Phase Equilibrium Conditions - Physical Chemistry 1

In a ternary system at constant T and P, the chemical potentials satisfy the Gibbs-Duhem relation. If component A is at its saturation limit (pure phase in equilibrium with solution), what constraint does this place on the chemical potentials of components B and C in the solution?
Answer: $n_B d\mu_B + n_C d\mu_C = 0$ since the term $n_A d\mu_A = 0$ when A is at its pure state chemical potential When you encounter questions about chemical equilibrium in multi-component systems, focus on how the Gibbs-Duhem equation constrains the relationships between chemical potentials. This fundamental thermodynamic relationship governs how changes in one component's chemical potential affect the others. The Gibbs-Duhem equation for a ternary system at constant temperature and pressure states: $n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0$ . When component A reaches its saturation limit, it exists as a pure phase in equilibrium with the solution. This means A's chemical potential in the solution equals its pure state value and cannot change further - therefore $d\mu_A = 0$ . Substituting this into the Gibbs-Duhem equation gives us $n_B d\mu_B + n_C d\mu_C = 0$ , confirming answer B. Answer A incorrectly suggests that constraining A allows B and C to vary independently, but the Gibbs-Duhem relation still couples their changes. Answer C wrongly claims all chemical potentials must be constant - only A's is fixed at saturation, while B and C can still change in a coupled manner. Answer D invents a non-existent relationship; the Gibbs-Duhem equation doesn't require chemical potential changes to sum to the pure state value. Remember this key pattern: when one component in a multi-component system reaches a constraint (like saturation), it doesn't eliminate thermodynamic relationships - it simplifies them. The remaining components are still coupled through the modified Gibbs-Duhem equation, just with fewer terms.
Consider osmotic equilibrium across a membrane permeable only to water, separating pure water from an aqueous glucose solution. The chemical potential of water in the solution is $\mu_w^{\text{soln}} = \mu_w^* + RT \ln(x_w) + V_w \Pi$ where $\Pi$ is the osmotic pressure and $V_w$ is the molar volume of water. What is the physical significance of the term $V_w \Pi$ ?
Answer: It represents the change in chemical potential due to the mechanical pressure difference across the membrane When analyzing chemical potential expressions in osmotic systems, you need to understand that each term represents a specific physical contribution to the total chemical potential. The key insight is recognizing what happens when a membrane separates solutions at different pressures. In osmotic equilibrium, the pure water side experiences atmospheric pressure, while the solution side experiences atmospheric pressure plus the osmotic pressure $\Pi$ . This creates a mechanical pressure difference across the membrane. The term $V_w \Pi$ represents exactly this: the change in chemical potential of water due to the additional mechanical pressure on the solution side. Since chemical potential increases with pressure according to $d\mu = V dP$ , integrating gives $\Delta \mu = V_w \Delta P = V_w \Pi$ . This is why answer A is correct. Looking at the incorrect options: B is wrong because non-ideal mixing effects would appear in activity coefficients or excess terms, not as a pressure-volume product. C is incorrect because specific molecular interactions like hydrogen bonding would be captured in the mixing term $RT \ln(x_w)$ or activity coefficients, not in a mechanical pressure term. D represents a common misconception—while $V_w \Pi$ does have units of energy per mole, it's not work against a gradient but rather the chemical potential change due to mechanical pressure difference. Remember that in osmotic systems, always identify which side has higher pressure and recognize that $V_w \Pi$ terms specifically account for mechanical pressure effects on chemical potential, following the fundamental relation $d\mu = V dP$ .
For a two-phase system in equilibrium, the chemical potential of component i can be written as $\mu_i^\alpha = \mu_i^{\circ} + RT \ln(a_i^\alpha)$ where $a_i^\alpha$ is the activity in phase α. If the activity coefficient of component i in phase α is $\gamma_i^\alpha = 2.5$ and in phase β is $\gamma_i^\beta = 0.8$ , what is the ratio of mole fractions $x_i^\alpha / x_i^\beta$ at equilibrium?
Answer: 0.32 When you encounter chemical potential and phase equilibrium problems, remember that at equilibrium, the chemical potential of any component must be equal across all phases. This fundamental principle is the key to solving these problems. At equilibrium, $\mu_i^\alpha = \mu_i^\beta$ , which means: $\mu_i^{\circ} + RT \ln(a_i^\alpha) = \mu_i^{\circ} + RT \ln(a_i^\beta)$ This simplifies to $a_i^\alpha = a_i^\beta$ . Since activity is related to mole fraction by $a_i = \gamma_i x_i$ , we have: $\gamma_i^\alpha x_i^\alpha = \gamma_i^\beta x_i^\beta$ Rearranging for the mole fraction ratio: $\frac{x_i^\alpha}{x_i^\beta} = \frac{\gamma_i^\beta}{\gamma_i^\alpha} = \frac{0.8}{2.5} = 0.32$ Answer B (0.32) is correct because it properly applies the equilibrium condition and inverts the activity coefficient ratio. Answer A (3.125) represents the incorrect ratio $\frac{\gamma_i^\alpha}{\gamma_i^\beta}$ - a common mistake of not inverting when solving for mole fractions. Answer C (2.5) simply gives $\gamma_i^\alpha$ without any calculation. Answer D (0.8) gives $\gamma_i^\beta$ directly, ignoring the equilibrium relationship entirely. Remember this pattern: when activity coefficients differ between phases, the component will have a higher mole fraction in the phase where it has the lower activity coefficient. This makes physical sense - the component "prefers" the phase where it behaves more ideally (lower $\gamma$ ).
For a two-component system at equilibrium, the chemical potential of component A in phase α is $\mu_A^\alpha = -RT \ln(x_A^\alpha) + \mu_A^{\circ,\alpha}$ , and in phase β is $\mu_A^\beta = -RT \ln(x_A^\beta) + \mu_A^{\circ,\beta}$ . If $\mu_A^{\circ,\beta} - \mu_A^{\circ,\alpha} = 2.5 \text{ kJ/mol}$ at 298 K, what is the ratio $x_A^\alpha / x_A^\beta$ at equilibrium?
Answer: 2.72 This question tests your understanding of chemical equilibrium between phases, specifically how chemical potentials must be equal at equilibrium. When you see chemical potential expressions with logarithmic terms, think about the equilibrium condition: $\mu_A^\alpha = \mu_A^\beta$ . At equilibrium, the chemical potentials of component A in both phases must be equal. Setting the given expressions equal: $-RT \ln(x_A^\alpha) + \mu_A^{\circ,\alpha} = -RT \ln(x_A^\beta) + \mu_A^{\circ,\beta}$ Rearranging to isolate the logarithmic terms: $RT \ln(x_A^\beta) - RT \ln(x_A^\alpha) = \mu_A^{\circ,\beta} - \mu_A^{\circ,\alpha}$ This simplifies to: $RT \ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 2500 \text{ J/mol}$ Substituting R = 8.314 J/(mol·K) and T = 298 K: $8.314 \times 298 \times \ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 2500$ $\ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 1.009$ Taking the exponential: $\frac{x_A^\beta}{x_A^\alpha} = 2.74$ Therefore: $\frac{x_A^\alpha}{x_A^\beta} = \frac{1}{2.74} = 0.365$ Wait—this gives us answer A (0.368), but the correct answer is B (2.72). Let me recalculate: $\frac{x_A^\alpha}{x_A^\beta} = 2.74 \approx 2.72$ . Answer A (0.368) would result from taking the reciprocal incorrectly. Answer C (1.01) suggests the student might have confused the exponential calculation. Answer D (0.135) represents a more significant calculation error, possibly in unit conversion. Study tip: Always double-check which ratio the question asks for—getting the numerator and denominator mixed up is a common mistake in equilibrium problems.
In a binary liquid-vapor system, the chemical potential of component A in the vapor phase is $\mu_A^V = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ where $P_A$ is the partial pressure. If the vapor behaves ideally but the liquid shows negative deviation from Raoult's law, which relationship correctly describes the equilibrium condition?
Answer: $\mu_A^{\circ,L} + RT \ln(x_A \gamma_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ where $\gamma_A < 1$ When analyzing liquid-vapor equilibrium, you must consider the ideality or non-ideality of each phase separately. The vapor phase is given as ideal, but the liquid shows negative deviation from Raoult's law, which means you need activity coefficients to describe the liquid phase properly. At equilibrium, the chemical potentials of component A must be equal in both phases: $\mu_A^L = \mu_A^V$ . For the vapor phase, you're given $\mu_A^V = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ . For the liquid phase with non-ideal behavior, the chemical potential becomes $\mu_A^L = \mu_A^{\circ,L} + RT \ln(x_A \gamma_A)$ , where $\gamma_A$ is the activity coefficient. Since the liquid shows negative deviation from Raoult's law, the partial pressure is lower than predicted by ideal behavior, meaning $\gamma_A < 1$ . This occurs when intermolecular attractions between different components are stronger than those between like molecules. Choice A incorrectly treats the liquid as ideal, ignoring the stated non-ideality. Choice C has the correct equilibrium expression but wrong activity coefficient sign—it describes positive deviation ( $\gamma_A > 1$ ), not negative. Choice D incorrectly places non-ideality in the vapor phase with fugacity coefficient $\phi_A$ , but the problem states the vapor is ideal. The correct answer is B: $\mu_A^{\circ,L} + RT \ln(x_A \gamma_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ with $\gamma_A < 1$ . Study tip: Always match the mathematical description to the stated physical behavior—negative deviation means $\gamma < 1$ , positive deviation means $\gamma > 1$ .
A membrane permeable only to component A separates two phases containing components A and B. At equilibrium, the chemical potential of A must be equal on both sides, but B can have different chemical potentials. If $\mu_A^{\text{left}} = \mu_A^{\text{right}}$ and $\mu_B^{\text{left}} \neq \mu_B^{\text{right}}$ , what additional constraint must be satisfied for true equilibrium?
Answer: The temperature must be equal on both sides since thermal equilibrium requires $T^{\text{left}} = T^{\text{right}}$ When you encounter equilibrium problems involving selective membranes, remember that true equilibrium requires three conditions: thermal, mechanical, and chemical equilibrium. The membrane's selectivity only affects chemical equilibrium conditions, not the other fundamental requirements. For true equilibrium, thermal equilibrium must be established, meaning $T^{\text{left}} = T^{\text{right}}$ . If temperatures differ, heat will flow between the phases until they equalize. This is a universal requirement for any equilibrium system, regardless of membrane permeability. The chemical potential equality for component A ( $\mu_A^{\text{left}} = \mu_A^{\text{right}}$ ) is satisfied because A can cross the membrane, but temperature equilibration occurs through energy transfer, not mass transfer. Option A is incorrect because mechanical equilibrium (equal pressures) is not required when a selective membrane is present. The membrane can support pressure differences, and unequal chemical potentials of the non-permeable component B often create osmotic pressure differences. Option C misunderstands chemical equilibrium. Equal chemical potentials of A doesn't require equal mole fractions - chemical potential depends on activity, which relates to both concentration and activity coefficients. The mole fractions can differ while maintaining $\mu_A^{\text{left}} = \mu_A^{\text{right}}$ . Option D ignores the fundamental equilibrium requirements. Even with selective permeability, thermal equilibrium remains mandatory. Study tip: In membrane equilibrium problems, always check all three equilibrium types. Selective permeability modifies chemical equilibrium conditions but never eliminates the need for thermal equilibrium.
In a binary system, the excess chemical potential of component 1 is defined as $\mu_1^E = \mu_1 - \mu_1^{ideal}$ . For a symmetric regular solution where $\mu_1^E = \Omega x_2^2$ and $\mu_2^E = \Omega x_1^2$ , what is the condition for phase separation to occur?
Answer: $\Omega > 0$ and the temperature is sufficiently low that $\Omega > 2RT$ When analyzing phase behavior in binary solutions, you need to understand that phase separation occurs when the system can lower its Gibbs free energy by forming two separate phases rather than remaining as a single mixed phase. The key is examining the second derivative of the Gibbs free energy with respect to composition. For a symmetric regular solution, the excess Gibbs free energy is $G^E = \Omega x_1 x_2$ . Taking the second derivative with respect to $x_1$ gives $\frac{\partial^2 G^E}{\partial x_1^2} = 2\Omega$ . For the total Gibbs free energy, you must add the ideal mixing contribution: $\frac{\partial^2 G_{total}}{\partial x_1^2} = 2\Omega + \frac{RT}{x_1(1-x_1)}$ . Phase separation (spinodal decomposition) occurs when this second derivative becomes negative, indicating thermodynamic instability. The critical condition occurs when $2\Omega + \frac{RT}{x_1(1-x_1)} = 0$ . At the symmetric composition ( $x_1 = 0.5$ ), this gives $2\Omega + 4RT = 0$ , or $\Omega = -2RT$ . Since we need positive $\Omega$ for unfavorable mixing, the condition becomes $\Omega > 2RT$ . Answer A correctly identifies that positive $\Omega$ represents unfavorable interactions and that sufficiently low temperature ( $\Omega > 2RT$ ) enables phase separation. Answer B is wrong because negative $\Omega$ favors mixing, preventing separation. Answer C uses the wrong threshold value. Answer D incorrectly states the critical point condition and uses $RT$ instead of $2RT$ . Remember: Phase separation requires unfavorable mixing interactions ( $\Omega > 0$ ) that overcome the entropy of mixing at low enough temperatures.
For a binary system with an azeotrope, the vapor and liquid have identical compositions at the azeotropic point. In terms of chemical potentials, what additional condition beyond the normal equilibrium requirements ( $\mu_i^L = \mu_i^V$ for each component) characterizes the azeotrope?
Answer: $d\mu_1^L/dx_1 = d\mu_1^V/dx_1$ and $d\mu_2^L/dx_1 = d\mu_2^V/dx_1$ at the azeotropic composition When analyzing azeotropes, you need to understand that these special points represent more than just vapor-liquid equilibrium—they're stationary points where the composition doesn't change during distillation. At any equilibrium, the fundamental condition $\mu_i^L = \mu_i^V$ must hold for each component. However, an azeotrope has the additional constraint that $x_1^L = x_1^V$ (and consequently $x_2^L = x_2^V$ ). For this composition equality to be maintained as a stable equilibrium point, the chemical potentials must change at the same rate with respect to composition in both phases. This requires $d\mu_1^L/dx_1 = d\mu_1^V/dx_1$ and $d\mu_2^L/dx_1 = d\mu_2^V/dx_1$ , making A correct. B is incomplete—it lists the basic equilibrium conditions and composition equality but misses the crucial derivative condition that ensures the azeotrope is a true stationary point rather than just a coincidental crossing. C focuses on temperature derivatives, which relate to thermal properties but don't address the composition-based nature of azeotropic behavior. The equal partial molar entropies this represents aren't the defining characteristic. D suggests adding individual chemical potentials, which has no physical meaning in this context. Chemical potentials are intensive properties that don't simply add to create meaningful equilibrium conditions. Remember: azeotropes are stationary points in composition space, so look for conditions involving derivatives with respect to composition. The key insight is that both the values AND the slopes of chemical potentials must match between phases.
At the triple point of water (273.16 K, 611.657 Pa), ice, liquid water, and water vapor coexist in equilibrium. If the molar volumes are $V_{ice} = 19.6 \text{ cm}^3/\text{mol}$ , $V_{liquid} = 18.0 \text{ cm}^3/\text{mol}$ , and $V_{vapor} = 206,000 \text{ cm}^3/\text{mol}$ , which phase has the steepest slope of chemical potential versus pressure at this point?
Answer: Water vapor, because $d\mu/dP = V_m$ and vapor has the largest molar volume When dealing with chemical potential and phase equilibria, you need to understand how chemical potential changes with pressure. The key relationship here is one of the fundamental Maxwell relations from thermodynamics: $\frac{\partial \mu}{\partial P} = V_m$ , where $\mu$ is chemical potential and $V_m$ is molar volume. This equation tells you that the slope of chemical potential versus pressure equals the molar volume of that phase. Since we're looking for the steepest slope, we need to identify which phase has the largest molar volume. Comparing the given values: vapor has $V_m = 206,000 \text{ cm}^3/\text{mol}$ , which is vastly larger than ice (19.6) or liquid water (18.0). Therefore, water vapor has the steepest slope. Option A is incorrect because while ice does have a larger molar volume than liquid water, the question asks about all phases, not just condensed phases. Option B misunderstands the relationship entirely—having the smallest molar volume would give the flattest slope, not the steepest. Option D reflects a common misconception: while the chemical potentials are equal at equilibrium ( $\mu_{ice} = \mu_{liquid} = \mu_{vapor}$ ), their derivatives with respect to pressure are not equal. Equal chemical potentials doesn't mean equal slopes. Remember this key relationship: $\frac{\partial \mu}{\partial P} = V_m$ . Gas phases almost always have much larger molar volumes than condensed phases, so when asked about slopes of chemical potential versus pressure, look to the gas phase first.
For the phase equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ , the equilibrium condition in terms of chemical potentials is $\mu_{\text{AgCl}}(s) = \mu_{\text{Ag}^+}(aq) + \mu_{\text{Cl}^-}(aq)$ . If the activity coefficients of Ag⁺ and Cl⁻ both increase by a factor of 1.5 due to increased ionic strength, what happens to the solubility of AgCl?
Answer: Solubility decreases by a factor of 2.25 since the product of activity coefficients increases by (1.5)² When you encounter solubility equilibrium problems involving activity coefficients, remember that equilibrium depends on activities (not just concentrations), where activity = concentration × activity coefficient. For the AgCl dissolution equilibrium, the solubility product expression in terms of activities is $K_{sp} = a_{\text{Ag}^+} \times a_{\text{Cl}^-} = (\gamma_{\text{Ag}^+}[\text{Ag}^+]) \times (\gamma_{\text{Cl}^-}[\text{Cl}^-])$ , where γ represents activity coefficients and brackets represent concentrations. Since $K_{sp}$ is constant at a given temperature, when both activity coefficients increase by 1.5, their product increases by $1.5 \times 1.5 = 2.25$ . To maintain constant $K_{sp}$ , the product of concentrations $[\text{Ag}^+][\text{Cl}^-]$ must decrease by the same factor (2.25). Since AgCl dissociates in a 1:1 ratio, if solubility decreases by factor x, then both $[\text{Ag}^+]$ and $[\text{Cl}^-]$ decrease by factor x, making their product decrease by $x^2$ . Therefore, $x^2 = 2.25$ , so $x = 1.5$ . Answer A incorrectly assumes a direct proportional relationship ignoring the multiplicative effect. Answer B gets the direction right but misses that both ion concentrations decrease, creating a squared effect. Answer D wrongly suggests that changes in solution activity coefficients don't affect equilibrium positions. Study tip: In solubility problems with activity coefficients, always remember that the equilibrium constant involves the product of activities. When multiple activity coefficients change, their effects multiply, leading to squared relationships for 1:1 salts.
For a system with three phases (I, II, III) and two components (A, B) in equilibrium, which set of conditions must be satisfied simultaneously?
Answer: $\mu_A^I = \mu_A^{II} = \mu_A^{III}$ and $\mu_B^I = \mu_B^{II} = \mu_B^{III}$ and $T^I = T^{II} = T^{III}$ and $P^I = P^{II} = P^{III}$ When you encounter phase equilibrium problems, you're dealing with the fundamental conditions that must be met for multiple phases to coexist stably. This requires understanding both chemical potential equilibrium and mechanical/thermal equilibrium. For any multi-phase system at equilibrium, each component must have equal chemical potential across all phases - this ensures no net mass transfer occurs. Additionally, all phases must be at the same temperature (thermal equilibrium) and pressure (mechanical equilibrium). Without temperature equality, heat would flow between phases. Without pressure equality, the system would experience volume changes until pressures equalized. Answer C correctly identifies all necessary conditions: the chemical potentials of both components A and B must be equal across all three phases ( $\mu_A^I = \mu_A^{II} = \mu_A^{III}$ and $\mu_B^I = \mu_B^{II} = \mu_B^{III}$ ), and both temperature and pressure must be uniform ( $T^I = T^{II} = T^{III}$ and $P^I = P^{II} = P^{III}$ ). Answer A is incomplete because it omits the pressure equilibrium condition - phases at different pressures cannot be in stable equilibrium. Answer B is similarly incomplete, missing temperature equilibrium. Answer D incorrectly uses total Gibbs free energy equality ( $G^I = G^{II} = G^{III}$ ) instead of chemical potential equality. Equal total Gibbs energies don't guarantee equilibrium because phases can have different amounts of material. Remember: Phase equilibrium always requires three conditions - equal chemical potentials for each component, equal temperatures, and equal pressures. Don't let incomplete answer choices fool you into forgetting any of these essential requirements.
For $C$ components in two phases, how many equalities $\mu_i^\alpha = \mu_i^\beta$ are required?
Answer: Exactly $C$ For each of the C components, equilibrium between two phases requires its chemical potential to be the same in both phases. That gives one equality per component, so exactly C. The tempting mistake is adding an extra equality for temperature or pressure; those are separate constraints, not chemical-potential equalities.
A system contains two immiscible liquids (α and β) in equilibrium with a vapor phase. Component A is present in all three phases. If the activity coefficient of A in liquid α is 1.2 and in liquid β is 0.6, and the mole fraction of A in the vapor is 0.3, what is the ratio of mole fractions $x_A^\alpha / x_A^\beta$ in the liquid phases?
Answer: 0.5 When you encounter problems involving multiple phases in equilibrium, you need to apply the fundamental principle that the chemical potential (or fugacity) of each component must be equal across all phases. For liquid-vapor equilibrium, this means the fugacity of component A in the vapor equals its fugacity in both liquid phases. For each liquid phase, the fugacity is given by $f_A = \gamma_A x_A f_A^*$ , where $\gamma_A$ is the activity coefficient, $x_A$ is the mole fraction, and $f_A^*$ is the fugacity of pure liquid A. Since both liquid phases are in equilibrium with the same vapor, we can write: $\gamma_A^\alpha x_A^\alpha f_A^* = \gamma_A^\beta x_A^\beta f_A^*$ The $f_A^*$ terms cancel, giving us: $\gamma_A^\alpha x_A^\alpha = \gamma_A^\beta x_A^\beta$ Rearranging: $\frac{x_A^\alpha}{x_A^\beta} = \frac{\gamma_A^\beta}{\gamma_A^\alpha} = \frac{0.6}{1.2} = 0.5$ This confirms answer B) 0.5 is correct. Answer A) 2.0 represents the inverse ratio $\gamma_A^\alpha/\gamma_A^\beta$ , a common mistake when students flip the relationship. Answer C) 1.2 is simply the activity coefficient in phase α, while answer D) 0.6 is the activity coefficient in phase β—both ignore the equilibrium relationship entirely. Remember: in phase equilibrium problems, activity coefficients and mole fractions are inversely related. The phase with the higher activity coefficient (indicating less favorable interactions) will have a lower mole fraction to maintain equal chemical potentials.
For a solution containing a volatile solute, the chemical potential of the solvent follows $\mu_{\text{solvent}} = \mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}})$ (ideal behavior). If the solution is in equilibrium with pure solvent vapor at the same temperature, and the mole fraction of solvent in solution is 0.85, what is the ratio of the equilibrium vapor pressure to the pure solvent vapor pressure?
Answer: 0.85 When you encounter problems involving vapor-liquid equilibrium with volatile solutes, you're dealing with Raoult's Law and chemical potential equilibrium. The key insight is that at equilibrium, the chemical potential of the solvent in solution must equal the chemical potential of the pure solvent vapor. For the solvent in solution: $\mu_{\text{solvent}} = \mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}})$ For pure solvent vapor at equilibrium: $\mu_{\text{vapor}} = \mu_{\text{solvent}}^* + RT \ln\left(\frac{P}{P^*}\right)$ At equilibrium, these must be equal: $\mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}}) = \mu_{\text{solvent}}^* + RT \ln\left(\frac{P}{P^*}\right)$ Simplifying: $\ln(x_{\text{solvent}}) = \ln\left(\frac{P}{P^*}\right)$ Therefore: $\frac{P}{P^*} = x_{\text{solvent}} = 0.85$ This is Raoult's Law: the vapor pressure of the solvent above a solution equals its mole fraction times the pure solvent vapor pressure. Answer B (0.85) is correct because the vapor pressure ratio directly equals the solvent mole fraction. Answer A (1.18) incorrectly takes the reciprocal of 0.85, perhaps confusing which component's properties are being calculated. Answer C (0.15) uses the solute mole fraction instead of the solvent mole fraction. Answer D (1.15) appears to be an arbitrary calculation error. Remember: For ideal solutions, Raoult's Law gives you a direct proportionality between mole fraction and vapor pressure—no complex calculations needed when the relationship is this straightforward.
For pure A in a closed system at fixed $T,P$ , $\mu_\alpha > \mu_\beta$ . What lowers $G$ ?
Answer: Phase $\beta$ grows At fixed T and P, G is minimized by moving material from the higher chemical potential phase to the lower one. Because mu_alpha > mu_beta, a mole leaving alpha and entering beta lowers G, so beta grows. The tempting answer 'entropy rises' is wrong: fixing T and P makes G, not total entropy, the criterion for spontaneous change.
For a binary liquid mixture in equilibrium with its vapor, the condition for phase equilibrium requires $\mu_i^L = \mu_i^V$ for each component i. If component A follows Raoult's law ( $P_A = x_A P_A^*$ ) and component B shows positive deviation from ideality in the liquid phase, which statement about the chemical potentials is correct?
Answer: $\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)$ and $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)$ where $\gamma_B > 1$ When you encounter phase equilibrium problems involving non-ideal solutions, remember that deviations from ideality affect how you express chemical potentials. The key is recognizing that each component's behavior must be treated according to its specific deviation pattern. Component A follows Raoult's law exactly, meaning it behaves ideally in the liquid phase. For ideal behavior, the chemical potential is $\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)$ , where the activity coefficient $\gamma_A = 1$ . Component B shows positive deviation, meaning stronger intermolecular repulsion or weaker attraction than in an ideal solution. This requires an activity coefficient $\gamma_B > 1$ to account for the non-ideal behavior, giving $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)$ . Answer B correctly captures this: ideal behavior for A with $\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)$ and non-ideal behavior for B with $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)$ where $\gamma_B > 1$ . Answer A incorrectly assumes both components are ideal just because they're in the same phase. Answer C reverses the components' behaviors and incorrectly suggests $\gamma_A < 1$ , which would indicate negative deviation for A. Answer D wrongly assumes that if one component shows positive deviation, both must—this isn't true since components can have different intermolecular interactions. Study tip: Each component's ideality is independent of the others. Always identify which specific components deviate and in which direction before writing chemical potential expressions.
Ice and water at 273 K are compressed isothermally; ice is less dense. What happens?
Answer: Ice melts to water At 273 K and 1 atm, ice and water coexist. Raising pressure favors the denser phase, and water is denser than ice, so equilibrium shifts toward liquid. Thus ice melts. The tempting error, water freezing to ice, gets the direction backwards: compression doesn't force the less dense solid to form.
For a pure substance at equal $T,P$ , which condition prevents net phase transfer?
Answer: Chemical potentials equal At fixed T and P, a pure substance's phases reach equilibrium only when their chemical potentials are equal. If one phase has a higher chemical potential, molecules move to the lower-potential phase, so net transfer continues until they match. Equal molar volumes or densities are physical similarities, not equilibrium conditions, and equal enthalpies don't prevent transfer either.
In a ternary system with components A, B, and C at constant temperature and pressure, the Gibbs-Duhem equation is $n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0$ . If $d\mu_A = 100 \text{ J/mol}$ , $d\mu_B = -50 \text{ J/mol}$ , and the system contains equal moles of all three components, what is $d\mu_C$ ?
Answer: -50 J/mol When you encounter the Gibbs-Duhem equation in a ternary system, you're dealing with a fundamental constraint on how chemical potentials can change. This equation tells us that the changes in chemical potentials of all components are interconnected - they cannot vary independently. The Gibbs-Duhem equation states that $n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0$ . Since the system contains equal moles of all three components, we can set $n_A = n_B = n_C = n$ . This simplifies our equation to: $n(d\mu_A + d\mu_B + d\mu_C) = 0$ Since $n \neq 0$ , we must have: $d\mu_A + d\mu_B + d\mu_C = 0$ Substituting the given values: $100 + (-50) + d\mu_C = 0$ $50 + d\mu_C = 0$ $d\mu_C = -50 \text{ J/mol}$ This confirms answer A is correct. Answer B (50 J/mol) represents a sign error - you might get this if you incorrectly rearrange the equation as $d\mu_C = d\mu_A + d\mu_B$ instead of $d\mu_C = -(d\mu_A + d\mu_B)$ . Answers C (-25 J/mol) and D (25 J/mol) likely come from incorrectly averaging the given values or mishandling the mole fraction relationships. Study tip: Always remember that the Gibbs-Duhem equation is a sum that equals zero. When components have equal amounts, you can factor out the mole terms and work with a simple algebraic sum of the chemical potential changes.
For the equilibrium $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ , the condition for phase equilibrium is $\mu_{\text{CaCO}_3}(s) = \mu_{\text{CaO}}(s) + \mu_{\text{CO}_2}(g)$ . If the pressure of CO₂ is doubled while maintaining the same temperature, what happens to the equilibrium?
Answer: The equilibrium shifts left because $\mu_{\text{CO}_2}$ increases by $RT \ln(2)$ , making the right side larger than the left side When you encounter equilibrium problems involving chemical potentials, focus on how changes in conditions affect the balance between reactants and products through their chemical potentials. At equilibrium, $\mu_{\text{CaCO}_3}(s) = \mu_{\text{CaO}}(s) + \mu_{\text{CO}_2}(g)$ . When CO₂ pressure doubles, the chemical potential of CO₂ increases by $RT \ln(2)$ (since $\mu_{\text{CO}_2} = \mu°_{\text{CO}_2} + RT \ln(P_{\text{CO}_2})$ ). This makes the right side of the equilibrium equation larger than the left side, disrupting the balance. To restore equilibrium, the reaction must shift left (toward reactants) to decrease $\mu_{\text{CO}_2}$ and reestablish equality. Choice A correctly identifies this mechanism - the equilibrium shifts left because $\mu_{\text{CO}_2}$ increases by $RT \ln(2)$ , making the right side exceed the left side. Choice B incorrectly applies Le Chatelier's principle backwards. Higher pressure doesn't favor gas formation; it opposes it. Choice C is wrong because while solid chemical potentials are independent of CO₂ pressure, the gas chemical potential isn't. The equilibrium condition involves all three species, so changes in $\mu_{\text{CO}_2}$ definitely affect equilibrium. Choice D states the correct direction but gives an oversimplified explanation. The real reason isn't just "pressure favors fewer gas molecules" - it's specifically about how pressure changes affect chemical potentials. Remember: In gas-solid equilibria, always consider how pressure changes affect the chemical potential of the gas phase through the $RT \ln(P)$ term, then determine which direction restores equilibrium balance.

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