Questão 2 (Química)

**Sketching a Diagram** I'm trying to create a diagram, using a table to identify numeric values and setting up an ASCII chart with specific spacing. The points I want to represent go from x0 to x6, equivalent to values mapped out on the y-axis, with a curve that rises steeply at first and eventually levels out into a plateau. This approximate growth is hyperbolic until it reaches a constant state. I need to ensure the lines don’t exceed 100 characters in the final formula. It feels important to capture all these details accurately!

**Analyzing osmotic pressure** I need to solve for osmotic pressure, dealing with geometry and displacement. The solution compartment has an area of 1 dm² and an initial height of 10 dm, giving a volume of 10 L. The pistons must be taken into account, with one applying atmospheric pressure plus osmotic pressure, while the other applies just atmospheric pressure. I’m considering how the pistons move and how the osmotic pressure balances with mechanical pressure differences in this equilibrated system.**Determining osmotic pressure** I need to calculate osmotic pressure using van't Hoff's formula, which involves concentration, temperature, and volume. As the piston displaces, the pressure could change due to the varying volumes in each compartment. I’m visualizing a diagram where one piston applies pressure (P_atm + π) on the left and the other (P_atm) on the right. Both pistons are movable, affecting the solution’s height and solvent volume as solvent transfers across the membrane. It seems I’ll need to consider how these changes impact the overall system.**Calculating concentrations and volumes** I’m figuring out the relationships between the volumes of the solution and the solvent. When the solvent is expelled, the solvent compartment volume increases by the same amount, causing the piston on the pure solvent to move up. So, the total volume should stay fixed at 20 L, while the heights shift. As I consider concentration, I find that with 0.2 mol of solute, the concentration can exceed solubility as the solution compresses.**Considering displacement definition** I'm thinking about the definition of displacement, specifically piston displacement, which refers to the movement from an initial position, in this case from x to 9 dm. I also noted that the left compartment volume started at 10 dm³. It's interesting to see how these two elements relate, and how understanding displacement can impact calculations and further analysis. I wonder how they might connect in a practical scenario.**Analyzing osmotic pressure** I’m looking at the relationship between substance concentration and molar mass without considering the van't Hoff factor. I’ll apply the ideal formula to create a graph that rises from 0 to 6, resulting in a convex curve. I need to calculate several plotting points along the way. I’ve got some table values mapped out for x from 0 to 6, noting that at x=6, precipitation begins. I also need to ensure that I account for pressure balance with osmotic pressure in the system.**Exploring volume exchange dynamics** I'm considering the reversible relationship involving volume exchange with a piston system. If the piston on the left lowers by x, and the area is 1 dm², that results in a volume reduction of x dm³, which is equal to x liters. The pure solvent then receives the same volume, causing its piston to rise by x dm. This setup highlights how the volume reduction on one side corresponds directly with the increase on the other side.**Exploring liquid pressure dynamics** I’m considering how the discrepancy in pressure within a liquid at a membrane relies on hydrostatic levels. The heights of the fluid columns can differ based on the positions of the pistons. Both compartments start at an equal height, but as one goes down, the other goes up, creating a height difference that affects pressure. This understanding is important to grasp the overall dynamics at play!**Investigating hydrostatic balance** I'm examining pressure differences due to liquid levels. Is hydrostatic head relevant here? The figure indicates a membrane as a vertical separator, possibly splitting compartments. The diagram shows outline lines at specific points with a narrow gap. There seems to be compartments separated by a membrane almost entirely. As I consider the osmotic pressure equilibrium, I realize I need to account for height differences and hydrostatic pressure in this standard problem. I'll need to carefully analyze the schematic.**Analyzing chamber configuration** I’m looking at an image showing two chambers separated by a membrane represented by a vertical line. The left chamber has boundaries from point x836 to x1055, and the right from x1077 to x1296. There seems to be a central piston mechanism. The membrane might form a separator at the bottom, with vertical lines indicating the sides. It appears there’s a flat membrane allowing only the solvent through, but I need to infer its exact orientation. This seems complex!**Examining membrane configuration** The thin blue lines likely indicate a central membrane that spans vertically down to y501 and from y550 to the bottom. There seems to be a gray or black rectangular piece at vertical positions suggesting it could be a membrane or piston on the solvent side. The figure features two fluid chambers separated by this vertical membrane, possibly with a movable element. The blue walls might correspond to a central rod that acts as a separator, with the semipermeable membrane at the bottom.**Exploring fluid displacement** I’m considering gravity, but it might not apply here. We're looking at a possible saturation point with a solubility of 4g/L, and I have 16g in 10L. There's also this compression aspect. I’m investigating how the solution piston displaces volume downward based on solvent movement across the membrane. It's tricky because as the solvent moves, it loses that amount, which affects the volume calculations. The total volumes might stay at 20L, but it's complex!**Constructing solubility graph** I'm using a solubility of 4 g/L at saturation. If x is 6, 4 g/L times 4L equals 16g, which is the saturation point; anything beyond that leads to precipitates. I want to present this in Portuguese and possibly include a table with values like 0, 2, 4, 6, and 9. The graph will show a plateau starting at x=6. I’ll create a rough ASCII graph and ensure the values in the table are clear, labeling everything accurately. I'll connect the points smoothly as a hyperbola and make adjustments for clarity!

I'm setting up the solution compartment problem: 10 L initial volume holding 0.2 mol (16 g) of a substance with 80 g/mol molar mass and 4 g/L solubility, so saturation occurs once the volume shrinks to 4 L. As the piston moves 0 to 6 dm, the volume drops from 10 L to 4 L and the concentration rises from 0.02 mol/L to 0.05 mol/L at saturation. Using RT = 0.0821 × 305 = 25.04 L·atm/mol, I calculate osmotic pressure as π = 5.008/(10−x) atm for 0 ≤ x ≤ 6 dm (giving values like 0.50 atm at x=0 rising to 1.25 atm at x=6), then π stays constant at 1.25 atm for 6 ≤ x ≤ 9 dm since excess solute precipitates and keeps the solution saturated. I'm assuming van't Hoff factor i = 1 since it's a non-electrolyte, noting the membrane position and that the applied piston pressure equals the osmotic pressure reversibly, and I'll present this as a table plus a sketched graph.

With the compartment holding 10 L initially and 0.2 mol of solute at 80 g/mol (16 g total), I note the solution stays saturated as long as volume is at least 4 L, meaning the piston can move up to 6 dm before undersaturation kicks in. Then I'm computing the osmotic pressure using π = cRT, giving π = 5.008/V. Working through specific displacement values, pressure rises from about 0.50 atm at x=0 to 1.252 atm at x=6, where the solution saturates, after which pressure stays constant at 1.252 atm as excess solute precipitates for x between 6 and 9. This gives a curve that rises hyperbolically then flattens into a plateau once saturation is reached.

The compartment has an initial volume of 10 L holding 0.2 mol solute (16 g, M=80 g/mol), with solubility 4 g/L giving saturation at 4 L. So precipitation begins once the volume drops below 4 L, corresponding to a piston displacement beyond 6 dm, after which concentration stays constant. Using π = cRT with i=1, I calculate the initial osmotic pressure as c=0.02 mol/L times R(0.0821) times T(305K), giving about 0.50 atm. As the piston displaces by x dm, the volume becomes (10−x) L, so π = 5.008/(10−x) atm, which I compute stepwise from x=0 through x=6: 0.50, 0.56, 0.63, 0.72, 0.84, 1.00, and 1.25 atm — the last value being reached exactly at saturation (V=4 L). Beyond x=6 up to x=9, the pressure stays fixed at this 1.25 atm plateau since the solute concentration no longer changes. The graph I'm picturing shows a hyperbolic rise from (0, 0.50) to (6, 1.25), followed by a flat horizontal segment at 1.25 atm for the remaining displacement.

The problem is in Portuguese about reverse osmosis. O problema descreve um processo de osmose reversa com compressão reversível. O pistão da solução sofre pressão π. O êmbolo do solvente puro também é móvel. A membrana é semipermeável e permite apenas a passagem do solvente. Os dois compartimentos têm volume inicial igual e base com área de 1 dm². A altura inicial do êmbolo móvel é 10 dm. A solução contém 0,2 mol de uma substância com massa molar de 80 g·mol⁻¹ e solubilidade de 4 g·L⁻¹. O processo ocorre à temperatura constante de 305 K. Preciso calcular a pressão osmótica (π) em atm como função do deslocamento do êmbolo (em dm), de 0 até 9 dm. A figura mostra o compartimento da solução com pressão P_atm + π e o do solvente puro com P_atm. A escala vai de 0 no topo até 9 dm na base. Analisando a escala da figura: o deslocamento do pistão é medido de 0 para baixo até 9 dm. Os pistões começam na posição inicial 0 no topo. A barra indicada na escala provavelmente representa o deslocamento total. O deslocamento do pistão ocorre no compartimento da solução. A compressão reversível é aplicada nesse lado para forçar o fluxo do solvente. A pressão π atua sobre esse pistão. O volume inicial de cada compartimento é 10 dm³. A massa da substância é 16 g, com concentração inicial de 0,02 mol/L. Isso está abaixo da solubilidade de 4 g/L, então a solução permanece totalmente dissolvida. A pressão osmótica será calculada com i = 1, assumindo solução ideal. Ao deslocar o pistão para baixo, o volume da solução diminui e o do solvente aumenta. O lado da solução fica com P_atm + π e o do solvente com P_atm. A compressão reversível exige que a pressão aplicada seja igual à pressão osmótica em cada instante. Ao sair do solvente, a concentração da solução aumenta e π cresce. Contudo, quando a concentração atinge a solubilidade de 4 g/L, a solução fica saturada. A concentração então se mantém constante até toda a substância se precipitar. Nesse ponto, π também permanece constante. A concentração da solução saturada permanece constante enquanto houver sólido presente. A pressão osmótica então se mantém estável até que toda a substância se precipite. O volume do lado do solvente puro aumenta conforme o líquido entra. O deslocamento do êmbolo indica o movimento do pistão da solução para baixo, de 0 a 9 dm. A altura inicial é 10 dm. Como ambos os pistões são móveis e compartilham a mesma base, o volume de cada compartimento varia simultaneamente. O volume da solução cai para 10 − d litros quando o pistão desce d dm. O pistão do solvente sobe e seu volume aumenta. A escala da figura vai de 0 a 9 dm. A concentração de saturação é 0,05 mol/L. O volume da solução é V_sol = (10 − x) L. A solução fica saturada quando o volume da solução atinge 4 L. Abaixo disso, a concentração permanece constante em 0,05 mol/L e a quantidade dissolvida é 0,05 × V mol. O volume do sólido é desprezível. A concentração da solução varia de 0,02 mol/L até 0,05 mol/L conforme o volume passa de 10 L a 4 L. Acima de 4 L a concentração permanece constante em 0,05 mol/L. O produto RT é calculado como 25,04 atm·L·K⁻¹·mol⁻¹. A pressão osmótica é π = 25,0405 c atm. Para a região não saturada, π = 5,008 / (10 − x) atm. O valor exato de RT é 25,0405 e não 25. Vou descrever o gráfico de π versus deslocamento do êmbolo. A função é π = 5,008 / (10 − x) atm para volumes entre 10 L e 4 L. Acima de 4 L a pressão se mantém constante em π = 1,252 atm. O gráfico terá uma região decrescente seguida de um trecho horizontal. O valor exato de RT é 25,0405. A pressão osmótica no início do processo vale π = 5,0081 / (10 − x) atm. No deslocamento inicial de 0 dm o valor é aproximadamente 0,5008 atm. No ponto de saturação a 6 dm o valor é exatamente 1,252 atm. O cálculo confirma que a concentração de saturação coincide com 0,2 / 4 = 0,05 mol/L. A pressão osmótica depende apenas da concentração da solução, que varia com o volume do compartimento da solução. O volume do solvente não altera π. A menção ao pistão móvel do solvente na descrição do problema parece desnecessária para o cálculo. A resposta final deve descrever o gráfico de π versus deslocamento. Não há um único valor numérico para preencher a linha solicitada. A questão parece ser de um vestibular. Há uma inconsistência na terminologia: o problema chama o compartimento da solução de "pistão", mas o gráfico pede o deslocamento do "êmbolo". A escala da figura parece estar no lado da solução, o que gera confusão sobre qual pistão deve ser considerado. A escala da figura está no lado esquerdo, indicando o deslocamento do compartimento da solução. A confusão entre "pistão" e "êmbolo" provavelmente vem de sinônimos usados na terminologia. A transferência de solvente equivale à redução do volume da solução. A pressão osmótica depende indiretamente do deslocamento, pois o volume trocado altera a concentração. A direção exata do movimento do pistão do solvente é confusa: a seta indica pressão para baixo, mas um aumento de volume deveria empurrar o pistão para cima. O deslocamento do pistão do solvente é positivo e para cima. A escala da figura indica movimento para baixo, de 0 a 9 dm. Essa divergência entre a direção real do movimento e a orientação da escala gera confusão sobre qual pistão deve ser considerado. A solubilidade de 4 g·L⁻¹ justifica a quebra no gráfico. Sem esse limite, a pressão seria apenas π = 5,008 / (10 − x) atm. Com ele, a função é piecewise: decrescente até o volume de 4 L, então constante em 1,252 atm. A menção ao êmbolo móvel do solvente indica que esse compartimento se expande livremente, sem pressão adicional. A questão exige atenção às figuras significativas. O valor de 8,21 × 10⁻² tem três algarismos exatos. O número 0,2 mol pode ser exato ou ter uma casa decimal. A solubilidade de 4 g/L também pode ser exata. O cálculo de RT resulta em 25,0405. A expressão π = 5,008 / (10 − x) atm precisa ser arredondada corretamente. O valor final de 1,252 atm também deve respeitar as regras de significância. A altura inicial do êmbolo é 10 dm. O gráfico para em 9 dm para evitar volume nulo. Com x = 9 dm o volume da solução fica em 1 L. A concentração de saturação permanece em 0,05 mol/L. A pressão osmótica continua constante em 1,252 atm. A diferença de altura entre os pistões gera pressão hidrostática, mas essa variação é desprezível. A pressão osmótica π é a parcela extra sobre P_atm. Com alturas de 0,1 m, a pressão hidrostática equivale a cerca de 0,01 atm, muito menor que o valor de π, que varia entre 0,5 e 1,25 atm. A solução começa com 10 L de volume. O desenvolvimento inicia listando os dados: área da base, altura inicial, número de mols do soluto, massa molar, solubilidade e temperatura. A resposta final será apresentada com vírgula decimal, conforme a notação dos dados do problema. A expressão π = 5,0081 / (10 − x) atm será escrita com vírgulas em vez de pontos. O valor de RT deve ser arredondado para 25,0, com três algarismos significativos. A pressão osmótica final fica então em 1,25 atm. O número 0,2 mol provavelmente é exato. O ITA costuma exigir cálculos com a precisão dos dados fornecidos, sem arredondamento excessivo. A resposta final deve ser apresentada com o valor exato de RT, ou aproximado apenas quando solicitado. A resposta final deve ser apenas o gráfico: curva decrescente de 0,501 a 1,252 atm entre 0 e 6 dm, seguida de segmento horizontal em 1,252 atm até 9 dm. Há uma possível inconsistência entre a escala da figura (0 a 9 dm) e a altura inicial de 10 dm. O problema define a posição inicial do êmbolo como zero. O deslocamento positivo reduz o volume da solução. A escala da figura indica movimento para baixo, o que contradiz essa definição. A transferência de solvente para o compartimento puro aumenta o volume do solvente e reduz o da solução em y dm. A função π versus deslocamento permanece π = 5,008 / (10 − x) atm. A escala da figura pode ser apenas ilustrativa. A ambiguidade sobre qual pistão se desloca é resolvida. O volume da solução e o do solvente variam com a mesma magnitude x, mas em sentidos opostos. A pressão osmótica depende apenas dessa mudança de volume, então π versus x é única e inambígua.

We need answer question in Portuguese, as candidate. Need solve. Let's parse. We have reverse osmosis process? Text: Consider reverse osmosis process as figure. Piston where solution is suffers pressure π in reversible compression. Piston containing pure solvent also movable. Two compartments separated by semipermeable membrane, allows only solvent passage; both same initial volume, base area 1 dm^2; initial height of movable piston relative to base = 10 dm. Aqueous solution contains 0.2 mol of a substance, molar mass 80 g/mol. Substance solubility 4 g/L. All process at constant T=305 K. From info, present calculations and construct graph of osmotic pressure π (atm) as function of displacement of piston (dm), per figure, from initial position, as zero, until value 9 dm. Need solve likely calculate osmotic pressure as function of piston displacement (volume change) due to reverse osmosis? Need derive graph points? We need include calculations. Let's understand. Setup: two compartments separated by semipermeable membrane. Left: solution, right: solvent. Both have movable pistons. Pure solvent side pressure P_atm, solution side pressure P_atm + π. Compressing solution side? Actually reverse osmosis: applied pressure greater than osmotic pressure to force solvent from solution to pure solvent (or from solution to solvent? Osmosis normally solvent flows from pure solvent to solution; reverse osmosis applies pressure > osmotic pressure on solution side to force solvent from solution to solvent side). Figure: left is Solução (solution), right is Solvente (solvent). Pistons on top. Left piston pressure P_atm + π, right P_atm. So solution side compressed, solvent side maybe expands? Initially same volume? The solution contains 0.2 mol solute in volume? Initial height 10 dm, area 1 dm² => volume = 10 dm³ = 10 L? Wait 1 dm² area, height 10 dm = 100 cm? 1 dm = 10 cm; area 1 dm² = 100 cm²; height 10 dm = 100 cm; volume = 10 dm³ = 10 L. Yes. But solution contains 0.2 mol, molar mass 80 g/mol => mass = 16 g. Solubility 4 g/L. Wait if mass 16 g and solubility 4 g/L, need volume at least 4 L to dissolve? Actually solubility 4 g/L means saturated solution concentration 4 g/L. But solution contains 0.2 mol =16 g; volume 10 L gives concentration 1.6 g/L, below solubility, so unsaturated? Wait maybe solubility used to determine if precipitation? Hmm. Maybe "substância" is solute, molar mass 80, solubility 4 g/L, so maximum dissolved = 4 g/L. Initial solution volume 10 L, solute mass 16 g => 1.6 g/L < 4 g/L, so all dissolved. As reverse osmosis removes solvent, solution volume decreases; concentration increases. At some point reaches solubility 4 g/L, then solute may precipitate? But only solvent passes through membrane, solute remains (if dissolved or precipitated?). They ask graph of osmotic pressure π as function of displacement of piston (dm), from initial position to 9 dm. Displacement of piston? Need understand sign. Piston on solution side moves down? It says "deslocamento do êmbolo (em dm), conforme a figura, da posição inicial, como sendo igual a zero, até o valor de 9 dm." So displacement variable maybe x = 0 at initial height 10 dm; until 9 dm. If piston moves downward by x dm? Initially at 10 dm above base; final height = 10 - x? If x=9 dm, height =1 dm. So solution volume decreases by x L (since area 1 dm², displacement x dm = volume x L), from 10 L to 1 L. Yes. But osmotic pressure depends on concentration of solute particles; as volume changes, concentration increases; if solubility exceeded, some solute precipitates and concentration remains saturated. Need calculate π = c R T? For dilute solutions, van't Hoff: π = (n_solute/V) R T or with i? Since substance non-electrolyte? likely nonelectrolyte. Mass molar 80 g/mol. Solubility 4 g/L. Need account. Initial: n = 0.2 mol. Volume = 10 L. Molar concentration = 0.02 mol/L. Osmotic pressure = c R T. R = 0.0821 atm L K^-1 mol^-1? Given 8,21 ×10^-2 atm L K^-1 mol^-1 = 0.0821. T=305 K. π = 0.02 * 0.0821 * 305 = 0.02 * 25.0405 = 0.50081 atm ≈ 0.501 atm. At x=0, π ≈ 0.501 atm. As solvent removed reversibly by applied pressure on solution greater by π, the osmotic pressure against which work is done? The graph π vs displacement: requires calculate π at various volumes after concentration, considering solubility. Solute amount n=0.2 mol remains unless precipitation? Solubility = 4 g/L = 4/80 = 0.05 mol/L saturated concentration. At initial 0.02 M < 0.05 M, unsaturated. When volume V < n/C_sat? Actually total solute n = 0.2 mol; saturated concentration 0.05 mol/L; volume at saturation = 0.2 / 0.05 = 4 L. So when solution volume reaches 4 L, saturated; for V < 4 L, dissolved solute amount = 0.05 mol/L * V = 0.05V mol; precipitated solute = 0.2 - 0.05V mol. But osmotic pressure depends only on dissolved solute particles, so π = C_sat R T = 0.05 * R * T for V ≤ 4 L (constant). Wait but if precipitation occurs, dissolved concentration remains 0.05 M (assuming solid present). So osmotic pressure becomes constant at max value: 0.05*0.0821*305 = 1.252025 atm ≈ 1.252 atm. Is that the plateau? Need graph up to x=9 dm, which corresponds V = 10 - x = 1 L. So from V=10 to 4 L unsaturated, π increases hyperbolically; from V=4 to 1 L saturated, π constant = 1.252 atm. But wait at x=6 dm, V=4 L, π=1.252 atm; at x >6? π same. However maybe "solubilidade 4 g/L" means maximum concentration; yes. But let's be careful: The question might expect graph π vs displacement, maybe calculate values at x=0, x=6, x=9? Need construct graph. It asks "A partir dessas informações, apresente os cálculos e construa um gráfico da pressão osmótica (π, em atm) em função do deslocamento do êmbolo (em dm), conforme a figura, da posição inicial, como sendo igual a zero, até o valor de 9 dm." So need provide calculations and graph. We can describe graph and maybe table. Need decide if pressure applied is exactly π, reversible. The pressure on solution side is P_atm + π, where π is osmotic pressure? In reverse osmosis, the applied pressure difference equals osmotic pressure for equilibrium/reversible? Wait if pressure difference equals osmotic pressure, no net solvent flow? But "processo de compressão que ocorre de modo reversível" maybe π is the additional pressure needed to prevent osmosis or to reverse? Actually if solution side pressure is P_atm + π, and π is osmotic pressure of solution, then at all times the chemical potential of solvent equal? The piston displacement changes volume, solvent passes through membrane? Let's model. There are two compartments: left solution, right pure solvent. Semipermeable membrane. Both pistons movable. The applied pressure on solution side exceeds solvent side by π. This is exactly the osmotic pressure of the solution at any moment, so system is at equilibrium? In reversible compression, as solution volume decreases (by applying external pressure), solvent molecules are forced from solution (left) to solvent (right) through membrane; external pressure does work. The osmotic pressure π depends on concentration. So π is the pressure difference needed to just prevent osmosis? Wait if pressure difference = π, net flow zero? But they say "processo de osmose reversa" so pressure difference > π? But they denote π as pressure action maybe osmotic pressure? The figure labels P_atm + π, P_atm. In reverse osmosis, applied pressure on solution side is greater than osmotic pressure by an amount to drive flow; here label π maybe applied pressure difference. But question asks "gráfico da pressão osmótica (π, em atm) em função do deslocamento do êmbolo" So π is osmotic pressure, not applied pressure. They probably assume reversible compression, i.e., at each displacement, external pressure difference equals osmotic pressure, so π computed by van't Hoff. Need perhaps graph π vs x. Now, there may be subtle issue: The solvent also moves? Initially both compartments same volume 10 L. As solution volume decreases from 10 L to 1 L? Wait if solution volume decreases by x L (piston moves down by x), where does solvent go? Through membrane into solvent compartment. But solvent compartment piston is movable, initially volume 10 L; it receives solvent, volume increases by x L, so piston rises by x dm (from 10 to 19 dm?). The problem says "O êmbolo que contém o solvente puro também é móvel." The graph x-axis is displacement of "êmbolo" (which one? Figure shows dm scale on left solution side? Actually the figure has scale on left with 0-9? Let's inspect: Figure: two compartments; left solution side with piston at top; right solvent side. There is a ruler from 0 to 9 dm? Wait scale labels: left side has 0,1,2,...,9 downward from top? The piston initial height is 10 dm, but scale maybe extends beyond? In image, left vertical has ticks 0-9, with a vertical arrow pointing down "dm" and labels: top is 0? Let's parse: The figure: left compartment with solution; right solvent. There are lines at top? Maybe a graduated cylinder? It shows numbers 0,1,...,9? The bottom? Let's zoom mentally: Left side has a vertical scale with numbers 0,1,2,3,4,5,6,7,8,9 descending from top to bottom? There is an arrow at bottom pointing down with "dm". The piston initial position maybe at 0? Wait text: "altura inicial do êmbolo móvel em relação à base é igual a 10 dm. ... construa um gráfico ... em função do deslocamento do êmbolo (em dm), conforme a figura, da posição inicial, como sendo igual a zero, até o valor de 9 dm." In figure, scale maybe from 0 to 9? Hmm The ruler labels 0 to 9 dm maybe not absolute height, but displacement from initial? Let's see: Figure has horizontal lines across compartments maybe 0 at top? Actually top of solution compartment has piston; above piston arrow P_atm+π. There is a vertical scale along left side with marks 0 to 9? If initial piston height is 10 dm, scale 0-9 could be below piston? Maybe they draw a ruler where 0 is initial position? Let's examine: left compartment's piston at top; below it numbers 0 to 9? The base is at 10 dm below piston? The numbers maybe mark depth of piston from initial position? The figure may show the piston can move downward from initial (0) to 9 dm. The scale 0 at top initial position, 1,2,...9 downward. Yes. The arrow dm indicates displacement. So x-axis is displacement downward of solution piston (left), from 0 to 9 dm. This decreases solution volume by x L. The pure solvent piston moves upward by x dm? But graph only π vs x for solution. Need calculate for x from 0 to 9 dm. At x=0 π = 0.5008 atm; for x from 0 to 6? Let's compute concentration. V_solution = 10 - x L (since area 1 dm², x dm³ = x L). n_total = 0.2 mol. If unsaturated (until saturation): c = n/V = 0.2 / (10 - x) mol/L. Saturation concentration = solubility / M = 4 g/L / 80 g/mol = 0.05 mol/L. Set c = 0.05 => 0.2/(10-x) = 0.05 => 10-x = 4 => x=6 dm. So for 0 ≤ x ≤ 6, π = c R T = [0.2/(10-x)] * 0.0821 * 305. Compute 0.0821*305 =? 0.0821*300 = 24.63; plus 0.0821*5=0.4105; total 25.0405. Then π = 0.2*25.0405/(10-x) = 5.0081/(10-x) atm? Let's compute: 0.2 * 25.0405 = 5.0081. Yes π = 5.0081/(10-x) atm. At x=0: 0.50081 atm. At x=6: 5.0081/4 = 1.252025 atm. Good. For x > 6, saturation: c = 0.05 mol/L; π = 0.05 * 25.0405 = 1.252025 atm (constant). Up to x=9, V=1 L; dissolved amount = 0.05 mol/L * 1 L = 0.05 mol, precipitated 0.15 mol? Actually total 0.2, dissolved 0.05, solid 0.15 mol? But if solid present, dissolved concentration fixed. Osmotic pressure same. However, at x=9, solution volume = 1 L; solid may precipitate. Need perhaps note solubility allows maximum 4 g/L; at V=1 L maximum dissolved mass = 4 g = 0.05 mol; so 0.15 mol precipitate. But osmotic pressure only dissolved. But wait: Is it possible that as volume decreases by compression, the solution becomes saturated at x=6, then further compression causes precipitation; but the osmotic pressure might also depend on pressure? Osmotic pressure definition for nonideal? But likely ideal dilute. Also solute molar mass 80 g/mol; mass total = 16 g; solubility 4 g/L => volume needed to dissolve all = 4 L; yes. Need maybe graph: x-axis 0 to 9 dm; y-axis π atm. Two regions: curve π = 5.0081/(10-x) for 0≤x≤6; at x=6 π=1.252; horizontal line from x=6 to x=9. Need maybe table of points. Let's compute points maybe at x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9? Could provide graph by plotting. Since cannot actually draw, describe. But user asks "construa um gráfico"; final answer should include explanation and graph perhaps in text? We need solve. We can include a rough graph with values. Since this is text response, we can give data table and describe shape; maybe "gráfico: Nos primeiros 6 dm, ramo de hipérbole; de 6 a 9 dm, reta horizontal." That's enough? But should present calculations. Let's verify all constants: R = 8.21×10^-2 atm L K^-1 mol^-1; T=305 K. Product = 25.0405 atm L/mol. Good. Initial volume: base area 1 dm², height 10 dm => 10 dm³ = 10 L. Wait 1 dm³ = 1 L. Yes. Moles = 0.2 mol; mass = 0.2*80 =16 g; initial concentration = 16g/10L=1.6 g/L < 4g/L. So unsaturated. Volume at saturation = mass/solubility = 16 g / 4 g/L = 4 L. So x_sat = 10 - 4 = 6 dm. Good. Osmotic pressure formula: π = (n/V) R T = (m/M)/V R T. Alternatively use mass concentration: π = (C/M) R T? Wait c = C/M. For unsaturated C = m/V =16/(10-x) g/L; c = 16/(80(10-x)) =0.2/(10-x) mol/L. Yes. At saturation: C=4 g/L; c=4/80=0.05 mol/L; π=0.05*25.0405=1.252 atm. But wait question says "substância apresenta uma solubilidade de 4 g·L⁻¹". Could solubility be in mol/L? But table gives 4 g/L. Good. Need consider "todo processo ocorre à temperatura constante de 305 K" and reversible; maybe need also include work? No, only graph. Let's consider if applied pressure difference π maybe osmotic pressure at each x, but if precipitation occurs, does the solute remain as solid? Yes. However, if solute is solid and membrane allows only solvent, no solute passes; solid may sediment but not affect osmotic pressure? But osmotic pressure of saturated solution is same as long as solid present. If all solid dissolved? It's saturated with excess solid. Need maybe consider volume change due to mixing? Solvent passing from solution to pure solvent: solution volume = solvent volume + solute volume? For dilute solutions, volume approx solvent volume. But maybe exact? Solute molar mass 80, density unknown; but likely assume volumes additive? In osmotic pressure problems, concentration molar in solution volume; volume is total solution volume. The statement "ambos possuem o mesmo volume inicial" and "base área 1 dm²" maybe simple. We used volume = height*area, ignore solute volume / precipitation volume. Good. Need maybe "pressão osmótica (π) em função do deslocamento do êmbolo" maybe applied pressure π? The pressure on solution side is P_atm + π, with π being the excess pressure applied. If the process is reversible, the applied pressure difference equals the osmotic pressure of solution. So π is indeed osmotic pressure. Good. Potential subtlety: In reverse osmosis, to cause solvent to flow from solution to pure solvent, the applied pressure must exceed osmotic pressure; if process reversible, at equilibrium infinitesimal difference, so π is osmotic pressure? Actually during reversible compression, the external pressure on solution side may be P_atm + π, where π is the osmotic pressure plus infinitesimal? In the limit, π equals osmotic pressure. The problem labels π as pressure. So yes. Need maybe compute pressure at x=0 and x=6; maybe also at x=9? At x=9 π=1.252 atm. But graph from 0 to 9. We can include table: x/dm: 0, 2, 4, 6, 7, 8, 9 π/atm: 0.501, 0.626, 0.835, 1.252, 1.252, 1.252, 1.252? Let's calculate more precise for table: At x=0: 5.0081/10 = 0.50081 ≈ 0.501 x=2: 5.0081/8 = 0.6260125 x=4: 5.0081/6 = 0.834683 x=6: 5.0081/4 = 1.252025 x=8? V=2 L saturated c=0.05 => 1.252 x=9: same 1.252. But wait at x=9 V=1 L, still saturated? Yes c=0.05 as long as solid present. At V=1, dissolved 0.05 mol, solid 0.15 mol; yes. But wait, if volume is 1 L, can 0.05 mol dissolve? Solubility 0.05 mol/L, yes. But total solute 0.2 mol; excess precipitates. So π constant. Good. Need perhaps "a partir da posição inicial, como sendo igual a zero, até o valor de 9 dm": x-axis x = 0 to 9. The graph should be for solution compartment piston displacement. It might be expected to show π increasing hyperbolically then plateau at 1.25 atm. Need indicate axes. Let's formulate final answer with calculations. Let's write in Portuguese: 1. Cálculo da concentração e da pressão osmótica inicial: - V0 = A h = 1 dm² × 10 dm = 10 dm³ = 10 L. - n = 0,2 mol; m = 0,2 × 80 = 16 g. - C0 = 16/10 = 1,6 g/L < 4 g/L (insaturada). - c0 = 0,2/10 = 0,02 mol/L. - π0 = c0 R T = 0,02 × 0,0821 × 305 = 0,50081 atm ≈ 0,501 atm. 2. Enquanto não saturar, V = (10 - x) L, x em dm. c = n/V = 0,2/(10 - x) mol/L. Sendo insaturada: - π(x) = [0,2/(10 - x)] × 0,0821 × 305 = 5,0081/(10 - x) atm, para 0 ≤ x ≤ x_sat. 3. Saturação: - concentração de saturação: c_sat = 4 g/L / 80 g/mol = 0,05 mol/L. - volume de saturação: 0,2 mol / 0,05 mol/L = 4 L. - x_sat = 10 - 4 = 6 dm. - π_sat = c_sat R T = 0,05 × 0,0821 × 305 = 1,252025 atm ≈ 1,25 atm. - Para 6 ≤ x ≤ 9 dm, π = 1,25 atm (constante). Could mention a partir de x=6, excesso de soluto precipita, mantendo a solução saturada; a pressão osmótica depende apenas da concentração dissolvida (0,05 mol/L). 4. Gráfico: - eixo x: deslocamento x (dm); eixo y: π (atm). - 0 a 6 dm: ramo de hipérbole y=5,0081/(10-x). Valores: x=0 →0,501; x=2→0,626; x=4→0,835; x=6→1,252. - 6 a 9 dm: reta horizontal y=1,25 atm (ou 1,252 atm). Need maybe include "conforme figura" maybe graph starts at x=0? It should be from 0 to 9. We can end with "Resposta final: ..." Probably need concise final line with result. The requested output: "Resposta final: <o resultado de cada item, de forma curta>". Since no explicit items a,b,c, only one question, maybe write "Resposta final: π = 0,501 atm no início; π = 5,0081/(10-x) atm para 0≤x≤6 dm; π = 1,252 atm para 6≤x≤9 dm; gráfico..." But "o resultado de cada item" maybe include graph data. It says "Se a questão tiver uma resposta só, escreva só o resultado." But there is a question with multiple calculations and graph. We'll include final response summarizing. Need ensure no mistakes with unit: area 1 dm², displacement 1 dm = 1 dm³ = 1 L. R = 0.0821 atm L mol^-1 K^-1. We use L. Let's double-check if maybe "substância" could be electrolyte? Not specified; molar mass given, maybe non-electrolyte. If it were electrolyte, need van't Hoff factor i. But no i. So assume molecular. Potential issue: The solubility is 4 g/L; initial solution volume 10 L and 16 g solute => 1.6 g/L. But as reverse osmosis removes solvent, concentration increases. At saturation volume 4 L, concentration 4 g/L. Then further removal causes precipitation of 12 g? Actually initial 16 g, at 4 L saturated solution holds 16 g? Wait 4 L *4 g/L =16 g, so all 16 g dissolved exactly at saturation. At lower volume, say 1 L, saturated solution holds 4 g, so 12 g precipitate. Yes. Wait 4 L * 4 g/L =16 g, yes. At 4 L exactly saturated, no solid? At V=4 L, c=4 g/L; if initially all 16 g dissolved, concentration 4 g/L; saturation; no excess solid? Actually if exactly saturated, no solid may be present; if further volume decrease, some solid appears. So plateau starts at x=6 exactly at saturation. At x=6, no solid? At saturation boundary. Good. Let's consider if at x=6 π =1.252, and for x>6 same. Since at exact saturation, solution saturated but no solid; still concentration 0.05 M. Good. Need maybe "pressão osmótica em atm" maybe use R=0.0821 atm L/mol/K. Product 25.0405 atm L/mol. Good. Could there be a correction for osmotic pressure due to applied pressure? In thermodynamics, osmotic pressure π = - (RT/V_solvent?) ln a_solvent. For dilute solution, π ≈ cRT, but under high applied pressure (P_atm + π) the solvent chemical potential changes; however constant T and pressure? Actually pressure on solution side is high, but osmotic pressure definition? The reversible work of compression? Let's reason more deeply. The problem may be from ITA; maybe they expect using concept of osmotic pressure and reverse osmosis; maybe "processo de osmose reversa" with semipermeable membrane; the applied pressure π is the osmotic pressure at each point. They ask graph π vs displacement. Could there be effect of pressure on osmotic pressure? Let's see: In equilibrium, chemical potential of solvent in solution under pressure P+π equals chemical potential of pure solvent under pressure P. For ideal dilute solution, μ_solvent(T,P+π) = μ_solvent^*(T,P+π) + RT ln x_solvent? Wait standard state at same pressure? The osmotic pressure equation for incompressible solvent: π = -(RT/V_m) ln x_solvent? Need use activity. For dilute solution, π = cRT. But if applied pressure P+π is high, the standard chemical potential depends on pressure; however solvent chemical potential in solution at pressure P+π: μ_solvent(T,P+π) = μ_solvent^*(T,P+π) + RT ln x_solvent (or activity). Pure solvent at pressure P: μ_solvent^*(T,P). Difference μ_solvent(T,P+π) - μ_solvent^*(T,P) = [μ_solvent^*(T,P+π)-μ_solvent^*(T,P)] + RT ln x_solvent = ∫_P^{P+π} V_m dP + RT ln x_solvent = π V_m + RT ln x_solvent = 0 => π = -(RT/V_m) ln x_solvent ≈ (RT/V_m) x_solute = cRT. The pressure dependence of V_m small; use molar volume. So same. But possible precipitation affects osmotic pressure? Yes. Let's perhaps derive from moles dissolved: n_diss = min(0.2, 0.05V). For V > 4 L, n_diss=0.2; for V≤4 L, n_diss=0.05V. π = n_diss RT / V. For V≤4 L, π = (0.05V)RT/V =0.05RT constant. Good. Let's compute RT more exactly maybe using R=8.21e-2 atm L/mol/K; 305 K = 25.0405. At x=0, π = (0.2/10)*25.0405 = 0.50081 atm. At x=6, π = (0.2/4)*25.0405 =1.252025. Constant after. Need maybe "pressão osmótica (π, em atm) em função do deslocamento do êmbolo (em dm), conforme a figura, da posição inicial, como sendo igual a zero, até o valor de 9 dm." The figure maybe has a scale already; graph should be plotted in the figure? Since they say "construa um gráfico" maybe they want draw on the exam? We can provide textual graph. In final, include a sketch description. Let's examine the image with scale: It has 0 to 9? Maybe the graph should be plotted for displacement from 0 to 9, but initial height 10 dm; maybe x-axis "deslocamento do êmbolo" not "volume". The values we computed match. Let's think if "deslocamento do êmbolo" could be positive upward? The figure maybe piston moves downward due to compression; displacement from initial position zero to 9 dm. We assumed downward. The scale in figure maybe 0 at top, 1-9 downward; yes. The graph has x from 0 to 9. If x increases, the solution piston moves downward; solvent piston moves upward? But question asks "deslocamento do êmbolo" maybe "do êmbolo" (which one? The solution piston? The scale is on left solution side). We'll state. Need maybe include "para x > 6 dm, o soluto em excesso precipita; a solução permanece saturada, de modo que a pressão osmótica não mais depende do volume." Good. Now, final answer should be in Portuguese, item by item. Let's compose with calculations. Keep not too verbose but sufficient. Potential final structure: "Como a área é 1 dm², cada dm de deslocamento corresponde a 1 L de variação de volume. Inicialmente V=10 L...". Let's include units and equations. Let's decide final line: "Resposta final: π(x) = 5,0081/(10-x) atm para 0 ≤ x ≤ 6 dm; π(x) = 1,25 atm para 6 ≤ x ≤ 9 dm. No gráfico: hipérbole até x=6 dm, seguida de patamar horizontal em 1,25 atm." Maybe include initial π: "π(0)=0,501 atm". The final line can contain this. The problem likely expects "gráfico" but final answer should summarize. Let's also maybe include table: x/dm | 0 | 2 | 4 | 6 | 9 π/atm | 0,501 | 0,626 | 0,835 | 1,25 | 1,25 But maybe too much? It asks construct graph; table is useful. In final answer, maybe not too overboard. But "Resposta final: <o resultado de cada item, de forma curta>" So final line concise. I can include graph description. Need ensure the final answer line format exactly: "Resposta final: ..." At end. Good. Let's write: "V0 = 1 dm² × 10 dm = 10 L. Soluto: n=0,2 mol, m=16 g, c0=0,02 mol/L; π0=0,02×0,0821×305=0,501 atm." Then "Para 0≤x≤6: V=10-x L; π=(0,2/(10-x))×0,0821×305=5,0081/(10-x) atm." Note: 0,2 mol constant because not precipitated yet. Then "Para x=6: V=4 L; C=16g/4L=4 g/L (saturação); c=0,05 mol/L; π=0,05×0,0821×305=1,252 atm." Wait at x=6, all 16 g dissolved exactly at saturation. Good. Then "Para 6≤x≤9: V≤4 L; precipita soluto; concentração dissolvida permanece 0,05 mol/L; π=1,252 atm." Then "Gráfico..." good. Let's consider if at x=6, "saturação" maybe maximum solubility but no precipitate; at x=6 the osmotic pressure on the unsaturated formula also gives 1.252; okay. Potential issue: At x=6, formula 5.0081/(10-6)=1.252. If all dissolved, no solid; plateau starts. At exactly x=6, both. Fine. Let's maybe compute "massa de soluto = 0,2 mol × 80 g/mol = 16 g; solubilidade = 4 g/L; volume de saturação = 16 g / 4 g/L =4 L; logo x=6 dm." Good. Let's now produce final answer in Portuguese. Ensure no "graph" impossible? We can describe axes. Maybe include "Gráfico: eixo vertical π (atm), eixo horizontal x (dm)." and "A curva é uma hipérbole y=5,0081/(10-x) no intervalo [0,6] e uma reta horizontal y=1,25 no intervalo [6,9]." Good. Let's final.

**Analyzing Osmosis Problem** I'm currently dissecting the provided image and its accompanying text, focusing on understanding the parameters of the reverse osmosis process described. My aim is to accurately interpret the question and formulate a solution strategy. **Calculating Osmotic Pressure** I'm working through the details of this reverse osmosis problem. I've identified the initial volume, the amount of solute and its molar mass, and the temperature. My next step is to calculate the initial osmotic pressure using the van't Hoff equation. I will then focus on how the piston displacement affects this pressure. **Calculating Initial Osmotic Pressure** I've confirmed the solution is unsaturated. My next step involves using the calculated initial concentration and temperature to determine the initial osmotic pressure. Following this, I will analyze how the piston's displacement impacts the system's volume and concentration. **Confirming Saturation Point** I've pinpointed the exact volume and piston position where saturation occurs. For displacements beyond this point, the dissolved solute concentration stabilizes, meaning the osmotic pressure will also become constant. I am now verifying this conclusion. **Verifying Osmotic Pressure Formula** I'm confirming the application of the van't Hoff equation and the value of the gas constant, $R$. I'm also re-evaluating the assumption of the van't Hoff factor, $i$, given the solute description, to ensure the osmotic pressure calculation is precise. **Calculating Osmotic Pressure Dependence** I'm refining my calculation for osmotic pressure as a function of piston displacement. I've confirmed the value of RT and am now plugging in varying volumes, which are directly dependent on x, to derive the pressure at different points. This is leading to the expression for $\pi(x)$. **Analyzing Membrane Dynamics** I am re-examining the role of the semipermeable membrane. My current focus is on how it dictates solvent passage and its interaction with the pressure gradient, especially concerning the pure solvent compartment. **Analyzing Piston Displacement** I'm re-interpreting the diagram and problem statement, particularly regarding the piston's initial position and the direction of displacement. It appears the displacement 'x' is measured downwards from an initial height of 10 dm, meaning the volume changes as x increases. My previous calculations assumed x was an upward displacement. This significantly impacts how I'll relate displacement to volume and thus osmotic pressure. **Revising Displacement Interpretation** I'm re-evaluating the piston's displacement based on the diagram and the statement that the initial piston height from the base is 10 dm. My understanding now is that the scale starts at 0 at the initial piston position, with the base at 10 dm. The displacement $x$ thus ranges from 0 to 9 dm. **Locating Membrane Position** I'm now focusing on the precise location of the semipermeable membrane. Examining the diagram, the channel containing the membrane appears to span between 7 dm and 9 dm on the scale, with the membrane itself situated within this region. **Determining Membrane Contact** I'm confirming that the piston's downward movement to 9 dm keeps the solution in contact with the semipermeable membrane located within the channel. This ensures the solvent can continue to pass through, which is crucial for the osmotic pressure calculations up to that point. **Revising Piston Interpretation** I'm re-evaluating the piston's displacement based on the diagram and the statement that the initial piston height from the base is 10 dm. My understanding now is that the scale starts at 0 at the initial piston position, with the base at 10 dm. The displacement $x$ thus ranges from 0 to 9 dm. **Confirming Solution Volume** I'm now focusing on whether the solute's volume is negligible. The problem statement implies that the solution's volume is determined by the cylinder's dimensions. My calculations of osmotic pressure are proceeding under this assumption, and I am meticulously verifying all numerical values derived from the given data. **Checking Calculation Steps** My current focus is on ensuring the accuracy of each step in my calculations. I've listed all the provided data, including the base area, initial height, initial solution volume, moles of solute, and molar mass. I'm meticulously reviewing these to prepare for the subsequent steps. **Revising Volume Calculation** I'm now recalculating the solution volume using the corrected interpretation of the piston's displacement. My focus is on how this revised volume impacts the solute concentration and, consequently, the osmotic pressure as the piston moves. **Confirming Initial Volume** I've confirmed the initial volume calculation using the stated initial height and base area, preparing to integrate this into the displacement-dependent volume function. **Revising Concentration Calculation** I'm recalculating the solute concentration based on the revised understanding of how piston displacement affects the solution volume. This is a critical step for accurately determining the osmotic pressure at different positions. **Calculating Osmotic Pressure** I'm now calculating the osmotic pressure in the unsaturated region. I've determined the expression for concentration as a function of displacement, $x$, and am substituting it into the van't Hoff equation along with the calculated value of $RT$ to get $\pi(x)$. **Calculating Osmotic Pressure** I've successfully calculated the osmotic pressure for the unsaturated region, deriving the expression $\pi(x) = \frac{5.01}{10-x}\text{ atm}$ for $0 \le x \le 6$. I'm now transitioning to analyzing the saturated region where the concentration and osmotic pressure stabilize. **Confirming Saturation Pressure** I've now determined the constant osmotic pressure in the saturated region to be approximately 1.25 atm, valid for piston displacements from $x=6$ to $x=9$ dm. I'm also verifying this is indeed the ITA 2024 problem and reviewing the official resolution details for confirmation. **Verifying Formula Values** I've confirmed the RT product and initial osmotic pressure calculation. My current focus is on ensuring the solute concentration and resultant osmotic pressure are accurately determined across both unsaturated and saturated regions for the final graphical representation. **Plotting Pressure Values** I've calculated and confirmed several points for both the unsaturated and saturated regions of the osmotic pressure versus piston displacement curve. I am now assembling these values to construct the final graph as requested, verifying the transition point and the constant pressure in the saturated zone. **Graphing Pressure Data** I've successfully calculated and confirmed the key pressure values for both unsaturated and saturated regions. Now, I'm compiling these to accurately plot the osmotic pressure against piston displacement, ensuring the transition at the saturation point is clearly represented.

**Analyzing Chemistry Problem** I am currently processing the details of a reverse osmosis problem from the ITA exam, carefully examining the provided diagram and the initial description concerning the piston. My focus is on understanding the fundamental principles at play and how they relate to the specific conditions presented. **Calculating Osmotic Pressure** I'm currently working through the ITA reverse osmosis problem, specifically focusing on deriving the osmotic pressure ($\pi$) as a function of piston displacement. I'm using the provided molar amount, molar mass, and solubility to determine the initial concentration and then applying the osmotic pressure equation, considering the change in volume as the piston moves. The goal is to plot $\pi$ versus displacement. **Deriving Osmotic Pressure** I'm now focused on calculating the osmotic pressure as the piston's position changes. My current step involves determining the concentration of the solution at various displacements, accounting for the solubility limit of the solute, and then applying the osmotic pressure equation to establish the relationship. **Calculating Saturation Volume** I'm now focusing on the saturation point. I've determined the initial concentration and compared it to the saturation concentration, confirming the solution is initially unsaturated. My next step is to calculate the volume at which saturation occurs as the piston compresses the solution. **Calculating Saturation Displacement** I have determined the saturation volume to be 4 L, which means saturation occurs when the piston displacement reaches 6 dm. My current focus is on calculating the osmotic pressure, and I've successfully derived the $R \cdot T$ value to be approximately 25 atm·L·mol⁻¹, a perfect fit for the given temperature. **Calculating Osmotic Pressure** I am now calculating the osmotic pressure for both unsaturated and saturated regions. For the unsaturated region, I've derived the formula $\pi(d) \approx \frac{5}{10 - d}$ atm and computed key pressure values at different displacements, which form a hyperbola. **Plotting Solution Curve** I am now focused on presenting the graphical representation of my findings. I will create an ASCII plot detailing the osmotic pressure's behavior across the different displacement intervals, ensuring all key points and the respective equations are clearly stated for an expert audience.