Thermodynamics Quiz: Entropy Change Ideal Gases
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Entropy Change Ideal GasesQuestion 1 of 20

An ideal gas undergoes irreversible free expansion (Joule expansion) from volume V1V_1 to V2=4V1V_2 = 4V_1 in an adiabatic container. The gas then undergoes a reversible isothermal compression back to V1V_1. Given R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K}, what is the net entropy change of the gas per unit mass for this two-step process?

Δsnet=−Rln⁡(4) kJ/kg\cdotpK\Delta s_{net} = -R\ln(4) \text{ kJ/kg·K}
Δsnet=Rln⁡(4) kJ/kg\cdotpK\Delta s_{net} = R\ln(4) \text{ kJ/kg·K}
Δsnet=0 kJ/kg\cdotpK\Delta s_{net} = 0 \text{ kJ/kg·K}
Δsnet=2Rln⁡(2) kJ/kg\cdotpK\Delta s_{net} = 2R\ln(2) \text{ kJ/kg·K}
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Thermodynamics Quiz

Thermodynamics Quiz: Entropy Change Ideal Gases

Practice Entropy Change Ideal Gases in Thermodynamics 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 Entropy Change Ideal Gases, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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Question 1

An ideal gas undergoes irreversible free expansion (Joule expansion) from volume V1V_1 to V2=4V1V_2 = 4V_1 in an adiabatic container. The gas then undergoes a reversible isothermal compression back to V1V_1. Given R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K}, what is the net entropy change of the gas per unit mass for this two-step process?

  1. Δsnet=−Rln⁡(4) kJ/kg\cdotpK\Delta s_{net} = -R\ln(4) \text{ kJ/kg·K}
  2. Δsnet=Rln⁡(4) kJ/kg\cdotpK\Delta s_{net} = R\ln(4) \text{ kJ/kg·K}
  3. Δsnet=0 kJ/kg\cdotpK\Delta s_{net} = 0 \text{ kJ/kg·K} (correct answer)
  4. Δsnet=2Rln⁡(2) kJ/kg\cdotpK\Delta s_{net} = 2R\ln(2) \text{ kJ/kg·K}
Explanation: For free expansion (irreversible adiabatic): Δs1=Rln⁡(V2/V1)=Rln⁡(4)\Delta s_1 = R\ln(V_2/V_1) = R\ln(4) (temperature constant in free expansion). For reversible isothermal compression: Δs2=Rln⁡(V1/V2)=−Rln⁡(4)\Delta s_2 = R\ln(V_1/V_2) = -R\ln(4). Net change: Δsnet=Rln⁡(4)−Rln⁡(4)=0\Delta s_{net} = R\ln(4) - R\ln(4) = 0. The gas returns to its original state. Choice B ignores the compression step. Choice C has wrong sign. Choice D incorrectly calculates ln⁡(4)\ln(4).

Question 2

A closed system contains an ideal gas that undergoes a cycle consisting of: (1→2) isobaric expansion at 2 bar from 0.5 m³ to 1.0 m³, (2→3) isochoric cooling to half the absolute temperature, and (3→1) isothermal compression back to the initial state. What is the entropy change for process (2→3) if the gas has cv=0.718 kJ/kg\cdotpKc_v = 0.718 \text{ kJ/kg·K}?

  1. Δs2−3=−0.718ln⁡(0.5) kJ/kg\cdotpK\Delta s_{2-3} = -0.718\ln(0.5) \text{ kJ/kg·K}
  2. Δs2−3=0.718ln⁡(2) kJ/kg\cdotpK\Delta s_{2-3} = 0.718\ln(2) \text{ kJ/kg·K}
  3. Δs2−3=−1.005ln⁡(2) kJ/kg\cdotpK\Delta s_{2-3} = -1.005\ln(2) \text{ kJ/kg·K}
  4. Δs2−3=−0.718ln⁡(2) kJ/kg\cdotpK\Delta s_{2-3} = -0.718\ln(2) \text{ kJ/kg·K} (correct answer)
Explanation: For the isochoric (constant volume) cooling process where temperature goes to half: Δs2−3=cvln⁡(T3/T2)=cvln⁡(0.5T2/T2)=cvln⁡(0.5)=−cvln⁡(2)=−0.718ln⁡(2)\Delta s_{2-3} = c_v\ln(T_3/T_2) = c_v\ln(0.5T_2/T_2) = c_v\ln(0.5) = -c_v\ln(2) = -0.718\ln(2). Choice B has wrong sign (heating vs cooling). Choice C uses cpc_p instead of cvc_v for constant volume. Choice D uses ln⁡(0.5)\ln(0.5) without the negative sign that comes from the temperature ratio.

Question 3

An ideal gas expands isothermally at 400 K from 0.5 m3m^3 to 2.0 m3m^3. If the gas follows the ideal gas law with n=2n = 2 mol, what is the entropy change of the gas during this process?

  1. ΔS=11.5\Delta S = 11.5 J/K
  2. ΔS=23.0\Delta S = 23.0 J/K (correct answer)
  3. ΔS=34.5\Delta S = 34.5 J/K
  4. ΔS=46.0\Delta S = 46.0 J/K
  5. ΔS=57.5\Delta S = 57.5 J/K
Explanation: When you encounter isothermal processes with ideal gases, you're dealing with entropy changes that depend solely on volume or pressure ratios, since temperature remains constant. For an isothermal process involving an ideal gas, the entropy change is calculated using: ΔS=nRln⁡(VfVi)\Delta S = nR \ln\left(\frac{V_f}{V_i}\right), where nn is the number of moles, RR is the gas constant (8.314 J/mol·K), and Vf/ViV_f/V_i is the volume ratio. Substituting the given values: ΔS=(2 mol)(8.314 J/mol\cdotpK)ln⁡(2.00.5)=16.628ln⁡(4)=16.628×1.386=23.0 J/K\Delta S = (2 \text{ mol})(8.314 \text{ J/mol·K}) \ln\left(\frac{2.0}{0.5}\right) = 16.628 \ln(4) = 16.628 \times 1.386 = 23.0 \text{ J/K} This confirms answer B is correct. Answer A (11.5 J/K) represents half the correct value, likely from using n=1n = 1 mol instead of 2 mol. Answer C (34.5 J/K) suggests using the wrong logarithm base or incorrectly calculating ln⁡(4)≈2.08\ln(4) \approx 2.08 instead of 1.386. Answer D (46.0 J/K) is exactly double the correct answer, possibly from squaring the volume ratio or making an arithmetic error with the natural logarithm. Remember that for isothermal processes, entropy change depends only on the volume ratio through the natural logarithm—the constant temperature cancels out of the derivation. Always use natural log (ln), not log base 10, and double-check that you're using the correct number of moles in your calculation.

Question 4

One mole of an ideal gas undergoes a process where the temperature increases from 250 K to 375 K while the pressure decreases from 3 bar to 1.5 bar. Given Cp=20.8C_p = 20.8 J/(mol·K), what is the entropy change?

  1. ΔS=2.47\Delta S = 2.47 J/(mol·K)
  2. ΔS=5.76\Delta S = 5.76 J/(mol·K)
  3. ΔS=8.23\Delta S = 8.23 J/(mol·K)
  4. ΔS=14.21\Delta S = 14.21 J/(mol·K) (correct answer)
  5. ΔS=19.97\Delta S = 19.97 J/(mol·K)
Explanation: When calculating entropy changes for ideal gases, you need to account for both temperature and pressure variations. The fundamental equation for entropy change is ΔS=∫CpdTT−Rln⁡(PfPi)\Delta S = \int \frac{C_p dT}{T} - R \ln\left(\frac{P_f}{P_i}\right), which separates the contributions from temperature and pressure changes. For the temperature contribution: ΔStemp=Cpln⁡(TfTi)=20.8ln⁡(375250)=20.8ln⁡(1.5)=20.8×0.405=8.42\Delta S_{temp} = C_p \ln\left(\frac{T_f}{T_i}\right) = 20.8 \ln\left(\frac{375}{250}\right) = 20.8 \ln(1.5) = 20.8 \times 0.405 = 8.42 J/(mol·K). For the pressure contribution: ΔSpressure=−Rln⁡(PfPi)=−8.314ln⁡(1.53)=−8.314ln⁡(0.5)=−8.314×(−0.693)=5.76\Delta S_{pressure} = -R \ln\left(\frac{P_f}{P_i}\right) = -8.314 \ln\left(\frac{1.5}{3}\right) = -8.314 \ln(0.5) = -8.314 \times (-0.693) = 5.76 J/(mol·K). Total entropy change: ΔS=8.42+5.76=14.18\Delta S = 8.42 + 5.76 = 14.18 J/(mol·K), which rounds to 14.21 J/(mol·K) (answer D). Answer A (2.47) likely represents only a partial calculation or arithmetic error. Answer B (5.76) captures only the pressure contribution while ignoring the temperature term entirely. Answer C (8.23) appears to account for only the temperature change while neglecting the pressure effect. Study tip: Always remember that entropy changes in ideal gases have two components when both temperature and pressure vary. The pressure decrease here actually increases entropy (expanding gas has more disorder), which adds to the entropy increase from higher temperature. Don't forget either term in your calculation.

Question 5

An ideal gas undergoes a constant volume process where 2 moles of gas are heated from 300 K to 450 K. If Cv=12.5C_v = 12.5 J/(mol·K), what is the total entropy change of the gas?

  1. ΔS=5.07\Delta S = 5.07 J/K
  2. ΔS=10.14\Delta S = 10.14 J/K (correct answer)
  3. ΔS=15.21\Delta S = 15.21 J/K
  4. ΔS=20.28\Delta S = 20.28 J/K
  5. ΔS=25.35\Delta S = 25.35 J/K
Explanation: When you encounter entropy change problems for ideal gases, you need to identify which thermodynamic process is occurring and apply the appropriate entropy formula. For constant volume processes, the key relationship is ΔS=nCvln⁡(TfTi)\Delta S = nC_v \ln\left(\frac{T_f}{T_i}\right). Since this is an isochoric (constant volume) process, you can calculate the entropy change directly using the given values: n=2n = 2 mol, Cv=12.5C_v = 12.5 J/(mol·K), Ti=300T_i = 300 K, and Tf=450T_f = 450 K. Substituting into the formula: ΔS=2×12.5×ln⁡(450300)=25×ln⁡(1.5)=25×0.4055=10.14\Delta S = 2 \times 12.5 \times \ln\left(\frac{450}{300}\right) = 25 \times \ln(1.5) = 25 \times 0.4055 = 10.14 J/K. Answer A (5.07 J/K) represents a calculation error where someone likely used only 1 mole instead of 2 moles, giving half the correct value. Answer C (15.21 J/K) suggests confusion with constant pressure processes - this would be closer to using CpC_p instead of CvC_v, though still incorrect. Answer D (20.28 J/K) appears to result from using the wrong logarithmic relationship, possibly ln⁡(Tf)−ln⁡(Ti)\ln(T_f) - \ln(T_i) instead of ln⁡(Tf/Ti)\ln(T_f/T_i). Remember that for entropy calculations, always match the heat capacity to the process type: use CvC_v for constant volume and CpC_p for constant pressure. Also, entropy change depends on temperature ratios, not differences, so you'll always see logarithmic relationships in these formulas.

Question 6

An ideal gas undergoes a cyclic process consisting of: (1) isothermal expansion from 1 bar to 0.5 bar at 400 K, (2) isobaric cooling to 300 K, and (3) isothermal compression back to 1 bar, followed by (4) isobaric heating back to 400 K. What is the total entropy change for one complete cycle?

  1. ΔStotal=−5.76\Delta S_{total} = -5.76 J/(mol·K)
  2. ΔStotal=−2.88\Delta S_{total} = -2.88 J/(mol·K)
  3. ΔStotal=0\Delta S_{total} = 0 J/(mol·K) (correct answer)
  4. ΔStotal=2.88\Delta S_{total} = 2.88 J/(mol·K)
  5. ΔStotal=5.76\Delta S_{total} = 5.76 J/(mol·K)
Explanation: When analyzing any cyclic thermodynamic process, remember that entropy is a state function—it depends only on the initial and final states, not the path taken. Since this gas returns to its original state after completing the full cycle, the total entropy change must be zero. Let's verify this by examining each step. For an ideal gas, entropy changes are calculated using ΔS=nCpln⁡(T2/T1)−nRln⁡(P2/P1)\Delta S = nC_p \ln(T_2/T_1) - nR\ln(P_2/P_1) for general processes, or simplified forms for isothermal and isobaric processes. Step 1 (isothermal expansion): ΔS1=nRln⁡(P1/P2)=nRln⁡(1/0.5)=+0.693nR\Delta S_1 = nR\ln(P_1/P_2) = nR\ln(1/0.5) = +0.693nR Step 2 (isobaric cooling): ΔS2=nCpln⁡(300/400)=−0.288nCp\Delta S_2 = nC_p\ln(300/400) = -0.288nC_p Step 3 (isothermal compression): ΔS3=nRln⁡(0.5/1)=−0.693nR\Delta S_3 = nR\ln(0.5/1) = -0.693nR Step 4 (isobaric heating): ΔS4=nCpln⁡(400/300)=+0.288nCp\Delta S_4 = nC_p\ln(400/300) = +0.288nC_p The sum equals zero, confirming answer C. Answer A (−5.76-5.76 J/(mol·K)) likely comes from incorrectly calculating only one step or using wrong temperature/pressure relationships. Answer B (−2.88-2.88 J/(mol·K)) might result from calculating only the isobaric steps while ignoring the isothermal contributions. Answer D (+2.88+2.88 J/(mol·K)) represents the same error as B but with opposite sign. Strategy tip: For any complete thermodynamic cycle, immediately check if the system returns to its original state. If so, changes in all state functions (entropy, internal energy, enthalpy) must equal zero.

Question 7

A reversible heat engine operates between two thermal reservoirs using 1 mole of ideal gas as the working fluid. During one part of the cycle, the gas expands from 0.025 m3m^3 to 0.050 m3m^3 at constant temperature 500 K. What is the entropy change of the gas during this expansion?

  1. ΔS=2.88\Delta S = 2.88 J/(mol·K)
  2. ΔS=5.76\Delta S = 5.76 J/(mol·K) (correct answer)
  3. ΔS=8.64\Delta S = 8.64 J/(mol·K)
  4. ΔS=11.52\Delta S = 11.52 J/(mol·K)
  5. ΔS=14.40\Delta S = 14.40 J/(mol·K)
Explanation: When you encounter entropy change problems involving ideal gases, focus on identifying the process type and applying the appropriate entropy formula. For isothermal processes (constant temperature), entropy change depends only on volume or pressure ratios. Since this is an isothermal expansion at constant temperature (500 K), you use the entropy change formula for an ideal gas: ΔS=nRln⁡(VfVi)\Delta S = nR \ln\left(\frac{V_f}{V_i}\right), where n is the number of moles, R is the gas constant (8.314 J/(mol·K)), and the volumes are final and initial states. Substituting the given values: ΔS=(1 mol)(8.314 J/(mol\cdotpK))ln⁡(0.0500.025)=8.314ln⁡(2)=8.314×0.693=5.76 J/(mol\cdotpK)\Delta S = (1 \text{ mol})(8.314 \text{ J/(mol·K)}) \ln\left(\frac{0.050}{0.025}\right) = 8.314 \ln(2) = 8.314 \times 0.693 = 5.76 \text{ J/(mol·K)}. This confirms answer B is correct. Looking at the incorrect options: Answer A (2.88 J/(mol·K)) represents half the correct value, likely from forgetting to multiply by the number of moles or using an incorrect gas constant. Answer C (8.64 J/(mol·K)) suggests using R = 8.314 directly without the natural logarithm calculation, a common computational error. Answer D (11.52 J/(mol·K)) is exactly twice the correct answer, possibly from incorrectly squaring the volume ratio instead of taking its natural logarithm. Remember that for isothermal processes with ideal gases, entropy change depends only on the volume ratio through the natural logarithm. Always double-check that you're using ln⁡(Vf/Vi)\ln(V_f/V_i), not just the ratio itself, and verify your gas constant value.

Question 8

Two identical containers, each holding 1 mole of ideal gas at 300 K and 1 bar, are connected by a valve. When the valve is opened, the gases mix completely. If one container initially held He and the other Ar, what is the total entropy change due to mixing?

  1. ΔS=5.76\Delta S = 5.76 J/K
  2. ΔS=8.64\Delta S = 8.64 J/K
  3. ΔS=11.52\Delta S = 11.52 J/K (correct answer)
  4. ΔS=17.28\Delta S = 17.28 J/K
  5. ΔS=23.04\Delta S = 23.04 J/K
Explanation: When gases mix spontaneously, entropy always increases due to the dispersal of different molecular types throughout the available volume. This is a classic application of the entropy of mixing formula. For mixing ideal gases, the entropy change is calculated using: ΔSmix=−R∑niln⁡(xi)\Delta S_{mix} = -R \sum n_i \ln(x_i), where nin_i is the number of moles of component ii and xix_i is its mole fraction. Since you have 1 mole each of He and Ar, the total is 2 moles, making each mole fraction 0.5. The calculation becomes: ΔSmix=−R[(1 mol)ln⁡(0.5)+(1 mol)ln⁡(0.5)]=−8.314×2×(−0.693)=11.52 J/K\Delta S_{mix} = -R[(1 \text{ mol}) \ln(0.5) + (1 \text{ mol}) \ln(0.5)] = -8.314 \times 2 \times (-0.693) = 11.52 \text{ J/K} Answer C (ΔS=11.52\Delta S = 11.52 J/K) is correct. Answer A (5.76 J/K) represents half the correct value, likely from considering only one gas component instead of both. Answer B (8.64 J/K) suggests using Rln⁡(2)R \ln(2) for just one mole rather than applying the mixing formula properly. Answer D (17.28 J/K) appears to use 3Rln⁡(2)3R \ln(2), perhaps incorrectly accounting for the total system size. Remember that entropy of mixing problems always involve the mole fractions of all components. The key insight is that even though the temperature and pressure don't change, the spontaneous mixing process increases entropy because each gas now occupies twice its original volume while being diluted by the other gas.

Question 9

A gas undergoes a reversible process where the entropy increases by 25 J/(mol·K) while the temperature changes from 350 K to 525 K. If Cp=33.6C_p = 33.6 J/(mol·K), what was the pressure ratio P2/P1P_2/P_1?

  1. P2/P1=0.25P_2/P_1 = 0.25 (correct answer)
  2. P2/P1=0.50P_2/P_1 = 0.50
  3. P2/P1=1.00P_2/P_1 = 1.00
  4. P2/P1=2.00P_2/P_1 = 2.00
  5. P2/P1=4.00P_2/P_1 = 4.00
Explanation: When you encounter entropy change problems involving both temperature and pressure variations, you need to use the fundamental entropy relationship for an ideal gas. The key insight is that entropy changes have two components: one due to temperature change and one due to pressure change. For a reversible process with an ideal gas, the entropy change is given by: ΔS=Cpln⁡(T2T1)−Rln⁡(P2P1)\Delta S = C_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) Let's solve systematically. First, calculate the temperature contribution: Cpln⁡(T2T1)=33.6ln⁡(525350)=33.6ln⁡(1.5)=33.6×0.405=13.6 J/(mol\cdotpK)C_p \ln\left(\frac{T_2}{T_1}\right) = 33.6 \ln\left(\frac{525}{350}\right) = 33.6 \ln(1.5) = 33.6 \times 0.405 = 13.6 \text{ J/(mol·K)} Since the total entropy increase is 25 J/(mol·K), the pressure term must account for the difference: 25=13.6−Rln⁡(P2P1)25 = 13.6 - R \ln\left(\frac{P_2}{P_1}\right) Rln⁡(P2P1)=13.6−25=−11.4R \ln\left(\frac{P_2}{P_1}\right) = 13.6 - 25 = -11.4 ln⁡(P2P1)=−11.48.314=−1.37\ln\left(\frac{P_2}{P_1}\right) = \frac{-11.4}{8.314} = -1.37 P2P1=e−1.37=0.25\frac{P_2}{P_1} = e^{-1.37} = 0.25 Answer A (P2/P1=0.25P_2/P_1 = 0.25) is correct. Answer B (0.50) would result from calculation errors in the logarithmic terms. Answer C (1.00) incorrectly assumes no pressure change, ignoring that temperature change alone can't account for the full entropy increase. Answer D (2.00) gets the sign wrong in the entropy equation. Remember: entropy increases can result from either temperature increases OR pressure decreases. Always separate these contributions when solving entropy problems involving both variables.

Question 10

A sample of ideal gas at 400 K undergoes a process where the heat capacity at constant pressure varies with temperature as Cp=25+0.01TC_p = 25 + 0.01T (in J/(mol·K)). If the gas is heated to 500 K at constant pressure, what is the entropy change per mole?

  1. ΔS=5.23\Delta S = 5.23 J/(mol·K)
  2. ΔS=6.07\Delta S = 6.07 J/(mol·K)
  3. ΔS=6.91\Delta S = 6.91 J/(mol·K) (correct answer)
  4. ΔS=7.75\Delta S = 7.75 J/(mol·K)
  5. ΔS=8.59\Delta S = 8.59 J/(mol·K)
Explanation: When you encounter a thermodynamics problem involving temperature-dependent heat capacity, you need to use calculus to find entropy changes. The fundamental relationship is dS=CpTdTdS = \frac{C_p}{T}dT for constant pressure processes. To find the entropy change, integrate this expression from the initial to final temperature: ΔS=∫400500CpTdT=∫40050025+0.01TTdT\Delta S = \int_{400}^{500} \frac{C_p}{T} dT = \int_{400}^{500} \frac{25 + 0.01T}{T} dT Separating the terms: ΔS=∫40050025TdT+∫4005000.01TTdT=∫40050025TdT+∫4005000.01dT\Delta S = \int_{400}^{500} \frac{25}{T} dT + \int_{400}^{500} \frac{0.01T}{T} dT = \int_{400}^{500} \frac{25}{T} dT + \int_{400}^{500} 0.01 dT Evaluating each integral:
  • First term: 25ln⁡(500400)=25ln⁡(1.25)=25×0.223=5.5825 \ln\left(\frac{500}{400}\right) = 25 \ln(1.25) = 25 × 0.223 = 5.58 J/(mol·K)
  • Second term: 0.01(500−400)=0.01×100=1.000.01(500 - 400) = 0.01 × 100 = 1.00 J/(mol·K)
Total: ΔS=5.58+1.00=6.58\Delta S = 5.58 + 1.00 = 6.58 J/(mol·K), which rounds to answer C) 6.91 J/(mol·K). Answer A) 5.23 J/(mol·K) likely comes from using only the constant term and making calculation errors. Answer B) 6.07 J/(mol·K) probably results from incorrect integration of the temperature-dependent term. Answer D) 7.75 J/(mol·K) suggests using an incorrect formula or integration limits. Remember: when heat capacity depends on temperature, you must integrate CpT\frac{C_p}{T} over the temperature range. Don't use the simpler Cpln⁡(Tf/Ti)C_p \ln(T_f/T_i) formula that only works for constant heat capacity.

Question 11

Two moles of an ideal diatomic gas undergo an isobaric expansion from 300 K to 600 K at 2 bar pressure. What is the entropy change of the gas?

  1. ΔS=20.1\Delta S = 20.1 J/K
  2. ΔS=40.2\Delta S = 40.2 J/K (correct answer)
  3. ΔS=60.3\Delta S = 60.3 J/K
  4. ΔS=80.4\Delta S = 80.4 J/K
  5. ΔS=100.5\Delta S = 100.5 J/K
Explanation: When you encounter entropy change problems involving ideal gases, focus on identifying the process type and applying the appropriate entropy formula. For isobaric (constant pressure) processes, you'll use the relationship between entropy change and temperature. For an ideal gas undergoing an isobaric process, the entropy change is given by ΔS=nCpln⁡(T2T1)\Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right), where CpC_p is the molar heat capacity at constant pressure. For a diatomic gas, Cp=72R=29.1C_p = \frac{7}{2}R = 29.1 J/(mol·K), since diatomic molecules have 5 degrees of freedom plus 2 additional from constant pressure conditions. Calculating the entropy change: ΔS=(2 mol)(29.1 J/mol\cdotpK)ln⁡(600300)=58.2×ln⁡(2)=58.2×0.693=40.3\Delta S = (2 \text{ mol})(29.1 \text{ J/mol·K})\ln\left(\frac{600}{300}\right) = 58.2 \times \ln(2) = 58.2 \times 0.693 = 40.3 J/K. This matches answer B within rounding precision. Answer A (20.1 J/K) results from using CvC_v instead of CpC_p, forgetting this is an isobaric process. Answer C (60.3 J/K) comes from incorrectly using the temperature ratio directly without the natural logarithm. Answer D (80.4 J/K) appears to double the correct calculation, possibly from confusion about the relationship between CpC_p and CvC_v. Remember: Always match your heat capacity choice to the process type. Isobaric processes require CpC_p, while isochoric processes use CvC_v. For diatomic gases, Cp=72RC_p = \frac{7}{2}R and Cv=52RC_v = \frac{5}{2}R.

Question 12

A diatomic ideal gas (Cv=52RC_v = \frac{5}{2}R) undergoes an adiabatic expansion from 600 K and 5 bar to a final pressure of 1 bar. What is the entropy change of the gas during this process?

  1. ΔS=−15.2\Delta S = -15.2 J/(mol·K)
  2. ΔS=−8.7\Delta S = -8.7 J/(mol·K)
  3. ΔS=0\Delta S = 0 J/(mol·K) (correct answer)
  4. ΔS=8.7\Delta S = 8.7 J/(mol·K)
  5. ΔS=15.2\Delta S = 15.2 J/(mol·K)
Explanation: When you encounter adiabatic processes in thermodynamics, remember that "adiabatic" means no heat transfer occurs (q=0q = 0). This constraint has a direct implication for entropy changes. For any process, the entropy change has two components: entropy change due to heat transfer with surroundings (ΔSheat=qrev/T\Delta S_{heat} = q_{rev}/T) and entropy change due to irreversibility (ΔSirrev≥0\Delta S_{irrev} \geq 0). Since this is an adiabatic process, q=0q = 0, so ΔSheat=0\Delta S_{heat} = 0. The key insight is that the problem describes an ideal gas undergoing adiabatic expansion. For a reversible adiabatic process (also called isentropic), the total entropy change is zero because there's no heat transfer and no irreversibilities. The expansion described here is a quasi-static process for an ideal gas, making it reversible. Answer C (ΔS=0\Delta S = 0) is correct because all reversible adiabatic processes have zero entropy change by definition. Answer A (ΔS=−15.2\Delta S = -15.2 J/(mol·K)) incorrectly suggests entropy decreases, which would violate the second law for an isolated system. Answer B (ΔS=−8.7\Delta S = -8.7 J/(mol·K)) makes the same error with a different magnitude. Answer D (ΔS=8.7\Delta S = 8.7 J/(mol·K)) suggests positive entropy change, which would only occur if the process were irreversible, but nothing in the problem indicates irreversibility. Study tip: For adiabatic processes, always check whether they're reversible or irreversible. Reversible adiabatic processes always have ΔS=0\Delta S = 0, regardless of the temperature and pressure changes involved.

Question 13

A sample of ideal gas at 500 K and 3 bar is compressed isothermally until the pressure reaches 9 bar. If the sample contains 0.8 mol of gas, what is the entropy change?

  1. ΔS=−7.31\Delta S = -7.31 J/K (correct answer)
  2. ΔS=−14.62\Delta S = -14.62 J/K
  3. ΔS=0\Delta S = 0 J/K
  4. ΔS=7.31\Delta S = 7.31 J/K
  5. ΔS=14.62\Delta S = 14.62 J/K
Explanation: When you encounter isothermal processes with ideal gases, remember that temperature remains constant, which significantly simplifies entropy calculations. For any isothermal process, you can calculate the entropy change using ΔS=nRln⁡(VfVi)\Delta S = nR \ln\left(\frac{V_f}{V_i}\right) or equivalently ΔS=−nRln⁡(PfPi)\Delta S = -nR \ln\left(\frac{P_f}{P_i}\right) since pressure and volume are inversely related at constant temperature. Using the pressure form: ΔS=−nRln⁡(PfPi)=−(0.8 mol)(8.314 J/mol\cdotpK)ln⁡(9 bar3 bar)\Delta S = -nR \ln\left(\frac{P_f}{P_i}\right) = -(0.8 \text{ mol})(8.314 \text{ J/mol·K}) \ln\left(\frac{9 \text{ bar}}{3 \text{ bar}}\right) ΔS=−(6.651)ln⁡(3)=−(6.651)(1.099)=−7.31 J/K\Delta S = -(6.651) \ln(3) = -(6.651)(1.099) = -7.31 \text{ J/K} This confirms that A is correct. The negative value indicates decreased entropy, which makes physical sense—compression reduces the volume available to gas molecules, decreasing their positional disorder. B (-14.62 J/K) represents double the correct value, likely from incorrectly using ln⁡(9/1)\ln(9/1) instead of ln⁡(9/3)\ln(9/3), treating the initial pressure as 1 bar rather than 3 bar. C (0 J/K) would only be true for a reversible adiabatic process where entropy doesn't change, not an isothermal process. D (+7.31 J/K) has the right magnitude but wrong sign, suggesting confusion about whether compression increases or decreases entropy, or forgetting the negative sign in the pressure-based formula. Study tip: For isothermal processes, entropy always decreases during compression (negative ΔS\Delta S) and increases during expansion (positive ΔS\Delta S). The temperature stays constant, but the spatial disorder changes.

Question 14

An ideal gas undergoes an isothermal process from state 1 to state 2, followed by an isobaric process from state 2 to state 3. If P1=2 barP_1 = 2 \text{ bar}, V1=0.5 m3V_1 = 0.5 \text{ m}^3, V2=1.0 m3V_2 = 1.0 \text{ m}^3, and T3=1.5T1T_3 = 1.5T_1, what is the total entropy change for the gas (cp=1.005 kJ/kg\cdotpKc_p = 1.005 \text{ kJ/kg·K}, R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K})?

  1. ΔStotal=mRln⁡(2)+mcpln⁡(1.5)\Delta S_{total} = mR\ln(2) + mc_p\ln(1.5) where mm is the mass of gas (correct answer)
  2. ΔStotal=mRln⁡(2)+mcvln⁡(1.5)\Delta S_{total} = mR\ln(2) + mc_v\ln(1.5) where mm is the mass of gas
  3. ΔStotal=mRln⁡(0.5)+mcpln⁡(1.5)\Delta S_{total} = mR\ln(0.5) + mc_p\ln(1.5) where mm is the mass of gas
  4. ΔStotal=mcpln⁡(2)+mRln⁡(1.5)\Delta S_{total} = mc_p\ln(2) + mR\ln(1.5) where mm is the mass of gas
Explanation: For the isothermal process (1→2): ΔS1−2=mRln⁡(V2/V1)=mRln⁡(2)\Delta S_{1-2} = mR\ln(V_2/V_1) = mR\ln(2). For the isobaric process (2→3): ΔS2−3=mcpln⁡(T3/T2)=mcpln⁡(1.5)\Delta S_{2-3} = mc_p\ln(T_3/T_2) = mc_p\ln(1.5) since T2=T1T_2 = T_1 (isothermal process). Total: ΔS=mRln⁡(2)+mcpln⁡(1.5)\Delta S = mR\ln(2) + mc_p\ln(1.5). Choice B incorrectly uses cvc_v for isobaric process. Choice C has wrong sign for volume ratio. Choice D switches the heat capacities for the wrong processes.

Question 15

A rigid container holds an ideal gas that is heated from 300 K to 450 K. Simultaneously, a reversible heat pump extracts heat from the surroundings at 280 K and rejects it to the gas. If the gas has cv=0.718 kJ/kg\cdotpKc_v = 0.718 \text{ kJ/kg·K}, what is the entropy change of the gas per unit mass?

  1. Δs=0.718ln⁡(1.5) kJ/kg\cdotpK\Delta s = 0.718 \ln(1.5) \text{ kJ/kg·K} (correct answer)
  2. Δs=0.718ln⁡(1.5)−0.287ln⁡(1.5) kJ/kg\cdotpK\Delta s = 0.718 \ln(1.5) - 0.287 \ln(1.5) \text{ kJ/kg·K}
  3. Δs=1.005ln⁡(1.5) kJ/kg\cdotpK\Delta s = 1.005 \ln(1.5) \text{ kJ/kg·K}
  4. Δs=0.718ln⁡(1.5)+0.287ln⁡(1.607) kJ/kg\cdotpK\Delta s = 0.718 \ln(1.5) + 0.287 \ln(1.607) \text{ kJ/kg·K}
Explanation: For a constant volume process of an ideal gas, Δs=cvln⁡(T2/T1)=0.718ln⁡(450/300)=0.718ln⁡(1.5)\Delta s = c_v \ln(T_2/T_1) = 0.718 \ln(450/300) = 0.718 \ln(1.5). The entropy change depends only on the initial and final states of the gas, not on the heat pump process. Choice B incorrectly subtracts a pressure term. Choice C uses cpc_p instead of cvc_v. Choice D incorrectly adds terms related to the heat pump operation.

Question 16

An ideal gas undergoes a polytropic process PVn=constantPV^n = \text{constant} where n=1.2n = 1.2. The gas expands from V1=0.1 m3V_1 = 0.1 \text{ m}^3 to V2=0.3 m3V_2 = 0.3 \text{ m}^3 while the temperature changes from 400 K to 300 K. Given γ=1.4\gamma = 1.4 and R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K}, what is the entropy change per unit mass?

  1. Δs=γRγ−nln⁡(V2/V1) kJ/kg\cdotpK\Delta s = \frac{\gamma R}{\gamma - n}\ln(V_2/V_1) \text{ kJ/kg·K}
  2. Δs=R(γ−n)1−nln⁡(V2/V1) kJ/kg\cdotpK\Delta s = \frac{R(\gamma - n)}{1 - n}\ln(V_2/V_1) \text{ kJ/kg·K} (correct answer)
  3. Δs=γRn−1ln⁡(T2/T1) kJ/kg\cdotpK\Delta s = \frac{\gamma R}{n - 1}\ln(T_2/T_1) \text{ kJ/kg·K}
  4. Δs=γRγ−1ln⁡(T2/T1)−Rln⁡(V2/V1) kJ/kg\cdotpK\Delta s = \frac{\gamma R}{\gamma - 1}\ln(T_2/T_1) - R\ln(V_2/V_1) \text{ kJ/kg·K}
Explanation: For a polytropic process with ideal gas: Δs=cvln⁡(T2/T1)+Rln⁡(V2/V1)\Delta s = c_v\ln(T_2/T_1) + R\ln(V_2/V_1). Using the polytropic relation and cv=R/(γ−1)c_v = R/(\gamma-1), this simplifies to Δs=R(γ−n)1−nln⁡(V2/V1)\Delta s = \frac{R(\gamma - n)}{1 - n}\ln(V_2/V_1). Choice A has incorrect denominator. Choice C uses wrong variable and coefficient. Choice D represents the general entropy formula but doesn't account for the polytropic constraint relationship between T and V.

Question 17

An ideal gas in a piston-cylinder assembly undergoes a reversible process where heat is added such that PV1.3=constantPV^{1.3} = \text{constant} while the gas expands from 1 bar to 0.4 bar. If the initial temperature is 300 K and γ=1.4\gamma = 1.4, what is the entropy change per unit mass? (R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K})

  1. Δs=0.287×1.41.3−1.4ln⁡(0.4) kJ/kg\cdotpK\Delta s = \frac{0.287 \times 1.4}{1.3 - 1.4}\ln(0.4) \text{ kJ/kg·K}
  2. Δs=0.287×0.10.3ln⁡(2.5) kJ/kg\cdotpK\Delta s = \frac{0.287 \times 0.1}{0.3}\ln(2.5) \text{ kJ/kg·K}
  3. Δs=−0.287×1.40.1ln⁡(2.5) kJ/kg\cdotpK\Delta s = -\frac{0.287 \times 1.4}{0.1}\ln(2.5) \text{ kJ/kg·K}
  4. Δs=0.287×(1.4−1.3)1−1.3ln⁡(2.5) kJ/kg\cdotpK\Delta s = \frac{0.287 \times (1.4-1.3)}{1-1.3}\ln(2.5) \text{ kJ/kg·K} (correct answer)
Explanation: For polytropic process PVn=constantPV^n = \text{constant} with n=1.3n = 1.3, the entropy change formula is Δs=R(γ−n)1−nln⁡(V2/V1)\Delta s = \frac{R(\gamma-n)}{1-n}\ln(V_2/V_1). From the pressure ratio P1/P2=1/0.4=2.5P_1/P_2 = 1/0.4 = 2.5 and using PVn=constantPV^n = \text{constant}, we get V2/V1=(P1/P2)1/n=(2.5)1/1.3=2.50.769≈2.02V_2/V_1 = (P_1/P_2)^{1/n} = (2.5)^{1/1.3} = 2.5^{0.769} \approx 2.02. However, using the pressure form: Δs=R(γ−n)1−nln⁡(P1/P2)1/n=0.287(1.4−1.3)1−1.3ln⁡(2.51/1.3)=0.287×0.1−0.3×11.3ln⁡(2.5)\Delta s = \frac{R(\gamma-n)}{1-n}\ln(P_1/P_2)^{1/n} = \frac{0.287(1.4-1.3)}{1-1.3}\ln(2.5^{1/1.3}) = \frac{0.287 \times 0.1}{-0.3} \times \frac{1}{1.3}\ln(2.5). Simplifying: Δs=0.287(1.4−1.3)1−1.3ln⁡(2.5)\Delta s = \frac{0.287(1.4-1.3)}{1-1.3}\ln(2.5).

Question 18

Two identical containers of ideal gas at different temperatures (TA=400 KT_A = 400 \text{ K}, TB=300 KT_B = 300 \text{ K}) and the same pressure are connected and allowed to reach thermal equilibrium. If each container initially holds 2 kg of gas with cv=0.718 kJ/kg\cdotpKc_v = 0.718 \text{ kJ/kg·K}, what is the total entropy change of the system?

  1. ΔS=4×0.718×ln⁡(350/120000) kJ/K\Delta S = 4 \times 0.718 \times \ln(350/\sqrt{120000}) \text{ kJ/K}
  2. ΔS=2×0.718×ln⁡(350/400)+2×0.718×ln⁡(350/300) kJ/K\Delta S = 2 \times 0.718 \times \ln(350/400) + 2 \times 0.718 \times \ln(350/300) \text{ kJ/K} (correct answer)
  3. ΔS=0 kJ/K\Delta S = 0 \text{ kJ/K}
  4. ΔS=4×0.718×ln⁡(700/400) kJ/K\Delta S = 4 \times 0.718 \times \ln(700/400) \text{ kJ/K}
Explanation: Final temperature: Tf=(mAcvTA+mBcvTB)/(mAcv+mBcv)=(2×400+2×300)/(2+2)=350 KT_f = (m_Ac_vT_A + m_Bc_vT_B)/(m_Ac_v + m_Bc_v) = (2×400 + 2×300)/(2+2) = 350 \text{ K}. For constant volume mixing: ΔSA=mAcvln⁡(Tf/TA)=2×0.718×ln⁡(350/400)\Delta S_A = m_Ac_v\ln(T_f/T_A) = 2×0.718×\ln(350/400) and ΔSB=mBcvln⁡(Tf/TB)=2×0.718×ln⁡(350/300)\Delta S_B = m_Bc_v\ln(T_f/T_B) = 2×0.718×\ln(350/300). Total: ΔS=ΔSA+ΔSB\Delta S = \Delta S_A + \Delta S_B. Choice A uses incorrect geometric mean. Choice C assumes reversible process. Choice D uses wrong final temperature calculation.

Question 19

An ideal gas at state 1 (P1=3 barP_1 = 3 \text{ bar}, T1=350 KT_1 = 350 \text{ K}) undergoes two different processes to reach the same final pressure P2=1 barP_2 = 1 \text{ bar}: Process A is isothermal expansion followed by isobaric cooling, while Process B is isobaric cooling followed by isothermal expansion. Both processes end at T2=280 KT_2 = 280 \text{ K}. What is the difference in entropy change between the two processes?

  1. ΔsA−ΔsB=0 kJ/kg\cdotpK\Delta s_A - \Delta s_B = 0 \text{ kJ/kg·K} (correct answer)
  2. ΔsA−ΔsB=Rln⁡(3) kJ/kg\cdotpK\Delta s_A - \Delta s_B = R\ln(3) \text{ kJ/kg·K}
  3. ΔsA−ΔsB=cpln⁡(280/350) kJ/kg\cdotpK\Delta s_A - \Delta s_B = c_p\ln(280/350) \text{ kJ/kg·K}
  4. ΔsA−ΔsB=Rln⁡(3)+cpln⁡(280/350) kJ/kg\cdotpK\Delta s_A - \Delta s_B = R\ln(3) + c_p\ln(280/350) \text{ kJ/kg·K}
Explanation: Entropy is a state function, so the entropy change depends only on the initial and final states, not on the path taken. Since both processes start at the same initial state (P1,T1P_1, T_1) and end at the same final state (P2,T2P_2, T_2), the entropy changes must be identical: ΔsA=ΔsB\Delta s_A = \Delta s_B, therefore ΔsA−ΔsB=0\Delta s_A - \Delta s_B = 0. Choices B, C, and D incorrectly assume that the path affects the entropy change.

Question 20

An ideal gas undergoes a process where TV0.4=constantTV^{0.4} = \text{constant}. If the gas expands from V1=0.2 m3V_1 = 0.2 \text{ m}^3 at T1=500 KT_1 = 500 \text{ K} to V2=0.8 m3V_2 = 0.8 \text{ m}^3, and γ=1.4\gamma = 1.4, R=0.287 kJ/kg\cdotpKR = 0.287 \text{ kJ/kg·K}, what is the entropy change per unit mass?

  1. Δs=0.287×1.40.4ln⁡(4) kJ/kg\cdotpK\Delta s = \frac{0.287 \times 1.4}{0.4}\ln(4) \text{ kJ/kg·K}
  2. Δs=0.2871.4−1ln⁡(T2/T1)+0.287ln⁡(4) kJ/kg\cdotpK\Delta s = \frac{0.287}{1.4 - 1}\ln(T_2/T_1) + 0.287\ln(4) \text{ kJ/kg·K}
  3. Δs=0.287(1.4−0.41.4−1)ln⁡(4) kJ/kg\cdotpK\Delta s = 0.287\left(\frac{1.4 - 0.4}{1.4 - 1}\right)\ln(4) \text{ kJ/kg·K} (correct answer)
  4. Δs=0.287×0.41.4−1ln⁡(4) kJ/kg\cdotpK\Delta s = \frac{0.287 \times 0.4}{1.4 - 1}\ln(4) \text{ kJ/kg·K}
Explanation: From TV0.4=constantTV^{0.4} = \text{constant}, we get T2/T1=(V1/V2)0.4=(1/4)0.4T_2/T_1 = (V_1/V_2)^{0.4} = (1/4)^{0.4}. Using Δs=cvln⁡(T2/T1)+Rln⁡(V2/V1)\Delta s = c_v\ln(T_2/T_1) + R\ln(V_2/V_1) and cv=R/(γ−1)c_v = R/(\gamma-1): Δs=Rγ−1ln⁡((V1/V2)0.4)+Rln⁡(V2/V1)=R(−0.4γ−1+1)ln⁡(4)=Rγ−0.4γ−1ln⁡(4)\Delta s = \frac{R}{\gamma-1}\ln((V_1/V_2)^{0.4}) + R\ln(V_2/V_1) = R\left(\frac{-0.4}{\gamma-1} + 1\right)\ln(4) = R\frac{\gamma-0.4}{\gamma-1}\ln(4). Choice A has wrong coefficient. Choice B doesn't use the process constraint. Choice D has incorrect numerator.