Questão 40

**Evaluating pressure contributions** I’m considering the PX + PZ interaction where B contributes all non-Y components. In B, the total initial pressure across the volume is 2P for a volume of 2V. Given that the amount proxy is 4PV, the final pressure contribution calculates to 4PV divided by 6V, simplifying to 2P/3. This gives me insight into how pressures can interplay in this context.**Considering gas mixtures** I'm thinking about the pressure involved in mixtures and the importance of ensuring that the initial pressure of a certain species is set properly. It looks like I need to ensure there's no unwanted contribution from another species. I should make sure to calculate the final partial pressure correctly using the ideal gas law, while keeping total volumes rigid at the same temperature. Let’s gather examples to clarify this and wrap it up with a conclusion!

I'm computing total moles weighted by each container's pressure and volume, finding the final mixed pressure comes out to 7P/3. Then I'm working through each container's partial pressure contribution to figure out how gases X, Y, Z are distributed among containers A and B. Now I'm checking the sum ranges for pairs: P_X+P_Y gives P/6 to 7P/3, which doesn't match option A's claimed range. P_X+P_Z gives exactly 2P/3 to 7P/3, matching option C perfectly. I'm now working out the P_Y+P_Z range using container C's full contribution plus the variable parts from A and B.

I'm computing total moles via PV proportionality for each vessel: 6V total volume, with A=PV, B=4PV, C=9PV giving 14PV total and final pressure 7P/3. Then I'm setting up equations for gas amounts, where A holds X+Y=1 unit, B holds X+Z=4 units, and C holds Y+Z=9 units, with final partial pressures given by each species' moles divided by 6V. Working through the ranges: X ranges 0 to 5 giving partial pressure 0 to 5P/6, Y ranges 0 to 10 giving 0 to 5P/3, and Z ranges 0 to 13 giving up to 13P/6. Checking combined sums against each answer option, I find that X+Z spans from 2P/3 to 7P/3, matching option C exactly, while the other combinations don't match their respective options.

I'm setting up three containers with volumes V, 2V, 3V and pressures P, 2P, 3P, computing the total "pV" sum as 14PV so the final mixed pressure comes to 7P/3. Now I'm assigning which two gases occupy each container and writing the partial pressure balance equations to solve for the final partial pressure of gas X across the combined volume. Working through the ranges, I find P_X can reach up to 5P/6, P_Y up to 5P/3, and P_Z up to 13P/6, and I'm testing each answer choice by substituting these bounds. Checking the sum P_X+P_Y I get a range from P/6 to 7P/3, ruling out option A, and for P_X alone the upper bound of 5P/6 exceeds the proposed 2P/3 limit, ruling out option B — so I'm now evaluating option C by computing the min and max of P_X+P_Z. The sum P_X+P_Z gives exactly 2P/3 to 7P/3, matching option C precisely. Checking the remaining options, P_Y's range of 0 to 5P/3 rules out D, and P_Y+P_Z gives 3P/2 to 7P/3, ruling out E — confirming C as the answer.

The problem involves three rigid containers A, B, and C with volumes V, 2V, and 3V. Each contains an ideal gas mixture of up to three inert gases X, Y, and Z at the same temperature. Initial total pressures are P, 2P, and 3P. Partial pressures after equilibrium are P_X, P_Y, and P_Z. Container A lacks gas Z. Container B lacks gas Y. Container C lacks gas X. The task is to find the strict limits on these partial pressures. The total gas amount is 14 PV. After connecting the containers, the total volume becomes 6V. The final total pressure is therefore 7P/3. This yields the constraint P_X + P_Y + P_Z = 7P/3. Gas X can only occupy containers A and B. Gas Y can only occupy containers A and C. Gas Z can only occupy containers B and C. Partial pressures may be zero if a gas is absent from a container. The initial pressure constraints are now explicit. Container A has partial pressures p_{A,X} and p_{A,Y} summing to P. Container B has p_{B,X} and p_{B,Z} summing to 2P. Container C has p_{C,Y} and p_{C,Z} summing to 3P. All partial pressures must be non-negative. The final partial pressures are now expressed as weighted averages of the initial partials. P_X equals (p_{A,X} + 2 p_{B,X}) / 6. P_Y equals (p_{A,Y} + 3 p_{C,Y}) / 6. P_Z equals (2 p_{B,Z} + 3 p_{C,Z}) / 6. The next step is to determine the tightest possible bounds for these expressions. The phrase "estritamente aplicáveis" means the tightest possible bounds that always hold. I am now parameterizing the initial partial pressures with variables a and b, and deriving their constraints. The parameterization is now complete with c for container C. The final partial pressures are P_X = (a + 2b)/6, P_Y = (P − a + 3c)/6, and P_Z = (13P − 2b − 3c)/6. Their sum is confirmed to be 7P/3. The range of P_X is from 0 to 5P/6, which is achievable when a = P and b = 2P. The variables remain independent. The range of P_Y is 0 to 5P/3. The range of P_Z is 0 to 13P/6. The next step is to determine the tightest constraints on combinations such as P_X + P_Y. Checking option (A): P/3 ≤ P_X + P_Y ≤ 5P/2. Since P_X + P_Y + P_Z = 7P/3, this is equivalent to bounding P_Z. The maximum value of P_Z is 13P/6, which occurs when b and c are both zero. I am now verifying whether this maximum is physically achievable. The maximum P_Z = 13P/6 is physically achievable when b = c = 0, leaving only gas Z in containers B and C. The minimum P_Z = 0 is also reachable when b = 2P and c = 3P, which places all X in B and all Y in C while eliminating Z entirely. Therefore P_X + P_Y ranges from P/6 to 7P/3. This contradicts option (A), which claims the upper bound is 5P/2. Option (A) is false because the lower bound P/3 is exceeded when P_X + P_Y equals P/6. Option (B) is also false since P_X can reach 5P/6, which exceeds 2P/3. The analysis now turns to option (C), where P_X + P_Z equals 7P/3 minus P_Y. Option (C) is confirmed as the correct answer. The range of P_Y is 0 to 5P/3, so P_X + P_Z ranges from 2P/3 to 7P/3. Both bounds are achievable. The next step is to verify whether any other inequalities also hold. Option (D) is false because P_Y can reach 5P/3, which exceeds P/2. Option (E) is also incorrect: the true range of P_Y + P_Z is 3P/2 to 7P/3, not the claimed 5P/3. The lower bound of P_X + P_Z = 2P/3 is confirmed as exactly reachable. With a = 0 and c = 3P, the expression simplifies to b/3 + (2P − b)/3 = 2P/3 for any valid value of b.

We need answer in Portuguese, solve MCQ. Need reason. Let's restate problem. Three rigid recipients A, B, C volumes V, 2V, 3V, same T, contain ideal gas mixture up to three inert gases X,Y,Z. Initially isolated, total pressures P, 2P, 3P respectively. Partial pressures unknown. After simultaneous opening valves connecting them, equilibrium at same T, final uniform partial pressures in whole system are P_X, P_Y, P_Z. Before opening, know: i. A does not contain gas Z. ii. B does not contain gas Y. iii. C does not contain gas X. Indicate which alternative presents strictly applicable limits to process, in terms of P_X,P_Y,P_Z. Options: (A) P/3 ≤ P_X + P_Y ≤ 5P/2 (B) 0 ≤ P_X ≤ 2P/3 (C) 2P/3 ≤ P_X + P_Z ≤ 7P/3 (D) 0 ≤ P_Y ≤ P/2 (E) P/3 ≤ P_Y + P_Z ≤ 5P/3 Need determine correct bounds. Need be careful with "limites estritamente aplicáveis" maybe means tightest possible bounds given constraints. This is a mixing problem. Let's derive. Let total moles n_A, n_B, n_C? Ideal gas: P V = n_A R T? For mixture, total pressure = sum partial pressures. Initially total pressures: A = P = sum partial pressures in A over gases present; B = 2P; C = 3P. After mixing all volumes total V+2V+3V = 6V. Same T. Final partial pressure of each gas i (X,Y,Z) = total moles of that gas across all vessels * R T / (6V)? Actually final uniform partial pressure p_i = n_i_total R T / V_total. Initially in vessel j with volume V_j, if gas i present, its partial pressure p_i^{(j)} = n_i^{(j)} R T / V_j. So n_i^{(j)} = p_i^{(j)} V_j / (RT). Total n_i = (1/RT) sum_j p_i^{(j)} V_j. Final p_i = n_i_total RT / 6V = (1/6V) sum_j p_i^{(j)} V_j. Also total initial pressure contribution? Let's express using mole fractions maybe. Let final partial pressures P_X, P_Y, P_Z. Since all gases inert and no reaction, total final pressure = P_X+P_Y+P_Z. We can relate to initial total pressures by conservation of total moles: total initial "PV" sum? For total moles, n_total = sum_j P_j V_j / RT = (P*V + 2P*2V + 3P*3V)/RT = (1 +4+9) PV/RT =14 PV/RT. Final total pressure = n_total RT /6V =14P/6 = 7P/3. So P_X + P_Y + P_Z = 7P/3. Check options maybe. Need constraints due to each vessel missing certain gas and initial total pressures. We need possible range of final partial pressures given unknowns. Let's define for each gas i and vessel j, partial pressure before opening, denote x_{i,j} ≥0, with condition: if gas i absent in vessel j then x_{i,j}=0. Also sum_i x_{i,j} = initial total pressure of vessel j: for A = P, B = 2P, C = 3P. And final P_i = (1/6)[V x_{i,A} + 2V x_{i,B} + 3V x_{i,C}] / V? Actually formula: P_i = (1/(6V)) [V x_{i,A} + 2V x_{i,B} + 3V x_{i,C}] = (1/6)(x_{i,A} + 2x_{i,B} + 3x_{i,C}). Good. Conditions: A: no Z => x_{Z,A}=0, but x_{X,A}+x_{Y,A}=P. B: no Y => x_{Y,B}=0, x_{X,B}+x_{Z,B}=2P. C: no X => x_{X,C}=0, x_{Y,C}+x_{Z,C}=3P. All x ≥0. We need possible ranges for linear combinations? Options maybe involve sums P_X+P_Y, P_X+P_Z, etc. Since total fixed 7P/3, some follow. Let's compute P_X, P_Y, P_Z in terms of independent variables. Let variables: A: x_X,A = a, x_Y,A = P - a (since a between 0,P), x_Z,A=0. B: x_X,B = b, x_Z,B = 2P - b (since b between 0,2P), x_Y,B=0. C: x_Y,C = c, x_Z,C = 3P - c (c between 0,3P), x_X,C=0. Then: P_X = (1/6)[a + 2b + 3*0] = (a + 2b)/6. P_Y = (1/6)[(P-a) + 0 + 3c] = (P - a + 3c)/6. P_Z = (1/6)[0 + 2(2P - b) + 3(3P - c)? Wait C: z_C = 3P - c; B z_B=2P-b; A z_A=0. Coefficient 1,2,3 => (1/6)[0 + 2*(2P - b) + 3*(3P - c)] = (1/6)[4P - 2b + 9P - 3c] = (13P - 2b - 3c)/6. Sum: (a+2b + P-a+3c + 13P -2b -3c)/6 =14P/6 =7P/3 yes. Variables: 0≤a≤P, 0≤b≤2P, 0≤c≤3P. Then P_X = (a+2b)/6. P_Y = (P - a + 3c)/6. P_Z = (13P -2b -3c)/6. Need identify correct bounds. Options: A: P/3 ≤ P_X + P_Y ≤ 5P/2. Compute P_X+P_Y = (a+2b + P-a+3c)/6 = (P + 2b +3c)/6 = P/6 + b/3 + c/2. Since b∈[0,2P], c∈[0,3P], min = P/6 +0+0 = P/6? Wait lower not P/3? Maybe option says P/3 ≤ P_X+P_Y ≤ 5P/2. Our min P/6 = 0.1667P < P/3=0.333P, so not strict lower? Maybe due other constraints? Let's check can b=c=0? Then a? Need A contains X/Y, no Z, a∈[0,P]; B contains X,Z only, b=0 means B all Z pressure 2P; C contains Y,Z only, c=0 means C all Z pressure 3P. Then P_X = a/6, P_Y = (P-a)/6. P_X+P_Y = P/6. So P/3 lower false. Option A maybe not. Option C: 2P/3 ≤ P_X+P_Z ≤ 7P/3. Compute P_X+P_Z = (a+2b + 13P -2b -3c)/6 = (a +13P -3c)/6 = 13P/6 + a/6 - c/2. Since a∈[0,P], c∈[0,3P]. Min =13P/6 +0 - (3P)/2 =13P/6 -9P/6 =4P/6=2P/3. Max =13P/6 + P/6 -0 =14P/6=7P/3. So option C exactly lower 2P/3 ≤ ≤7P/3. This seems true. But we need check "estritamente aplicáveis" maybe means tight bounds attainable? Attainable: min at a=0,c=3P; max at a=P,c=0. So yes. Option E: P/3 ≤ P_Y+P_Z ≤ 5P/3. Compute P_Y+P_Z = (P - a +3c +13P -2b -3c)/6 = (14P - a -2b)/6 = 7P/3 - (a+2b)/6 = 7P/3 - P_X. Since P_X ∈ [0,?]. What bounds P_X? Option B says 0 ≤ P_X ≤ 2P/3. Is that true? P_X=(a+2b)/6, a∈[0,P], b∈[0,2P], so max = (P+4P)/6=5P/6, not 2P/3. So option B false? Wait maybe due B cannot contain Y but total pressure 2P, yes b can be 2P, P_X max= (P+4P)/6=5P/6. Option B says ≤2P/3. Not tight. So E would then P_Y+P_Z = 7P/3 - P_X, with P_X max 5P/6 gives min = 7P/3 -5P/6 =14P/6 -5P/6=9P/6=3P/2=1.5P, not P/3. So E false. Option D: 0 ≤ P_Y ≤ P/2. Compute P_Y=(P - a +3c)/6. Max when a=0,c=3P: (P+9P)/6=10P/6=5P/3 > P/2. Min when a=P,c=0:0. So D false. But question: "Indique qual das alternativas a seguir apresenta os limites estritamente aplicáveis ao processo, em termos de P_X, P_Y e P_Z." Maybe there is only one correct? Let's verify all. Our derivation found C true. But need ensure no other hidden constraints? Let's investigate thoroughly. Because problem likely expects one answer. Let's analyze all possible constraints due to missing gases. Maybe there is nuance: The initial total pressures P,2P,3P; "contêm uma mistura gasosa ideal formada por até três gases inertes X,Y,Z" So each vessel may contain subset of gases, with partial pressures possibly zero. Conditions specify absent certain gases, but not necessarily other gases present? They may contain "até três" but conditions say "não contém" one. It could contain the other two or one? But total pressure given; partial pressures nonnegative; if absent set zero. We included variables. But perhaps question asks "limites estritamente aplicáveis" meaning inequalities that always hold for any possible initial distribution satisfying given absences. We found C. Need test if option A/B/D/E maybe also always hold? Let's test. Option A: P/3 ≤ P_X+P_Y ≤ 5P/2. We saw min P/6 (if b=c=0,a=P gives P_X+P_Y=P/6? Wait if b=0,c=0,a=P => P_X=P/6, P_Y=0? Actually P_Y=(P-P+0)/6=0, sum P/6). So A lower fails. But maybe b=0,c=0,a=P impossible? Check B all Z pressure 2P, C all Z pressure 3P; A all X pressure P. Then final P_X=(P +0+0)/6=P/6, P_Z=(0+4P+9P?) Wait B Z=2P, C Z=3P => P_Z=(2*2P+3*3P)/6=13P/6. Total P_X+P_Y = P/6 < P/3. So A not. Option B: 0 ≤ P_X ≤ 2P/3. We saw max 5P/6? Let's test possible distribution: A: P_X=P, P_Y=0,Z=0; B: P_X=2P (all X), P_Z=0; C: no X, all Z=3P. Then P_X=(P +2*2P +0)/6=5P/6? Wait B x_X=2P, yes a=P, b=2P => (P+4P)/6=5P/6. So B false. Option D: 0 ≤ P_Y ≤ P/2. Max maybe? A all Y? Let's max P_Y: A a=0 (all Y=P), B b=0? B no Y, all Z=2P; C c=3P (all Y=3P, Z=0). Then P_Y=(P +9P)/6=10P/6=5P/3. So D false. Option E: P/3 ≤ P_Y+P_Z ≤ 5P/3. Since P_Y+P_Z =7P/3 - P_X. Min when P_X max 5P/6 gives 7P/3 -5P/6 =14P/6 -5P/6=9P/6=3P/2=1.5P > P/3. So lower false. Max when P_X min 0? P_X can be 0, then P_Y+P_Z=7P/3 ≈2.333P >5P/3=1.667P. So upper false too. So only C. But let's ensure "estritamente aplicáveis" maybe they want range for each final partial pressure? But options are inequalities. C seems correct. Need produce explanation. Let's craft final answer in Portuguese with equations. Need be mindful: The problem statement says "os limites estritamente aplicáveis ao processo" maybe "strictly applicable" means "válidos" (tight). Our derivation for C: 2P/3 ≤ P_X + P_Z ≤ 7P/3. We should show via variables and highlight min/max. Let's produce final with notation: Sejam a = p_{X,A}, b = p_{X,B}, c = p_{Y,C}. Then constraints: A: p_{X,A}=a, p_{Y,A}=P-a, p_{Z,A}=0, 0≤a≤P. B: p_{X,B}=b, p_{Z,B}=2P-b, p_{Y,B}=0, 0≤b≤2P. C: p_{Y,C}=c, p_{Z,C}=3P-c, p_{X,C}=0, 0≤c≤3P. Então: P_X=(a+2b)/6 P_Y=(P-a+3c)/6 P_Z=(13P-2b-3c)/6 Daí: P_X+P_Z = (a+2b +13P -2b -3c)/6 = (a +13P -3c)/6. Como 0≤a≤P e 0≤c≤3P: Mínimo: 13P/6 + 0 - 3*(3P)/6 = 13P/6 - 9P/6 =4P/6=2P/3. Máximo: 13P/6 + P/6 - 0 =14P/6=7P/3. Logo 2P/3 ≤ P_X+P_Z ≤7P/3. Check if other options? Maybe not needed but mention "as demais não são estritamente aplicáveis". Final answer C. Let's ensure no mistake in P_Z formula: A z=0; B z=2P-b; C z=3P-c. Weighted sum: 1*0 +2*(2P-b)+3*(3P-c)=4P-2b+9P-3c=13P -2b -3c. Yes. Potential issue: "as pressões parciais finais uniformes em todo o sistema são P_X, P_Y, P_Z" after equilibrium. Conservation of total moles: final total pressure =? Initial total moles: for vessel i, n_i_total = P_i V_i/(RT). Total initial moles = (P*V +2P*2V+3P*3V)/(RT)=14PV/RT. Final volume=6V => total pressure=14P/6=7P/3. So P_X+P_Y+P_Z=7P/3. Option C not involving sum all? It involves P_X+P_Z. Good. Let's finalize. Need "Resposta: C" last line exactly. Include explanation.

**Analyzing Thermodynamics Problem** I've begun processing an image containing a thermodynamics/ideal gas question from the IME exam. My current focus is on meticulously reading and understanding the problem statement, which involves three rigid containers, A, B, and C, with varying volumes, holding a mixture. **Deconstructing Thermodynamics Setup** I'm now delving into the specifics of the three rigid containers and their contents. My attention is on noting the initial conditions: container A lacks gas Z, B lacks Y, and C lacks X. The system starts with total pressures P, 2P, and 3P respectively, all at the same temperature. **Calculating Initial Pressures** I'm now calculating the initial $(PV)$ products for each component in containers A, B, and C, based on their total initial pressures and volumes. I'm setting up equations that represent the partial pressures of the gases present, keeping in mind which components are absent in each container. **Deriving Partial Pressures** I'm now calculating the final partial pressures of X, Y, and Z. I've established the total $(PV)$ contributions for each gas across the combined containers, leading to a total final pressure of $\frac{7}{3}P$. I am currently determining the valid ranges for each individual partial pressure. **Refining Partial Pressure Ranges** I'm now refining the bounds for the partial pressures of X and Y, considering their combined total $(PV)$ contributions. I've established that their sum must be at least $P/6$ and at most $\frac{7}{3}P$, based on the initial conditions and potential $(PV)$ values of components B and C. **Verifying Partial Pressure Bounds** I've confirmed that option (C) accurately reflects the calculated bounds for $P_X + P_Z$, ranging from $2P/3$ to $7P/3$. This aligns perfectly with my derived inequalities. **Confirming Option C** My calculations for the partial pressure sum $P_X + P_Z$ have yielded the exact range presented in option (C). This precise match strongly suggests that option (C) is the correct answer. **Validating $P_X+P_Z$ Range** I have rigorously derived the bounds for $P_X + P_Z$, establishing them to be within $[2P/3, 7P/3]$. This range precisely matches option (C), providing strong validation for its correctness. **Analyzing $P_X + P_Z$** My calculations for the sum $P_X + P_Z$ have confirmed the range $[2P/3, 7P/3]$. This matches option (C) precisely, increasing confidence in that choice. **Calculating $P_Y + P_Z$ Bounds** I've just calculated the bounds for $P_Y + P_Z$, finding it lies between $3P/2$ and $7P/3$. This contradicts option (E). I'm now examining option (B) for $P_X$. **Confirming Final Option** My calculations have confirmed that option (C) is indeed the correct choice. I've cross-referenced all derived bounds and partial pressure relationships, and they consistently point to this answer.