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Atkins’ Physical Chemistry (11th Ed.) – Complete Practice Test (2026)
Questions with correct answers and Rationale 2026
Instructions: Detailed rationales are provided to reinforce advanced physical
chemistry concepts.
Domain 1: Thermodynamics (Laws, Entropy, Gibbs Free Energy, Equilibrium)
1. Which of the following is a correct mathematical statement of the First Law of
Thermodynamics for a closed system?
A) ΔU = q + w
B) ΔH = ΔU + PΔV
C) ΔG = ΔH - TΔS
D) ΔS = q_rev / T
Answer: A
Rationale: The First Law of Thermodynamics states that the change in internal
energy (ΔU) of a closed system is equal to the heat added to the system (q) plus
the work done on the system (w). The sign convention is crucial: q > 0 for heat
added to the system, w > 0 for work done on the system. Option B is the definition
of enthalpy, C is the Gibbs free energy equation, and D is the definition of entropy
for a reversible process.
2. A gas undergoes a reversible isothermal expansion from volume V₁ to V₂ at
constant temperature T. Which of the following expressions correctly represents
the work done by the gas?
A) w = -nRT ln(V₂/V₁)
B) w = -P_ext ΔV
C) w = nC_vΔT
D) w = -nRT ln(V₁/V₂)
Answer: A
Rationale: For a reversible isothermal expansion of an ideal gas, the work done by
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the gas is given by: w = -nRT ln(V₂/V₁). The negative sign indicates work done by
the system. If the gas expands (V₂ > V₁), the natural logarithm is positive, so w is
negative (work is done by the gas). Option B is for irreversible expansion against
constant external pressure.
3. A 2.00 mol sample of an ideal gas undergoes a reversible adiabatic expansion.
Which of the following is true for this process?
A) ΔU = 0
B) q = 0
C) ΔH = 0
D) ΔS = 0
Answer: B
Rationale: Adiabatic means "no heat transfer." Therefore, q = 0. For an adiabatic
expansion, the gas does work (w < 0) and its internal energy decreases (ΔU < 0), so
ΔU ≠ 0. Enthalpy (ΔH) also changes because ΔH = ΔU + Δ(PV). The process is
reversible, so ΔS = 0 for the system, but the question asks for the general truth,
which is q = 0.
4. The entropy change for a reversible isothermal expansion of an ideal gas is
given by which of the following expressions?
A) ΔS = nR ln(V₂/V₁)
B) ΔS = nC_v ln(T₂/T₁)
C) ΔS = ΔH / T
D) ΔS = q_rev / T = nR ln(V₂/V₁)
Answer: D
Rationale: For a reversible isothermal process, the entropy change is ΔS = q_rev /
T. For an ideal gas isothermal expansion, q_rev = nRT ln(V₂/V₁). Substituting this
into the entropy definition gives ΔS = nR ln(V₂/V₁).
5. A reaction has ΔH = -50 kJ/mol and ΔS = +100 J/mol·K at 300 K. What is the
value of ΔG for this reaction?
A) -20 kJ/mol
B) -80 kJ/mol
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C) +20 kJ/mol
D) +80 kJ/mol
Answer: B
Rationale: The Gibbs free energy equation is ΔG = ΔH - TΔS. Substituting the
values: ΔG = -50,000 J/mol - (300 K × 100 J/mol·K) = -50,000 - 30,000 = -80,000
J/mol = -80 kJ/mol. The negative ΔG indicates the reaction is spontaneous at this
temperature.
6. For a chemical reaction at equilibrium, which of the following must be true?
A) ΔG° = 0
B) ΔG = 0
C) ΔH = 0
D) ΔS = 0
Answer: B
Rationale: At equilibrium, the reaction quotient Q equals the equilibrium constant
K (Q = K), and the free energy change for the reaction under those conditions is
zero (ΔG = 0). ΔG° is the standard free energy change and is related to K by ΔG° = -
RT ln K, but it is not zero unless K = 1.
7. The relationship between the equilibrium constant K and the standard Gibbs
free energy change ΔG° is given by:
A) ΔG° = RT ln K
B) ΔG° = -RT ln K
C) ΔG° = -RT ln Q
D) ΔG° = ΔH° - TΔS° + RT ln K
Answer: B
Rationale: The correct relationship is ΔG° = -RT ln K. If ΔG° is negative, ln K is
positive, meaning K > 1 (products favored). If ΔG° is positive, K < 1 (reactants
favored).
8. The van't Hoff equation describes the temperature dependence of the
equilibrium constant. Which of the following is the integrated form of the van't
Hoff equation?
Atkins’ Physical Chemistry (11th Ed.) – Complete Practice Test (2026)
Questions with correct answers and Rationale 2026
Instructions: Detailed rationales are provided to reinforce advanced physical
chemistry concepts.
Domain 1: Thermodynamics (Laws, Entropy, Gibbs Free Energy, Equilibrium)
1. Which of the following is a correct mathematical statement of the First Law of
Thermodynamics for a closed system?
A) ΔU = q + w
B) ΔH = ΔU + PΔV
C) ΔG = ΔH - TΔS
D) ΔS = q_rev / T
Answer: A
Rationale: The First Law of Thermodynamics states that the change in internal
energy (ΔU) of a closed system is equal to the heat added to the system (q) plus
the work done on the system (w). The sign convention is crucial: q > 0 for heat
added to the system, w > 0 for work done on the system. Option B is the definition
of enthalpy, C is the Gibbs free energy equation, and D is the definition of entropy
for a reversible process.
2. A gas undergoes a reversible isothermal expansion from volume V₁ to V₂ at
constant temperature T. Which of the following expressions correctly represents
the work done by the gas?
A) w = -nRT ln(V₂/V₁)
B) w = -P_ext ΔV
C) w = nC_vΔT
D) w = -nRT ln(V₁/V₂)
Answer: A
Rationale: For a reversible isothermal expansion of an ideal gas, the work done by
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the gas is given by: w = -nRT ln(V₂/V₁). The negative sign indicates work done by
the system. If the gas expands (V₂ > V₁), the natural logarithm is positive, so w is
negative (work is done by the gas). Option B is for irreversible expansion against
constant external pressure.
3. A 2.00 mol sample of an ideal gas undergoes a reversible adiabatic expansion.
Which of the following is true for this process?
A) ΔU = 0
B) q = 0
C) ΔH = 0
D) ΔS = 0
Answer: B
Rationale: Adiabatic means "no heat transfer." Therefore, q = 0. For an adiabatic
expansion, the gas does work (w < 0) and its internal energy decreases (ΔU < 0), so
ΔU ≠ 0. Enthalpy (ΔH) also changes because ΔH = ΔU + Δ(PV). The process is
reversible, so ΔS = 0 for the system, but the question asks for the general truth,
which is q = 0.
4. The entropy change for a reversible isothermal expansion of an ideal gas is
given by which of the following expressions?
A) ΔS = nR ln(V₂/V₁)
B) ΔS = nC_v ln(T₂/T₁)
C) ΔS = ΔH / T
D) ΔS = q_rev / T = nR ln(V₂/V₁)
Answer: D
Rationale: For a reversible isothermal process, the entropy change is ΔS = q_rev /
T. For an ideal gas isothermal expansion, q_rev = nRT ln(V₂/V₁). Substituting this
into the entropy definition gives ΔS = nR ln(V₂/V₁).
5. A reaction has ΔH = -50 kJ/mol and ΔS = +100 J/mol·K at 300 K. What is the
value of ΔG for this reaction?
A) -20 kJ/mol
B) -80 kJ/mol
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C) +20 kJ/mol
D) +80 kJ/mol
Answer: B
Rationale: The Gibbs free energy equation is ΔG = ΔH - TΔS. Substituting the
values: ΔG = -50,000 J/mol - (300 K × 100 J/mol·K) = -50,000 - 30,000 = -80,000
J/mol = -80 kJ/mol. The negative ΔG indicates the reaction is spontaneous at this
temperature.
6. For a chemical reaction at equilibrium, which of the following must be true?
A) ΔG° = 0
B) ΔG = 0
C) ΔH = 0
D) ΔS = 0
Answer: B
Rationale: At equilibrium, the reaction quotient Q equals the equilibrium constant
K (Q = K), and the free energy change for the reaction under those conditions is
zero (ΔG = 0). ΔG° is the standard free energy change and is related to K by ΔG° = -
RT ln K, but it is not zero unless K = 1.
7. The relationship between the equilibrium constant K and the standard Gibbs
free energy change ΔG° is given by:
A) ΔG° = RT ln K
B) ΔG° = -RT ln K
C) ΔG° = -RT ln Q
D) ΔG° = ΔH° - TΔS° + RT ln K
Answer: B
Rationale: The correct relationship is ΔG° = -RT ln K. If ΔG° is negative, ln K is
positive, meaning K > 1 (products favored). If ΔG° is positive, K < 1 (reactants
favored).
8. The van't Hoff equation describes the temperature dependence of the
equilibrium constant. Which of the following is the integrated form of the van't
Hoff equation?
