Physical Chemistry 1 Quiz: Osmotic Pressure And Chemical Potential
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Osmotic Pressure And Chemical PotentialQuestion 1 of 15

A polymer solution exhibits an osmotic pressure that deviates from ideal behavior according to π=cRT(1+Bc+Cc2)\pi = cRT(1 + Bc + Cc^2) where B and C are virial coefficients. If B = 0.02 L/mol and C = -0.001 L²/mol², at what concentration will the osmotic pressure equal that predicted by the ideal van't Hoff equation?

When the concentration equals 0.02 mol/L, making the virial correction terms cancel exactly
When the concentration equals 0.05 mol/L, where the positive B term balances the negative C term
When the concentration approaches zero, where all virial terms become negligible compared to the ideal term
When the concentration equals 20 mol/L, where the quadratic term dominates and restores ideal behavior
When the concentration equals -B/C = 20 mol/L, making the correction term (Bc + Cc²) equal to zero
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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Osmotic Pressure And Chemical Potential

Practice Osmotic Pressure And Chemical Potential in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Osmotic Pressure And Chemical Potential, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

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Question 1

A polymer solution exhibits an osmotic pressure that deviates from ideal behavior according to π=cRT(1+Bc+Cc2)\pi = cRT(1 + Bc + Cc^2) where B and C are virial coefficients. If B = 0.02 L/mol and C = -0.001 L²/mol², at what concentration will the osmotic pressure equal that predicted by the ideal van't Hoff equation?

  1. When the concentration equals 0.02 mol/L, making the virial correction terms cancel exactly
  2. When the concentration equals 0.05 mol/L, where the positive B term balances the negative C term
  3. When the concentration approaches zero, where all virial terms become negligible compared to the ideal term
  4. When the concentration equals 20 mol/L, where the quadratic term dominates and restores ideal behavior
  5. When the concentration equals -B/C = 20 mol/L, making the correction term (Bc + Cc²) equal to zero (correct answer)
Explanation: When analyzing non-ideal polymer solutions, you need to understand that the virial equation describes deviations from ideal osmotic pressure behavior. The ideal van't Hoff equation is simply π=cRT\pi = cRT, while the given equation includes correction terms that account for molecular interactions. For the osmotic pressure to equal the ideal prediction, the virial correction terms must sum to zero: Bc+Cc2=0Bc + Cc^2 = 0. This means 0.02c+(−0.001)c2=00.02c + (-0.001)c^2 = 0, which factors to c(0.02−0.001c)=0c(0.02 - 0.001c) = 0. The non-trivial solution occurs when 0.02−0.001c=00.02 - 0.001c = 0, giving c=20c = 20 mol/L. However, this mathematical result reveals a critical flaw in the question setup. At 20 mol/L concentration, we're far beyond the validity range of the virial expansion, which only applies at low to moderate concentrations where polymer chains don't significantly overlap. Option A incorrectly assumes the terms cancel at 0.02 mol/L, but substituting shows they don't. Option B's calculation for 0.05 mol/L is mathematically wrong - the terms don't balance there. Option C misunderstands the question entirely; at zero concentration, both ideal and non-ideal equations give zero pressure, but that's not when they're equal in a meaningful sense. Option D correctly identifies 20 mol/L as the mathematical answer but wrongly claims this "restores ideal behavior" - it doesn't. Remember: virial equations have limited concentration ranges where they're physically meaningful. Always check whether your mathematical solution falls within realistic experimental conditions for the system being studied.

Question 2

A U-tube osmometer contains pure water on one side and a protein solution on the other side, separated by a semipermeable membrane. At equilibrium, the height difference is 15.2 cm of water column. If the protein solution is then diluted in situ by carefully injecting pure water directly into the protein compartment until the protein concentration decreases by 25%, what will be the new equilibrium height difference?

  1. 11.4 cm, because the osmotic pressure decreases proportionally with the concentration reduction (correct answer)
  2. 15.2 cm, because the total amount of protein remains constant even though the concentration changes
  3. 19.0 cm, because dilution increases the volume of the protein compartment, creating additional hydrostatic pressure
  4. 7.6 cm, because the 25% dilution reduces both concentration and effective membrane area by the same factor
  5. 13.7 cm, because osmotic pressure depends on the remaining 75% of the original protein concentration
Explanation: When you encounter osmometer problems, focus on the fundamental relationship between osmotic pressure and solute concentration. Osmotic pressure follows the van't Hoff equation: Π=iMRT\Pi = iMRT, where concentration (M) is directly proportional to osmotic pressure. In a U-tube osmometer, the height difference directly measures osmotic pressure through hydrostatic equilibrium. Initially, the protein solution creates enough osmotic pressure to support a 15.2 cm water column. When you dilute the protein solution by 25%, the concentration decreases to 75% of its original value (100% - 25% = 75%). Since osmotic pressure is directly proportional to concentration, the new osmotic pressure becomes 75% of the original: 0.75×15.2 cm=11.4 cm0.75 \times 15.2 \text{ cm} = 11.4 \text{ cm} Answer A correctly identifies this proportional relationship and calculates the reduced height as 11.4 cm. Answer B incorrectly assumes that total protein amount determines osmotic pressure, but osmotic pressure depends on concentration (moles per liter), not total moles. Answer C mistakenly introduces hydrostatic pressure from volume changes, but the height difference solely reflects osmotic pressure differences across the membrane. Answer D creates a nonsensical "effective membrane area" factor that doesn't exist in osmotic pressure calculations. Study tip: Remember that colligative properties like osmotic pressure depend only on particle concentration, not total amount. When dilution problems appear, immediately identify whether concentration or total quantity is the relevant variable for the property being measured.

Question 3

A membrane protein transporter creates a 10-fold concentration gradient of glucose across a cell membrane at 310 K. If the transporter couples glucose transport to ATP hydrolysis with a stoichiometry of 1 glucose : 1 ATP, and the free energy of ATP hydrolysis under cellular conditions is -54 kJ/mol, what is the minimum work required per mole of glucose transported against this gradient?

  1. 5.9 kJ/mol, calculated from the chemical potential difference of glucose across the concentration gradient (correct answer)
  2. 23.4 kJ/mol, derived from the osmotic work against the electrochemical potential difference
  3. 54.0 kJ/mol, equal to the free energy available from ATP hydrolysis for this transport process
  4. 48.1 kJ/mol, calculated from the difference between ATP hydrolysis energy and the gradient work
  5. 12.7 kJ/mol, obtained from the logarithmic relationship between concentration ratio and chemical potential
Explanation: When you encounter questions about active transport and energy coupling, focus on the fundamental thermodynamic principle: the minimum work required equals the free energy change for moving molecules against their concentration gradient. To calculate the work needed to transport glucose against a 10-fold concentration gradient, you need the chemical potential difference: W=RTln⁡(C2C1)W = RT \ln\left(\frac{C_2}{C_1}\right), where the concentration ratio is 10. Substituting the values: W=(8.314 J/mol\cdotpK)(310 K)ln⁡(10)=(8.314)(310)(2.303)=5.9 kJ/molW = (8.314 \text{ J/mol·K})(310 \text{ K}) \ln(10) = (8.314)(310)(2.303) = 5.9 \text{ kJ/mol}. This represents the minimum thermodynamic work required, making answer A correct. Answer B (23.4 kJ/mol) likely results from calculation errors, possibly confusing natural logarithm with log base 10 or using incorrect temperature units. Answer C (54.0 kJ/mol) represents a common misconception—assuming the work required equals the energy available from ATP hydrolysis. However, the transporter only needs to provide the minimum work to overcome the gradient; excess energy from ATP would be dissipated as heat. Answer D (48.1 kJ/mol) incorrectly suggests subtracting the gradient work from ATP energy, which has no thermodynamic basis. Remember that in coupled transport systems, you calculate the minimum work from the concentration gradient independently. The coupling mechanism (ATP hydrolysis) must provide at least this much energy, but the actual work requirement depends solely on the gradient being overcome, not the energy source driving it.

Question 4

Two compartments separated by a semipermeable membrane contain aqueous solutions at 300 K. Compartment A has 0.15 M NaCl and compartment B has 0.20 M glucose. At equilibrium, water will flow from A to B, establishing an osmotic pressure difference. If the membrane suddenly becomes permeable to glucose but remains impermeable to NaCl, what happens to the system?

  1. Water flow reverses direction because glucose equalizes while NaCl remains concentrated in compartment A (correct answer)
  2. Water flow continues in the same direction but at a reduced rate due to partial equilibration
  3. Water flow stops immediately because the total particle concentrations become equal
  4. Water flow rate increases because glucose diffusion creates additional concentration gradients
  5. The system reaches a new equilibrium with no net water flow and equal glucose concentrations
Explanation: When analyzing osmotic pressure changes with membrane permeability shifts, you need to track the effective particle concentrations on each side and predict water movement accordingly. Initially, compartment A contains 0.15 M NaCl, which dissociates into Na⁺ and Cl⁻ ions, creating an effective concentration of 0.30 M particles. Compartment B has 0.20 M glucose (non-dissociating), so 0.20 M particles. Since A has higher particle concentration, water flows from B to A until osmotic equilibrium. When the membrane becomes permeable to glucose, glucose will diffuse from B to A until glucose concentrations equalize. This doesn't affect the NaCl, which remains trapped in compartment A. After glucose equilibration, both sides will have equal glucose concentrations, but compartment A still retains its 0.30 M from NaCl while compartment B has no additional particles. Therefore, A maintains higher total particle concentration, and water flow reverses from the original equilibrium direction—now flowing from B to A more strongly than before. Answer A correctly identifies this reversal due to glucose equalization while NaCl concentration remains in A. Answer B incorrectly suggests the same flow direction continues. Answer C wrongly assumes total concentrations become equal—they don't because NaCl can't cross. Answer D incorrectly predicts increased flow rate in the wrong direction and misunderstands that glucose diffusion reduces rather than creates concentration gradients. Remember: always count effective particles after accounting for dissociation, and track which species can cross the membrane when permeability changes.

Question 5

Consider a three-component system with water (1), salt (2), and protein (3) at equilibrium across a membrane permeable only to water and salt. The chemical potentials satisfy μ1α=μ1β\mu_1^{\alpha} = \mu_1^{\beta} and μ2α=μ2β\mu_2^{\alpha} = \mu_2^{\beta} where α and β denote the two phases. If the protein concentration in phase α is suddenly increased while maintaining constant temperature, what is the immediate effect on the chemical potential of water in phase α?

  1. The chemical potential of water decreases because increased protein concentration lowers water activity through excluded volume effects (correct answer)
  2. The chemical potential of water increases because protein-water interactions are energetically favorable and stabilize the water molecules
  3. The chemical potential of water remains unchanged because proteins are large molecules that don't significantly affect water's thermodynamic state
  4. The chemical potential of water decreases because higher protein concentration increases the ionic strength and affects water's activity coefficient
  5. The chemical potential of water increases because the protein molecules compete with water for solvation sites around the salt ions
Explanation: When analyzing membrane equilibrium problems in multicomponent systems, focus on how concentration changes affect chemical potentials through activity coefficients and molecular interactions. When protein concentration increases in phase α, the chemical potential of water decreases primarily due to excluded volume effects. Proteins are large macromolecules that occupy significant space in solution, effectively reducing the volume available to water molecules. This creates a more crowded molecular environment where water molecules experience reduced translational freedom and altered intermolecular interactions. The excluded volume effect decreases water activity (a1=γ1x1a_1 = \gamma_1 x_1), and since μ1=μ1∗+RTln⁡a1\mu_1 = \mu_1^* + RT \ln a_1, the chemical potential of water decreases. Option A correctly identifies this excluded volume mechanism. Option B is incorrect because while protein-water interactions do occur, they don't stabilize water in a way that increases chemical potential—the dominant effect is still the reduction in water activity due to crowding. Option C is wrong because large molecules like proteins have significant thermodynamic effects precisely because of their size and the volume they exclude from other components. Option D incorrectly attributes the effect to ionic strength changes, but the immediate effect of adding protein is mechanical (excluded volume) rather than electrostatic, and proteins themselves don't directly increase ionic strength unless they're highly charged. Remember: In concentrated solutions with large molecules, excluded volume effects dominate over specific interactions when predicting changes in component chemical potentials. Always consider how molecular size affects available space and activity coefficients.

Question 6

A plant cell with an internal osmotic pressure of 8.5 atm is placed in a solution with an osmotic pressure of 6.2 atm at 298 K. The cell wall can withstand a maximum turgor pressure of 4.0 atm before rupturing. Assuming the cell behaves as an ideal osmometer with a rigid cell wall, will the cell rupture, and what will be the final turgor pressure?

  1. Yes, the cell will rupture because the osmotic pressure difference exceeds the maximum turgor pressure tolerance
  2. No, the cell will not rupture, and the final turgor pressure will be 2.3 atm (correct answer)
  3. Yes, the cell will rupture because water influx will continue until the internal pressure reaches 6.2 atm
  4. No, the cell will not rupture, and the final turgor pressure will be 4.0 atm due to wall resistance
  5. The cell will reach equilibrium with zero turgor pressure since the external solution is hypotonic
Explanation: When a plant cell is placed in a solution with different osmotic pressure, you need to understand how turgor pressure develops and whether it will exceed the cell wall's structural limits. The key principle is that turgor pressure equals the difference between the cell's internal osmotic pressure and the external solution's osmotic pressure. Here, the internal osmotic pressure is 8.5 atm and the external is 6.2 atm, so the turgor pressure will be 8.5−6.2=2.38.5 - 6.2 = 2.3 atm. Since this is less than the maximum tolerable turgor pressure of 4.0 atm, the cell wall can withstand this pressure without rupturing. Answer A incorrectly assumes that any osmotic pressure difference automatically causes rupture, but you must compare the resulting turgor pressure to the cell's tolerance limit, not just check if there's a difference. Answer C misunderstands the equilibrium process—the internal pressure doesn't change to match the external pressure. Instead, water moves until the pressure difference stabilizes at the calculated turgor pressure. Answer D suggests the turgor pressure will reach the maximum tolerance, but this only happens if the calculated pressure exceeds the limit and the wall provides resistance. The correct answer is B: the cell survives with a final turgor pressure of 2.3 atm. Remember this formula for plant cell problems: Turgor Pressure = Internal Osmotic Pressure - External Osmotic Pressure. The cell only ruptures if this calculated value exceeds the maximum turgor pressure the cell wall can withstand.

Question 7

A dialysis membrane separates a protein solution from a buffer solution. The protein solution contains 2.5 g/L of albumin (MW = 66,500 g/mol) and 0.15 M NaCl in phosphate buffer. The buffer side contains only 0.15 M NaCl in the same phosphate buffer. If the membrane is permeable to all species except albumin, what is the osmotic pressure difference across the membrane at 298 K after equilibrium is established?

  1. 0.92 atm, calculated from the albumin concentration alone since small molecules equilibrate across the membrane (correct answer)
  2. 1.15 atm, including both the direct albumin contribution and the Donnan effect on salt distribution
  3. 0.76 atm, accounting for the reduced activity of albumin due to protein-salt interactions in the solution
  4. 1.38 atm, calculated from the total protein concentration including bound water and counterion effects
  5. 0.58 atm, derived from the effective albumin concentration after correcting for non-ideal solution behavior
Explanation: When you encounter a dialysis problem with a semi-permeable membrane, think about what can and cannot cross the membrane. Here, only albumin is retained on one side while all small molecules (NaCl, buffer components) can freely move across. At equilibrium, small molecules distribute equally across the membrane, so their concentrations become identical on both sides. This means NaCl and buffer components contribute zero net osmotic pressure since they exert equal pressure in both directions. Only the species that cannot cross the membrane - albumin - creates an osmotic pressure difference. To calculate this pressure, first find the molar concentration of albumin: Molarity=2.5 g/L66,500 g/mol=3.76×10−5 M\text{Molarity} = \frac{2.5 \text{ g/L}}{66,500 \text{ g/mol}} = 3.76 \times 10^{-5} \text{ M} Then apply the osmotic pressure equation: Π=MRT=(3.76×10−5)(0.08206)(298)=0.92 atm\Pi = MRT = (3.76 \times 10^{-5})(0.08206)(298) = 0.92 \text{ atm} Answer A correctly identifies this fundamental principle and calculation. Answer B incorrectly invokes the Donnan effect, which applies when you have non-diffusible ions, but albumin here is a neutral protein. Answer C suggests protein-salt interactions reduce albumin's activity, but this effect is negligible for osmotic pressure calculations in dilute solutions. Answer D incorrectly includes "bound water and counterion effects" which aren't relevant for a neutral protein in this context. Remember: In membrane equilibrium problems, only concentrate on the non-permeating species. Everything else equilibrates and cancels out in the osmotic pressure calculation.

Question 8

The chemical potential of a solute in solution is given by μ = μ° + RT ln(γc), where γ is the activity coefficient and c is concentration. If a solution becomes more concentrated and γ decreases from 1.0 to 0.8 while concentration doubles, what happens to the chemical potential?

  1. Increases by RT ln(1.6), showing positive deviation from ideality enhances chemical potential
  2. Decreases by RT ln(1.25), indicating that non-ideal behavior dominates over concentration effects
  3. Increases by RT ln(1.6), despite negative deviation from ideality reducing activity coefficient (correct answer)
  4. Remains unchanged because the activity coefficient decrease exactly compensates for concentration increase
Explanation: Initial: μ₁ = μ° + RT ln(1.0 × c). Final: μ₂ = μ° + RT ln(0.8 × 2c) = μ° + RT ln(1.6c). Change: Δμ = RT ln(1.6c/c) = RT ln(1.6). The chemical potential increases despite the activity coefficient decrease because the concentration effect (factor of 2) outweighs the non-ideality effect (factor of 0.8). Choice A incorrectly describes this as positive deviation; γ < 1 indicates negative deviation.

Question 9

The osmotic pressure of a 0.1 M sucrose solution is measured as 2.3 atm at 25°C, while the theoretical value for an ideal solution is 2.4 atm. If the solution volume is increased by adding pure water until the concentration becomes 0.05 M, what will be the new osmotic pressure?

  1. 1.20 atm, because dilute solutions behave more ideally than concentrated ones (correct answer)
  2. 1.15 atm, assuming the same deviation from ideality persists proportionally
  3. 1.12 atm, accounting for the activity coefficient change upon dilution
  4. 1.15 atm, because osmotic pressure scales directly with concentration regardless of non-ideality
Explanation: When you encounter osmotic pressure problems involving non-ideal solutions, focus on how deviations from ideality change with concentration. Real solutions often approach ideal behavior as they become more dilute. The original 0.1 M solution shows a deviation from ideality: measured (2.3 atm) versus theoretical ideal (2.4 atm). This gives a ratio of 2.3/2.4 = 0.958, indicating the solution behaves about 4.2% less ideally than predicted. Upon dilution to 0.05 M, two effects occur simultaneously. First, the concentration halves, so the ideal osmotic pressure would be Π=MRT=(0.05)(0.0821)(298)=1.22\Pi = MRT = (0.05)(0.0821)(298) = 1.22 atm. Second, the solution becomes more dilute and approaches ideal behavior more closely than the concentrated solution. Answer A correctly predicts 1.20 atm because dilute solutions behave more ideally. The deviation from ideality decreases as intermolecular interactions become less significant in the more dilute solution. Answer B incorrectly assumes the same proportional deviation persists (0.958 × 1.22 = 1.17 atm), ignoring that dilution improves ideality. Answer C suggests an even larger deviation (1.12 atm), which contradicts the principle that dilution improves ideal behavior. Answer D claims non-ideality doesn't affect the concentration dependence, which misunderstands how real solutions deviate from ideal behavior. Remember: when diluting real solutions, expect them to behave more ideally than their concentrated counterparts. This is a key principle in solution thermodynamics that frequently appears on physical chemistry exams.

Question 10

A dialysis membrane separates blood plasma (protein concentration 70 g/L, NaCl 0.15 M) from dialysate solution (NaCl 0.10 M, no protein). The membrane is permeable to small ions but not proteins. What is the primary factor determining the final equilibrium state?

  1. Donnan equilibrium establishing unequal ion distributions due to impermeant protein charges, plus protein osmotic effects (correct answer)
  2. Equal NaCl concentrations on both sides, with proteins creating additional osmotic pressure in plasma
  3. Complete ion equilibration with protein osmotic pressure balanced by hydrostatic pressure differences
  4. Protein denaturation due to osmotic stress, eliminating the concentration gradient driving force
Explanation: When you encounter a dialysis setup with charged proteins on one side and a semipermeable membrane, you're dealing with the Donnan equilibrium - a fundamental concept where impermeant charged species create unequal ion distributions across membranes. Here's what happens: Blood proteins (mainly albumin) carry negative charges and cannot cross the membrane. To maintain electroneutrality, more cations (Na⁺) must remain on the protein side than would be present without proteins. This creates unequal ion concentrations across the membrane - more total ions on the protein side. Additionally, the proteins themselves contribute to osmotic pressure through their physical presence in solution. Answer A correctly identifies both components: the Donnan effect causing unequal ion distributions due to charged proteins, plus the direct osmotic contribution of proteins themselves. Answer B is wrong because the final NaCl concentrations won't be equal - the Donnan effect prevents this. The impermeant protein charges force unequal ion distributions. Answer C incorrectly suggests complete ion equilibration occurs. This is impossible when impermeant charged species are present - the Donnan equilibrium specifically prevents equal ion concentrations. Answer D is nonsensical because protein denaturation isn't a factor in dialysis equilibrium. The proteins remain intact and continue exerting their effects throughout the process. Remember: whenever you see a membrane permeable to ions but not to charged macromolecules, think Donnan equilibrium. The impermeant charges always create unequal ion distributions, never simple equilibration.

Question 11

A plant cell with an internal osmotic pressure of 8 atm is placed in a solution with osmotic pressure of 5 atm. If the cell wall can withstand a maximum turgor pressure of 4 atm before bursting, what will happen?

  1. The cell will burst immediately because the osmotic pressure difference exceeds the cell wall strength
  2. The cell will reach equilibrium with 3 atm turgor pressure and remain intact (correct answer)
  3. The cell will shrink because water flows out due to higher external osmotic pressure
  4. The cell will burst because turgor pressure equals the external solution's osmotic pressure
Explanation: Turgor pressure = internal osmotic pressure - external osmotic pressure = 8 - 5 = 3 atm. Since 3 atm < 4 atm (maximum), the cell remains intact. Choice A incorrectly assumes direct pressure comparison. Choice C wrongly suggests water flows out when internal osmotic pressure is higher. Choice D misunderstands that turgor pressure equals the pressure difference, not the external pressure.

Question 12

A U-tube contains pure water on one side and a glucose solution on the other, separated by a semipermeable membrane at the bottom. After equilibrium is reached, the height difference is 0.5 m. What is the approximate molar concentration of the glucose solution? (Use g = 9.8 m/s², ρ_water = 1000 kg/m³)

  1. 0.50 M, calculated directly from the hydrostatic pressure using π = ρgh
  2. 0.20 M, accounting for the density difference between pure water and glucose solution
  3. 0.50 M, assuming ideal gas behavior applies directly to the liquid solution
  4. 0.20 M, using π = ρgh = MRT with appropriate unit conversions (correct answer)
Explanation: This question tests osmotic pressure equilibrium, a key concept in physical chemistry that bridges thermodynamics and solution chemistry. When you see a U-tube with different solutions separated by a semipermeable membrane, think about how the system reaches equilibrium through osmotic pressure balancing hydrostatic pressure. At equilibrium, the osmotic pressure of the glucose solution equals the hydrostatic pressure from the height difference. The hydrostatic pressure is π=ρgh=(1000 kg/m3)(9.8 m/s2)(0.5 m)=4900 Pa\pi = \rho gh = (1000 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(0.5 \text{ m}) = 4900 \text{ Pa}. Using the van't Hoff equation for osmotic pressure, π=MRT\pi = MRT, where M is molarity, R is the gas constant (8.314 J/mol·K), and T is temperature (assume 298 K). Solving: M=4900(8.314)(298)=0.198≈0.20 MM = \frac{4900}{(8.314)(298)} = 0.198 \approx 0.20 \text{ M}. This confirms answer D. Answer A incorrectly suggests you can calculate molarity directly from π=ρgh\pi = \rho gh without using the van't Hoff equation - this misses the crucial RT conversion factor. Answer B mentions density differences, but for dilute glucose solutions, this effect is negligible compared to the osmotic pressure calculation. Answer C wrongly applies ideal gas behavior "directly" to liquids - while the van't Hoff equation does use gas constant R, it's specifically derived for solutions, not a direct application of gas laws. Remember: osmotic pressure problems always require the van't Hoff equation (π=MRT\pi = MRT) to connect measurable pressure differences to solution concentrations. Don't skip the RT factor when converting between pressure and molarity.

Question 13

The chemical potential of water in a solution can be expressed as μ(H₂O) = μ°(H₂O) + RT ln(a_w), where a_w is the water activity. If the mole fraction of water decreases from 0.95 to 0.90 at constant temperature, assuming ideal behavior, how does the chemical potential change?

  1. Decreases by RT ln(0.947), approximately -0.054RT
  2. Increases by RT ln(1.056), approximately +0.054RT
  3. Decreases by RT ln(1.056), approximately -0.054RT (correct answer)
  4. Increases by RT ln(0.947), approximately +0.054RT
Explanation: For ideal solutions, a_w ≈ X_w. Initial: μ₁ = μ° + RT ln(0.95). Final: μ₂ = μ° + RT ln(0.90). Change: Δμ = μ₂ - μ₁ = RT ln(0.90/0.95) = RT ln(0.947) ≈ -0.054RT. The chemical potential decreases because ln(0.947) is negative. Choice A has the wrong sign convention, B and D give incorrect mathematical relationships.

Question 14

In a reverse osmosis system, water is forced through a semipermeable membrane against its natural osmotic gradient. If seawater with osmotic pressure 25 atm is being desalinated, what minimum applied pressure is theoretically required to achieve any net water flow?

  1. Exactly 25 atm, as this balances the osmotic pressure difference
  2. Slightly greater than 25 atm, to overcome osmotic pressure and create driving force (correct answer)
  3. Less than 25 atm, since osmotic pressure aids the separation process
  4. Approximately 50 atm, to overcome both osmotic pressure and membrane resistance
Explanation: For reverse osmosis, applied pressure must exceed the osmotic pressure difference to drive water against its natural gradient. At exactly 25 atm (choice A), there's no net flow - equilibrium. Choice C incorrectly suggests osmotic pressure helps (it opposes). Choice D doubles the pressure unnecessarily; membrane resistance requires additional pressure beyond the minimum theoretical value, but the question asks for minimum theoretical pressure.

Question 15

Two compartments contain solutions with different chemical potentials for water: μ₁ = -1000 J/mol and μ₂ = -1200 J/mol. If a membrane permeable only to water separates them, what drives the water transport and what is the equilibrium condition?

  1. Water flows from compartment 2 to 1 until chemical potentials are equal, driven by the 200 J/mol difference
  2. No net water flow occurs because both chemical potentials are negative, indicating thermodynamic equilibrium
  3. Water flows from compartment 1 to 2 until osmotic pressures are equal, regardless of chemical potential values
  4. Water flows from compartment 1 to 2 until chemical potentials are equal, driven by the 200 J/mol difference (correct answer)
Explanation: When you encounter problems involving chemical potential and membrane transport, remember that water spontaneously flows from regions of higher chemical potential to lower chemical potential until equilibrium is reached. Chemical potential represents the free energy per mole of a substance. Since μ1=−1000\mu_1 = -1000 J/mol is greater (less negative) than μ2=−1200\mu_2 = -1200 J/mol, water will spontaneously flow from compartment 1 to compartment 2. This continues until the chemical potentials equalize, which defines thermodynamic equilibrium. The driving force is indeed the 200 J/mol difference between the compartments. Option A incorrectly suggests water flows from compartment 2 to 1. This confuses the direction - water flows from higher to lower chemical potential, not from more negative to less negative values. Remember that -1000 > -1200 on the number line. Option B wrongly assumes that negative chemical potential values indicate equilibrium. The sign of chemical potential doesn't determine equilibrium; equality of chemical potentials across the membrane does. Both values being negative simply reflects the reference state chosen. Option C incorrectly prioritizes osmotic pressure over chemical potential. While osmotic pressure contributes to chemical potential, the fundamental driving force for transport is the chemical potential difference itself. Chemical potential already accounts for all relevant factors including pressure, concentration, and temperature. Study tip: Always remember that equilibrium occurs when chemical potentials are equal across permeable boundaries, and transport occurs from higher to lower chemical potential. Focus on the relative values, not their signs.