m)g/A. This yields h_f (call h_max) relation: \[ \frac32(m h_max - (M+m)h_i) = -m(h_max-h_i) \] => \(\frac32 m h_max - \frac32 (M+m)h_i = -m h_max + m h_i\) => \(\frac52 m h_max = \frac32(M+m)h_i + m h_i = [\frac32 M + \frac32 m + m]h_i = (\frac32 M + \frac52 m)h_i\) => \(h_max = h_i \frac{3M+5m}{5m}\). This is the same as the irreversible constant external pressure final equilibrium. But that's not maximum; it's the equilibrium point after oscillations. However some might confuse "altura máxima" with final equilibrium? No, because if sand removed instantaneously, the piston overshoots. Let's test with M=0 (no sand): Then external pressure initially mg/A; removing no sand, no motion. Formula gives h_max = h_i (0+5m)/(5m)=h_i, correct. For M>0, h_max > h_i. Compare with energy+adiabatic equation for M large: our implicit equation gives h_max ~ (3M/2m) h_i? Let's approximate: For large M, from adiabatic equation: (3/2)M (1 - x^{-2/3}) ≈ m x => (3/2)M ≈ m x => x≈1.5 M/m. The equilibrium formula: x≈(3M)/(5m)=0.6 M/m, which is much smaller and physically impossible because energy from gas could at least convert to potential; equilibrium height should be less than max. So the equilibrium formula is wrong for max. Good. Thus our adiabatic max equation is physically plausible. But let's verify with exact energy for M large: Initially gas internal energy ~ (3/2)M g h_i. If all converted to potential, h_max ~ (3/2)(M/m) h_i. Our equation for large M gives x ~1.5 M/m? Let's solve asymptotically: (3/2)(M+m)(1 - x^{-2/3}) = m(x-1). For large M, left ≈ (3/2)M, right ≈ m x, so x≈(3/2)M/m. Yes. Good. If we had used constant pressure work only (irreversible final equilibrium), ΔU = U_f - U_i = (3/2)(m g h_f - (M+m)g h_i) = -mg(h_f-h_i), giving h_f ≈ (3M)/(5m)h_i, much smaller; but max should be larger? Actually if external pressure is mg/A constant, the gas does work mg(h_f-h_i) only, so internal energy decrease would be only that; but if gas also had initial high pressure, the piston could gain kinetic energy? Wait if external pressure is constant mg/A, the work done by gas equals mg(h_f-h_i) at maximum as well? But if gas pressure drops below mg/A, then the gas is doing negative work? Actually at max, the gas may have pressure less than external; the piston's kinetic energy? Let's analyze energy: The gas does work on piston only while pressure > mg/A? During motion, gas pressure may fall below mg/A near end; then the piston still moves upward due to inertia, and the gas actually does negative work (piston compresses gas? Actually if gas pressure less than external, the net force is downward, so piston decelerates; the piston still moves upward, gas expands, but work done by gas is negative? Wait work done by gas = ∫ P dV. If P < mg/A and piston moves upward (dV>0), work is positive? No, force on piston by gas = P A upward; displacement upward; work by gas positive regardless of external. The external force (gravity) does negative work. So gas still does positive work, but less than external work; kinetic energy decreases. At max, total work by gas equals increase in potential energy plus kinetic? Actually work-energy theorem for piston: Work by gas + Work by gravity = ΔK =0 => W_gas - mgΔh =0 => W_gas = mgΔh. Yes, same. So even if P < mg/A, gas does positive work as long as it expands. The gas internal energy decreases by W_gas, which equals mgΔh. So the final internal energy depends on total work, which depends on path. The adiabatic relation path gives W_gas. The constant external pressure path (if P=mg/A throughout?) would give W=mgΔh, but that's only if gas pressure equals external throughout (quasi-static with no initial excess? Not possible because initial pressure higher). Actually if external pressure instantaneously drops to mg/A, the gas pressure will initially be higher than mg/A, so work done by gas is greater than mgΔh at the equilibrium point? Wait at equilibrium point, K maybe not zero? At equilibrium, acceleration zero but velocity max maybe; work done by gas = potential + kinetic; not equal potential. At max, work = potential. So W_gas = mgΔh always at max, independent of path! Wait yes: By work-energy theorem for piston, at max K=0, W_gas = mgΔh. This is exact. Therefore internal energy change ΔU = -W_gas = -mgΔh. So regardless of path, ΔU = -mgΔh. But to find ΔU, we need final state; but maybe we can determine final state from another relation? Hmm. This is interesting: Energy conservation gives ΔU = -mgΔh exactly. If we also know initial U_i, we need U_f. But perhaps we can use the fact that at maximum height, the gas has expanded adiabatically and no heat; but the final gas state might not be equilibrium. However if we assume the gas is ideal and the process is adiabatic, the change in internal energy equals work done by gas, which equals -mgΔh. This doesn't require adiabatic relation. But still one equation with two unknowns (U_f, h_max). Need another relation. The only additional physics is the equation of motion and gas dynamics. In the absence of dissipation, the system may conserve entropy? No, irreversible. However maybe the total energy of gas + piston is conserved; kinetic energy of piston at max zero; but gas may have kinetic energy (internal motions). If we assume the gas remains in equilibrium throughout, then P V^γ relation gives U_f. That's the standard assumption. Let's see if there is a more rigorous way: For a reversible adiabatic process, entropy constant. But sudden removal is irreversible, so entropy increases. The final state after damping would have same entropy? No, entropy increases. The maximum height state may have entropy? Hmm. But maybe the gas expansion is fast enough that it's reversible? No. Maybe the problem expects simply using conservation of energy with the gas considered as a spring with adiabatic law. Many high school problems do that. We'll provide that. Let's also consider if the equation in b should include the number of moles and R explicitly? Since n and R are given but not used? Maybe part a uses n,R? We didn't need because pressure determined by mechanics. But the problem gives n and R; maybe they are red herrings or needed if temperature? Let's see: Could part a ask final height as function of m,M,h_i and maybe not n,R? Since n and R cancel; but maybe they expect using \(T_i = P_i V_i/(nR)\) and adiabatic \(T V^{γ-1}=const\). Same result. They included n,R for completeness. Part b maybe could be solved using energy and \(U = (3/2)nRT\): Initial \(T_i = (M+m)g h_i/(nR)\). At max, use adiabatic \(T_i (A h_i)^{γ-1}=T_f (A h_max)^{γ-1}\) to get \(U_f = (3/2)nR T_f\). Then energy. Same. Let's write final answer with equation. Let's also consider maybe part b expects equation involving \(h_{\max}\) and \(m, M, h_i\) only, maybe: \[ m g h_{\max} + \frac32 (M+m)g h_i \left(\frac{h_i}{h_{\max}}\right)^{2/3} = \frac32 (M+m)g h_i + mg h_i \] or something. We'll provide the clean form. Let's ensure no mistakes in internal energy expression: \(U_f = \frac32 P_f V_f\). If \(P_f V_f = (M+m)g h_i (h_i/h_max)^{2/3}\). Yes. Then \(U_i - U_f = \frac32(M+m)g h_i [1 - (h_i/h_max)^{2/3}]\). Set = mg(h_max - h_i). Yes. Let's maybe present b as: \[ U_i - U_f = mg(h_{\max}-h_i) \] \[ \frac32(M+m)g h_i - \frac32(M+m)g h_i\left(\frac{h_i}{h_{\max}}\right)^{2/3} = mg(h_{\max}-h_i) \] \[ \frac32(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right] = m\left(\frac{h_{\max}}{h_i}-1\right). \] Now final line: Resposta final: a) \(h_f = h_i\left(\frac{M+m}{m}\right)^{3/5}\); b) \(\frac32(M+m)\left[1-\left(\frac{h_i}{h_{\max}}\right)^{2/3}\right]=m\left(\frac{h_{\max}}{h_i}-1\right)\). Need maybe include "onde \(h_{\max}\) é a altura máxima" no final? Short. Let's craft final response in Portuguese, with steps. Since user says "Resolva a questão como um candidato faria na prova: mostre o desenvolvimento em português, item por item". We'll do. Let's also mention \(\gamma=5/3\) for monoatômico. Potential issue: In part b, energy conservation for entire system includes the gas internal energy and gravitational potential energy of piston. But the piston also has mass m; the sand removed instantaneous; the sand mass M initially at height h_i and final maybe? The sand removed no longer part of system; but initially system included sand? The energy equation for gas+piston after removal: initial state includes sand on piston but sand is removed; the sand's potential energy removed? The gas does work against piston+sand initially? Actually before removal, the system is in equilibrium with sand; after removal, the sand is no longer coupled. The work done by gas after removal is only against piston weight mg. The initial internal energy of gas includes energy due to pressure from sand. So energy conservation for gas+piston: The gas internal energy initially U_i (with sand pressure), final U_f, plus piston potential energy change m g(h_f-h_i), equals zero? The sand's potential energy initial/final? Sand is removed; it may have been lifted? But removed at initial height? It's removed instantaneously; no work done by gas on sand after removal. So yes. But wait, if sand is removed instantaneously, the piston may move while sand is already gone; the gas initially has high pressure due to sand, but the sand is no longer there; the gas expands against only piston weight. So energy equation as above. Let's maybe consider if the piston's initial potential energy should be included? It cancels in ΔU + ΔE_p = 0 if the system is gas + piston; but the gas initial internal energy includes work done to support sand; okay. Let's verify with a numerical example for part b maybe: M=m, h_i=1. Equation: (3/2)*2 [1 - x^{-2/3}] = 1*(x-1) => 3[1 - x^{-2/3}] = x-1 => 4 - x = 3 x^{-2/3}. Solve x? x= ? Try x=2: 4-2=2; 3*2^{-2/3}=3*0.63=1.89; not. x=2.1: 1.9 vs 3*2.1^{-0.666}=3*0.59=1.77. x=1.9:2.1 vs 3*1.9^{-0.666}=3*0.653=1.96. x≈1.93. So h_max≈1.93 h_i. Initial internal energy = (3/2)(2m)g h_i=3mg h_i; potential increase ~0.93mg h_i, remaining internal energy ~2.07mg h_i. Final temperature lower. Seems plausible. Part a for M=m: h_f = h_i (2)^{3/5}=1.515 h_i. So quasi-static final equilibrium height 1.515; sudden max height 1.93; overshoot. Good. Let's ensure adiabatic relation for part a: γ=5/3. \(h_f = h_i [(M+m)/m]^{1/γ} = h_i (2)^{3/5}\). Yes. Now final answer should not be too long. We'll provide. Let's also note "Quando necessário, use ... senθ ≈ θ" not relevant. Let's craft final with equations neatly. Potential nuance: "cilindro vertical, de seção transversal de área A ... fechado na parte superior por um pistão ... Considerando que há vácuo acima do pistão" So above piston vacuum; external pressure atmospheric none. Good. Let's write final in Portuguese. Use \(P V^\gamma\) notation. For monoatomic ideal gas, \(C_V = 3R/2\), \(\gamma = C_P/C_V = 5/3\). Initial pressure \(P_i = (M+m)g/A\). Volume \(V_i=A h_i\). Final pressure in item a \(P_f = mg/A\). Let's include in part a: \[ P_i V_i^\gamma = P_f V_f^\gamma \Rightarrow \frac{(M+m)g}{A}(A h_i)^{5/3} = \frac{mg}{A}(A h_f)^{5/3} \] ... done. Part b: Use energy. We can mention "No instante de altura máxima, a energia cinética do pistão é nula." Then equation. Let's final with boxed equations. Final line: Resposta final: a) \(h_f=h_i\left(\dfrac{M+m}{m}\right)^{3/5}\); b) \(\dfrac32(M+m)\left[1-\left(\dfrac{h_i}{h_{\max}}\right)^{2/3}\right]=m\left(\dfrac{h_{\max}}{h_i}-1\right)\). Let's ensure no need to solve b for h_max. "Determine a equação que relaciona" so equation is fine. Now produce final.

**Analyzing Physics Problem** I'm currently focused on understanding the nuances of this ITA physics problem. My primary task is a thorough reading of the discursive exam text. I've noted the specific instruction regarding the small angle approximation for sine, which will likely be crucial. **Calculating Final Height** I've been working through part (a) of the physics problem. My main focus has been on the hydrostatic and thermodynamic equilibrium before and after the sand is removed. I'm deriving the final height $h_f$ based on the initial conditions and the change in mass on the piston. **Deriving New Relation** I'm now focusing on part (b) of the problem, aiming to derive the equation relating maximum height $h_{max}$ to initial parameters. This involves analyzing the subsequent motion after all sand is removed, considering the gas expansion and energy transformation. **Calculating Final Height** I have successfully derived the final height $h_f$ for the adiabatic reversible process. The key was equating the initial and final states, accounting for the masses and volumes, and utilizing the monoatomic gas property $\gamma = 5/3$, resulting in $h_f = h_i (1 + M/m)^{3/5}$. **Analyzing Gas Expansion** I'm re-evaluating the process of gas expansion after the sand removal. I'm considering whether the expansion is reversible, given the frictionless piston and adiabatic walls, and how this impacts energy conservation for the system. **Considering Reversibility** I'm questioning the reversibility of the piston's rapid expansion. While the walls are adiabatic and the piston frictionless, a fast movement might imply irreversibility, affecting the work calculation. I'm exploring the conservation of total energy instead, looking at changes in internal energy, kinetic, and potential energy. **Revisiting Expansion Dynamics** I'm reassessing the nature of the gas expansion. My current focus is on whether the process is isentropic by examining the relationship between work done by the gas and the change in its internal energy, and comparing this with the work done against gravity. I'm confirming if the derived condition for maximum height holds true under isentropic assumptions. **Confirming Reversibility Assumptions** I'm confirming that the standard assumption in such physics problems is indeed quasi-static equilibrium, treating the gas expansion as adiabatic and reversible (isentropic). This aligns with the negligible gas mass and high speed of sound relative to the piston. **Checking Isentropic Relation** I'm verifying the isentropic assumption. I'm deriving the internal energy change based on the ideal gas law and comparing it to the work done against gravity to check if the derived condition for maximum height holds. **Revising Irreversibility Concerns** I'm currently grappling with the implications of potential irreversibility during the piston's rapid expansion after sand removal. My thoughts are focused on how this would affect the calculation of $h_{max}$, particularly if dissipation mechanisms are at play. However, I'm reconfirming that in typical problem contexts, an isentropic expansion is assumed to yield a unique solution for the maximum height. **Revising Final Equilibrium Understanding** I'm re-evaluating the concept of final equilibrium in this problem. I realize that for the state where the piston *eventually* comes to rest, the process is indeed irreversible, and the work done against gravity is directly related to the change in internal energy. **Analyzing Oscillation Behavior** I'm now focusing on the oscillation dynamics after sand removal. The phrasing "maximum height reached" implies an overshoot and turning point where velocity is zero, not simply a final equilibrium. I'm also realizing that the problem likely asks for a relation, not an explicit solution for $h_{max}$, due to the complex, non-linear equation it generates. **Deriving Oscillation Equation** I'm now focused on deriving the equation relating the maximum height $h_{max}$ to the initial parameters, as the problem statement explicitly asks for the equation, not a direct solution. I'm also confirming that the small angle approximation clue is irrelevant here, as no angles are involved. **Revising Problem Interpretation** I'm re-examining part (b), realizing the request is for an equation relating $h_{max}$ to initial parameters, not a direct solution. The "instantaneously removed" phrasing might suggest irreversible work, but the vacuum above the piston simplifies forces to $mg + ma$. I'm checking if a reversible, adiabatic expansion ($P_{gas}V^\gamma = \text{constant}$) is still the correct framework. **Revising Gas Work Understanding** I'm reconsidering how to model the work done by the gas. With a vacuum above the piston, the external pressure is zero, so the gas work directly translates to changes in the piston's kinetic and gravitational potential energy, not work against an external pressure. **Deriving Final Relation** I've confirmed that total energy conservation, $\Delta U_{gas} + mg(h_{max} - h_i) = 0$, holds for the gas and piston system regardless of reversibility, given no dissipation. The key now is relating $U_{gas}$ to $h_{max}$, which requires the isentropic assumption, $T \propto h^{-\frac{2}{3}}$, to determine gas temperature at maximum height. **Confirming Isentropic Process** I'm confirming that the gas expansion after sand removal is indeed isentropic. This means the process is both adiabatic and reversible. I've established that total energy conservation dictates $\Delta U_{gas} + mg(h_{max} - h_i) = 0$, and now I'm focusing on relating internal energy to $h_{max}$ via the isentropic relation $T \propto h^{-2/3}$ to derive the final equation. **Deriving Isentropic Relation** I'm now focusing on the derivation using the isentropic relationship $T \propto h^{-2/3}$ and energy conservation $\Delta U_{gas} + mg(h_{max} - h_i) = 0$. I'm substituting the internal energy expression $U(h) = U_i (h_i/h)^{2/3}$ into the energy equation to formulate the final relationship for $h_{max}$. **Formulating Final Equation** I am now directly substituting the isentropic relation $T \propto h^{-2/3}$ into the energy conservation equation. My focus is on manipulating these expressions to derive the relationship for $h_{max}$ in terms of $m$, $M$, and $h_i$. I am also exploring alternative forms of this equation, including expressing it in terms of the ratio $h_{max}/h_i$. **Deriving Adiabatic Relation** I'm now applying the adiabatic reversible process conditions to derive the final height $h_f$. Using $P V^{5/3} = \text{const}$ and the initial and final pressures and volumes, I've arrived at $h_f = h_i (\frac{M+m}{m})^{3/5}$, which represents the final equilibrium height. **Formulating Oscillation Equation** I'm now focused on deriving the equation relating maximum height $h_{max}$ to initial parameters. I realize the phrasing implies an overshoot and turning point. I'm confirming that this involves total energy conservation for the gas and piston, relating internal energy change to gravitational potential energy change, assuming an isentropic expansion. **Analyzing Gas Expansion Process** I am now analyzing the expansion process of the gas after the load is instantaneously removed. My current focus is on whether this expansion can be treated as reversible and adiabatic (isentropic) given the problem's setup, particularly in the context of energy conservation and the resulting dynamics of the piston's movement. **Considering Alternative Processes** I am re-evaluating the core assumption regarding the gas expansion. While Possibility 1 assumes an isentropic process, I am exploring if there's a valid alternative scenario, particularly given the vacuum above the piston and the explicit absence of external atmospheric pressure, which seems to rule out a constant external pressure scenario. **Formulating Motion Equation** I am now constructing the equation of motion for the piston. My current focus is on integrating this equation with respect to time, starting from the initial state to the point of maximum height, to derive the relationship between the work done by the gas and the change in potential and kinetic energy. **Confirming Energy Balance** I'm confirming the energy balance for the gas and piston system, ensuring $\Delta U_{gas} + mg(h_{max} - h_i) = 0$. My current focus is on correctly substituting the derived expression for $\int P dV$ into this energy conservation equation to formulate the relationship for $h_{max}$. **Testing Equilibrium Point** I am now testing the derived equilibrium height $h_f$ as the center of oscillation. I'm confirming that $h_{max}$ should be greater than $h_{eq}$ since the piston starts at $h_i < h_{eq}$ with zero velocity, and I'm checking if the oscillation is symmetric around this point by examining edge cases and small perturbations. **Confirming Harmonic Oscillation** I've confirmed that for small deviations from equilibrium, the system behaves as a harmonic oscillator. The derived relationship for the maximum height overshoot is exactly twice the equilibrium displacement, matching the behavior of a simple harmonic oscillator released from rest. **Validating Final Equation** I've successfully confirmed my derived equation for $h_{max}$ through multiple analytical checks, including comparing it to a known limit and checking its consistency with harmonic oscillator behavior. The equation appears robust and accurate across various scenarios. **Exploring Equation Forms** I am now exploring different equivalent algebraic presentations of the equation relating $h_{max}$ to initial parameters. This involves manipulating the equation to simplify it and present it in various forms, which may offer different insights or ease of application. **Exploring Equation Forms** I am currently exploring various algebraic manipulations of the derived equation, aiming to present it in its cleanest and most insightful forms. My focus is on ensuring the equation clearly relates $h_{max}$ to the initial parameters $m$, $M$, and $h_i$. **Revising Gas Work Understanding** I'm reconsidering how to model the work done by the gas. With a vacuum above the piston, the external pressure is zero, so the gas work directly translates to changes in the piston's kinetic and gravitational potential energy, not work against an external pressure. **Analyzing Gas Expansion Dynamics** I'm analyzing the gas expansion after instantaneous sand removal. My focus is on the initial moment where the pressure drops significantly, causing the piston to accelerate upwards. I'm evaluating the energy transfer and potential overshoot beyond the new equilibrium point. **Formulating Oscillation Equation** I'm now focused on deriving the equation relating maximum height $h_{max}$ to initial parameters. I realize the phrasing implies an overshoot and turning point. I'm confirming that this involves total energy conservation for the gas and piston, relating internal energy change to gravitational potential energy change, assuming an isentropic expansion. **Confirming Energy Conservation** I am confirming the energy balance for the gas and piston system, ensuring $\Delta U_{gas} + mg(h_{max} - h_i) = 0$. My current focus is on correctly substituting the derived expression for the internal energy, based on the isentropic relation $T \propto h^{-2/3}$, into this energy conservation equation to formulate the relationship for $h_{max}$. **Deriving Motion Equation** I am currently focusing on the work-energy theorem for the piston. I've established that the work done by the gas must equal the negative of the work done against gravity, $W_{gas} = mg(h_{max} - h_i)$, and I'm now equating this to the internal energy change of the gas, $W_{gas} = -\Delta U_{gas}$, to finalize the equation.

**Analyzing Physics Problem** I am currently focused on understanding the initial conditions and constraints of the ITA physics problem involving a vertical adiabatic cylinder and a piston. My immediate goal is to fully grasp the physical setup and the properties of the gas and piston. **Processing Adiabatic Expansion** I'm currently working through the quasi-static removal of sand in part (a). My focus is on applying principles of mechanical equilibrium to relate the initial and final states, accounting for the decreasing mass on the piston as sand is removed. I will then move to analyze the instantaneous removal scenario in part (b). **Calculating Final Height** I have successfully derived the final height $h_f$ for the quasi-static process using the adiabatic expansion formula. I confirmed that the process is reversible adiabatic and that $\gamma = 5/3$ for a monatomic ideal gas. The expression $h_f = h_i (1 + M/m)^{3/5}$ matches the requirements of part (a). **Evaluating Instantaneous Process** I am now considering the instantaneous sand removal in part (b). I am debating whether this process is reversible or irreversible and how that impacts energy conservation. My current understanding is that it's an adiabatic expansion, but I need to clarify the specific constraints on the gas and piston's internal equilibrium. **Analyzing Gas Behavior** I am currently examining the state of the gas at maximum piston height. I am exploring two distinct interpretations: one where the gas expansion is internally reversible and follows $PV^\gamma = \text{constant}$, and another where the gas mass is negligible. I need to carefully assess the implications of each for energy conservation. **Clarifying Maximum Height** I am re-evaluating the concept of maximum piston height. I've realized that regardless of reversibility, the energy balance leads to $\Delta U + mg(h_{\text{max}} - h_i) = 0$. The key difference lies in how $U_{\text{max}}$ is determined; in a reversible scenario, it's governed by adiabatic relations, but in irreversible cases, the nature of the external force becomes critical. I need to ascertain the precise scenario described by the problem. **Investigating Equation Form** I am now considering why the problem asks for an "equation that relates" maximum height rather than a direct expression. This suggests the relationship might be transcendental or a complex algebraic equation, not easily solved for $h_{max}$ explicitly. My current focus is on understanding this potential complexity. **Confirming Equation Form** I've confirmed that the question's phrasing, "equation that relates," is indeed due to the mixture of an adiabatic process and linear gravitational potential energy, preventing explicit algebraic isolation of $h_{max}$. This insight perfectly clarifies the problem's intent. **Deriving Isentropic Relation** I'm now meticulously detailing the physics for part (b), specifically the application of $PV^\gamma = \text{const}$. My focus is on justifying this relation by considering the ideal gas, adiabatic nature, and the assumption of near-instantaneous internal equilibrium, ensuring a clear and rigorous derivation for the graders. **Applying Energy Conservation** I am now applying the first law of thermodynamics, $\Delta U + \Delta E_p = 0$, to the instantaneous process. I've established the initial internal energy and potential energy, and I'm working on the final internal energy using the isentropic relation $PV^{5/3} = \text{constant}$, relating it to the maximum height. **Simplifying Energy Equation** I am now simplifying the derived energy equation for the instantaneous process. My current effort focuses on algebraically manipulating the expression to isolate terms and present it in its most concise and clear form, ensuring it accurately reflects the physics of the system's maximum height.

**Analyzing hydrodynamic equations** I'm considering using the Euler hydrodynamic virial theorem with height-varying pressure. However, I realize there might not be a unique solution given finite speed. I need to decide how to solve this correctly while acknowledging the distinction between quasistatic and sudden processes. It seems I should treat sudden load removal carefully, as it may not be irreversible. The assumption of uniform gas properties at each instant complicates things, and I must ensure that the equation accounts for internal equilibrium.**Assessing thermodynamic relations** I need to evaluate whether an exact relationship holds using the adiabatic law, especially if the process is irreversible. If there are no friction and insulated walls, the process could be reversible, assuming an ideal gas without dissipation. I'm pondering if acceleration leads to thermodynamic irreversibility given the spatially non-uniform pressure and speeds. The reversible continuum Euler equations suggest that entropy might be conserved locally. At maximum height, while the piston is at rest, the gas continues to move in bulk.**Analyzing collisionless molecules** I'm considering collisionless specular molecules, which are noninteracting point particles in an ideal gas. In a rapidly changing piston, the initial stop velocities aren't distributed in a Maxwellian way, and it seems the moments correlate differently. This setup leads me to think about how the ideal gas behaves under initial equilibrium conditions, without taking adiabatic processes into account, alongside entropy considerations. It's an interesting but complex scenario!**Considering thermodynamic identities** I'm thinking there might be a thermodynamic identity that relates to deriving the adiabatic pressure law, even in non-equilibrium conditions. The classic ideal gas law is pV = nRT, but when the gas isn't uniform, things get tricky. I should be cautious not to overcomplicate my final explanation, as the user may want a clear formula. It seems important to include a conditional result along with a caveat about these complexities.