Entropy Change for Ideal Gases
1 / 10
One mole of an ideal gas (γ = 1.4) is compressed reversibly from 1 bar to 2 bar at a constant temperature of 300 K. The entropy change of the gas is:
-5.76 J/K
5.76 J/K
0 J/K
-11.52 J/K
For a reversible isothermal compression, ΔS = nR ln(V₂/V₁). Since V ∝ 1/P for an isothermal process, ΔS = nR ln(P₁/P₂). Given n = 1 mol, R = 8.314 J/mol·K, P₁ = 1 bar, P₂ = 2 bar:
ΔS = 1 × 8.314 × ln(1/2) = 8.314 × (-0.693) ≈ -5.76 J/K.
2 / 10
For an ideal gas undergoing a reversible process, the entropy change is zero when:
The process is isothermal
The process is adiabatic and reversible
The process is isobaric
The process is isochoric
A reversible adiabatic process is isentropic, meaning ΔS = 0. Other processes (isothermal, isobaric, isochoric) generally result in non-zero entropy changes unless specific conditions balance the terms.
3 / 10
An ideal gas undergoes a free expansion (no external work, no heat transfer). The entropy change of the gas is:
Zero
Positive
Negative
Cannot be determined
Free expansion is an irreversible process for an ideal gas. The entropy change is ΔS = nR ln(V₂/V₁), which is positive since V₂ > V₁, even though no heat is transferred.
4 / 10
The entropy change of an ideal gas during a constant-volume process is given by:
ΔS = nCₚ ln(T₂/T₁)
ΔS = nCᵥ ln(T₂/T₁)
ΔS = nR ln(V₂/V₁)
ΔS = nR ln(P₂/P₁)
For a constant-volume process, ΔS = nCᵥ ln(T₂/T₁), as derived from the differential form dS = Cᵥ dT/T for constant volume.
5 / 10
Two moles of an ideal gas (R = 8.314 J/mol·K) expand isothermally and reversibly from 1 L to 10 L at 300 K. The entropy change of the gas is:
38.29 J/K
19.15 J/K
57.44 J/K
For a reversible isothermal expansion, ΔS = nR ln(V₂/V₁). Given n = 2 mol, R = 8.314 J/mol·K, V₂/V₁ = 10/1:
ΔS = 2 × 8.314 × ln(10) ≈ 2 × 8.314 × 2.303 = 38.29 J/K.
6 / 10
An ideal gas undergoes a reversible polytropic process with polytropic index n = 1.5. The entropy change of the gas will be:
Always zero
Always positive
Always negative
Depends on the initial and final states
For a polytropic process (P Vⁿ = constant), the entropy change depends on the initial and final temperatures and pressures. It can be calculated as ΔS = nCₚ ln(T₂/T₁) – nR ln(P₂/P₁), which may be positive, negative, or zero depending on the states.
7 / 10
One mole of an ideal gas (Cₚ = 29.1 J/mol·K, Cᵥ = 20.8 J/mol·K) is heated from 300 K to 400 K at constant pressure. The entropy change of the gas is:
6.93 J/K
8.30 J/K
9.67 J/K
For a constant-pressure process, ΔS = nCₚ ln(T₂/T₁). Given n = 1 mol, Cₚ = 29.1 J/mol·K, T₂ = 400 K, T₁ = 300 K:
ΔS = 1 × 29.1 × ln(400/300) = 29.1 × ln(1.333) ≈ 29.1 × 0.285 = 8.30 J/K.
8 / 10
The entropy change of an ideal gas undergoing a reversible process can be expressed as:
ΔS = nCₚ ln(V₂/V₁) + nCᵥ ln(T₂/T₁)
ΔS = nCₚ ln(T₂/T₁) – nR ln(P₂/P₁)
ΔS = nCᵥ ln(P₂/P₁) + nR ln(V₂/V₁)
ΔS = nR ln(T₂/T₁) + nCₚ ln(V₂/V₁)
For an ideal gas, the entropy change can be written as ΔS = nCₚ ln(T₂/T₁) – nR ln(P₂/P₁), derived from the differential form dS = Cₚ dT/T – R dP/P.
9 / 10
An ideal gas undergoes a reversible adiabatic process. The entropy change of the gas is
Depends on the final state
A reversible adiabatic process is isentropic, meaning the entropy change of the system (ΔS) is zero.
10 / 10
For an ideal gas undergoing a reversible isothermal expansion, the entropy change of the system is given by:
ΔS = nR ln(T₂/T₁)
ΔS = 0
For a reversible isothermal process, the temperature is constant (T₂ = T₁), and the entropy change for an ideal gas is ΔS = nR ln(V₂/V₁), where n is the number of moles, R is the universal gas constant, and V₂/V₁ is the volume ratio.
Your score is
The average score is 0%
Restart quiz
No products in the cart.
