Thermodynamics Quiz: Isentropic Relations Ideal Gases
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Isentropic Relations Ideal GasesQuestion 1 of 17

A monatomic ideal gas (γ=5/3\gamma = 5/3) undergoes two different processes from the same initial state: Process A is isentropic compression to pressure P2P_2, and Process B is isothermal compression to the same pressure P2P_2. If the pressure ratio is P2/P1=27P_2/P_1 = 27, what is the ratio of final volumes V2A/V2BV_{2A}/V_{2B}?

1.671.67
0.600.60
3.003.00
0.330.33
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Thermodynamics Quiz

Thermodynamics Quiz: Isentropic Relations Ideal Gases

Practice Isentropic Relations Ideal Gases in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Isentropic Relations Ideal Gases, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A monatomic ideal gas (γ=5/3\gamma = 5/3) undergoes two different processes from the same initial state: Process A is isentropic compression to pressure P2P_2, and Process B is isothermal compression to the same pressure P2P_2. If the pressure ratio is P2/P1=27P_2/P_1 = 27, what is the ratio of final volumes V2A/V2BV_{2A}/V_{2B}?

  1. 1.671.67
  2. 0.600.60
  3. 3.003.00 (correct answer)
  4. 0.330.33
Explanation: For the isentropic process: V2A/V1=(P1/P2)1/γ=(1/27)3/5=1/9V_{2A}/V_1 = (P_1/P_2)^{1/\gamma} = (1/27)^{3/5} = 1/9. For the isothermal process: V2B/V1=P1/P2=1/27V_{2B}/V_1 = P_1/P_2 = 1/27. Therefore, V2A/V2B=(1/9)/(1/27)=27/9=3V_{2A}/V_{2B} = (1/9)/(1/27) = 27/9 = 3. The isentropic compression results in a larger final volume than isothermal compression to the same pressure. Choice A uses γ\gamma instead of 1/γ1/\gamma. Choice B inverts the correct ratio. Choice D uses the wrong pressure relationship.

Question 2

An ideal gas with γ=1.3\gamma = 1.3 undergoes an isentropic process. If the temperature ratio T2/T1=2.5T_2/T_1 = 2.5 and the initial specific volume is v1=0.8 m3/kgv_1 = 0.8 \text{ m}^3/\text{kg}, what is the final specific volume?

  1. 0.245 m3/kg0.245 \text{ m}^3/\text{kg} (correct answer)
  2. 0.320 m3/kg0.320 \text{ m}^3/\text{kg}
  3. 0.186 m3/kg0.186 \text{ m}^3/\text{kg}
  4. 0.128 m3/kg0.128 \text{ m}^3/\text{kg}
Explanation: For an isentropic process, the temperature-volume relation is T1v1γ−1=T2v2γ−1T_1 v_1^{\gamma-1} = T_2 v_2^{\gamma-1}. Rearranging: v2/v1=(T1/T2)1/(γ−1)=(1/2.5)1/0.3=(0.4)10/3=0.306v_2/v_1 = (T_1/T_2)^{1/(\gamma-1)} = (1/2.5)^{1/0.3} = (0.4)^{10/3} = 0.306. Therefore, v2=0.8×0.306=0.245 m3/kgv_2 = 0.8 \times 0.306 = 0.245 \text{ m}^3/\text{kg}. Choice B uses the wrong exponent relationship. Choice C applies the temperature ratio directly without proper exponentiation. Choice D uses γ\gamma instead of γ−1\gamma-1 in the denominator.

Question 3

A gas turbine operates with air entering at 1200 K1200 \text{ K} and 1000 kPa1000 \text{ kPa} and expanding isentropically to 100 kPa100 \text{ kPa}. After expansion, the air is further expanded in a second stage to 50 kPa50 \text{ kPa} through another isentropic process. What is the temperature after the second expansion stage for air with γ=1.4\gamma = 1.4?

  1. 578 K578 \text{ K}
  2. 520 K520 \text{ K}
  3. 461 K461 \text{ K} (correct answer)
  4. 403 K403 \text{ K}
Explanation: This is a two-stage isentropic expansion. The overall pressure ratio is P3/P1=50/1000=0.05P_3/P_1 = 50/1000 = 0.05. For the complete isentropic process: T3/T1=(P3/P1)(γ−1)/γ=(0.05)0.4/1.4=(0.05)2/7=0.384T_3/T_1 = (P_3/P_1)^{(\gamma-1)/\gamma} = (0.05)^{0.4/1.4} = (0.05)^{2/7} = 0.384. Therefore, T3=1200×0.384=461 KT_3 = 1200 \times 0.384 = 461 \text{ K}. Choice B calculates only the first stage expansion to 100 kPa. Choice C uses an incorrect exponent. Choice D applies the pressure ratio linearly without proper isentropic relations.

Question 4

During an isentropic expansion of air (γ=1.4\gamma = 1.4) in a piston-cylinder device, the volume increases by a factor of 6. If the initial pressure is 800 kPa800 \text{ kPa}, what is the ratio of final to initial density?

  1. 0.1670.167 (correct answer)
  2. 0.1020.102
  3. 0.0730.073
  4. 0.2440.244
Explanation: For an isentropic process, P1/ρ1γ=P2/ρ2γP_1/\rho_1^\gamma = P_2/\rho_2^\gamma. Since ρ=1/v\rho = 1/v and v2/v1=6v_2/v_1 = 6, we have ρ2/ρ1=v1/v2=1/6=0.167\rho_2/\rho_1 = v_1/v_2 = 1/6 = 0.167. The density ratio depends only on the volume ratio, not the pressure directly. Choice B incorrectly applies the isentropic pressure-volume relation. Choice C uses the pressure ratio formula incorrectly. Choice D assumes an isothermal process relationship.

Question 5

An ideal gas undergoes an isentropic compression from an initial state where P1=100P_1 = 100 kPa and T1=300T_1 = 300 K to a final pressure of P2=800P_2 = 800 kPa. If the specific heat ratio γ=1.4\gamma = 1.4, what is the final temperature T2T_2?

  1. T2=546T_2 = 546 K (correct answer)
  2. T2=600T_2 = 600 K
  3. T2=2400T_2 = 2400 K
  4. T2=514T_2 = 514 K
  5. T2=450T_2 = 450 K
Explanation: When you encounter isentropic processes with ideal gases, you're dealing with adiabatic changes where entropy remains constant. The key relationship connects pressure and temperature through the specific heat ratio. For an isentropic process, the pressure-temperature relationship is: P1P2=(T1T2)γγ−1\frac{P_1}{P_2} = \left(\frac{T_1}{T_2}\right)^{\frac{\gamma}{\gamma-1}} Rearranging to solve for T2T_2: T2=T1(P2P1)γ−1γT_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} First, calculate the exponent: γ−1γ=1.4−11.4=0.41.4=0.286\frac{\gamma-1}{\gamma} = \frac{1.4-1}{1.4} = \frac{0.4}{1.4} = 0.286 Then substitute the values: T2=300 K×(800 kPa100 kPa)0.286=300×(8)0.286T_2 = 300 \text{ K} \times \left(\frac{800 \text{ kPa}}{100 \text{ kPa}}\right)^{0.286} = 300 \times (8)^{0.286} Computing (8)0.286=1.82(8)^{0.286} = 1.82, so T2=300×1.82=546T_2 = 300 \times 1.82 = 546 K, confirming answer A. Answer B (600 K) likely comes from incorrectly using γ\gamma instead of γ−1γ\frac{\gamma-1}{\gamma} in the exponent. Answer C (2400 K) results from using a linear pressure-temperature relationship, ignoring the exponential nature of isentropic processes. Answer D (514 K) suggests a calculation error, possibly in computing the fractional exponent. Remember: isentropic processes always use the exponent γ−1γ\frac{\gamma-1}{\gamma} for pressure-temperature relationships. Double-check your exponent calculation—it's where most errors occur in these problems.

Question 6

An ideal gas at T1=400T_1 = 400 K and V1=0.5V_1 = 0.5 m³ undergoes isentropic compression to V2=0.1V_2 = 0.1 m³. If γ=1.35\gamma = 1.35, what is the work done by the gas?

  1. W=−P1V1γ−1[5γ−1−1]W = -\frac{P_1 V_1}{\gamma - 1}[5^{\gamma - 1} - 1]
  2. W=−P1V1ln⁡(5)W = -P_1 V_1 \ln(5)
  3. W=−P1V1γ−1[1−5γ−1]W = -\frac{P_1 V_1}{\gamma - 1}[1 - 5^{\gamma - 1}] (correct answer)
  4. W=−P1V1(5γ−1)W = -P_1 V_1(5^\gamma - 1)
  5. W=P1V1γ−1[5γ−1−1]W = \frac{P_1 V_1}{\gamma - 1}[5^{\gamma - 1} - 1]
Explanation: When you encounter isentropic processes, remember that these are adiabatic processes where entropy remains constant. The key relationship is PVγ=constantPV^\gamma = \text{constant}, which leads to the work formula for isentropic processes. For an isentropic process, work is calculated using: W=P1V1−P2V2γ−1W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}. Since we need P2P_2, we use the isentropic relation: P1V1γ=P2V2γP_1V_1^\gamma = P_2V_2^\gamma, so P2=P1(V1V2)γ=P1(5)γP_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = P_1(5)^\gamma. Substituting this into the work equation: W=P1V1−P1(5)γV2γ−1=P1V1−P1(5)γ(V1/5)γ−1=P1V1[1−5γ−1]γ−1W = \frac{P_1V_1 - P_1(5)^\gamma V_2}{\gamma - 1} = \frac{P_1V_1 - P_1(5)^\gamma (V_1/5)}{\gamma - 1} = \frac{P_1V_1[1 - 5^{\gamma-1}]}{\gamma - 1} Therefore: W=−P1V1γ−1[1−5γ−1]W = -\frac{P_1V_1}{\gamma - 1}[1 - 5^{\gamma-1}], which matches option C. Option A has the terms reversed in the brackets, giving 5γ−1−15^{\gamma-1} - 1 instead of 1−5γ−11 - 5^{\gamma-1}. This would give a positive work value, but compression work should be negative when done on the gas. Option B uses ln⁡(5)\ln(5), which suggests confusion with isothermal processes where W=−nRTln⁡(V2/V1)W = -nRT\ln(V_2/V_1). Option D omits the crucial (γ−1)(\gamma-1) denominator and uses 5γ5^\gamma instead of 5γ−15^{\gamma-1}, indicating dimensional errors. Remember: isentropic work always involves the (γ−1)(\gamma-1) factor in the denominator, and compression work is negative since work is done on the gas.

Question 7

Two identical containers of ideal gas at different initial conditions undergo separate isentropic processes. Container A starts at PA=200P_A = 200 kPa, TA=300T_A = 300 K and expands to PA′=50P_A' = 50 kPa. Container B starts at PB=400P_B = 400 kPa, TB=600T_B = 600 K and expands to PB′=100P_B' = 100 kPa. If γ=1.4\gamma = 1.4 for both gases, which statement is correct?

  1. Both containers reach the same final temperature since pressure ratios are identical
  2. Container A reaches a lower final temperature because it started at lower pressure
  3. Container B reaches a higher final temperature because it started at higher temperature (correct answer)
  4. The final temperatures are equal because both gases have the same γ\gamma value
  5. Container A reaches a higher final temperature due to greater volume expansion ratio
Explanation: When you encounter isentropic processes with ideal gases, remember that the relationship between pressure and temperature follows: T2T1=(P2P1)γ−1γ\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}. This means the final temperature depends on both the initial temperature and the pressure ratio. Let's calculate the final temperatures for both containers. For Container A: TA′=300×(50200)0.41.4=300×(0.25)0.286=300×0.668=200.4T_A' = 300 \times \left(\frac{50}{200}\right)^{\frac{0.4}{1.4}} = 300 \times (0.25)^{0.286} = 300 \times 0.668 = 200.4 K. For Container B: TB′=600×(100400)0.41.4=600×(0.25)0.286=600×0.668=400.8T_B' = 600 \times \left(\frac{100}{400}\right)^{\frac{0.4}{1.4}} = 600 \times (0.25)^{0.286} = 600 \times 0.668 = 400.8 K. Container B indeed reaches a higher final temperature because it started at a higher initial temperature, making answer C correct. Answer A is wrong because identical pressure ratios don't guarantee identical final temperatures when initial temperatures differ. The final temperature scales with the initial temperature through the isentropic relationship. Answer B incorrectly suggests that starting pressure alone determines the outcome, ignoring the crucial role of initial temperature in the calculation. Answer D misses the point entirely—while both gases have the same γ\gamma, this only affects the exponent in the relationship, not the final result when initial conditions differ. Study tip: In isentropic problems, always remember that final states depend on both initial conditions and the process path. Don't assume identical ratios lead to identical final states unless all initial conditions match.

Question 8

An ideal gas undergoes an isentropic process where the density increases by a factor of 3.2. If the initial temperature is T1=350T_1 = 350 K and γ=1.25\gamma = 1.25, what is the final temperature?

  1. T2=623T_2 = 623 K (correct answer)
  2. T2=1120T_2 = 1120 K
  3. T2=545T_2 = 545 K
  4. T2=700T_2 = 700 K
  5. T2=467T_2 = 467 K
Explanation: When you encounter isentropic processes with density changes, you need to connect the density ratio to temperature using the relationships for adiabatic processes in ideal gases. For an isentropic process, you can use the relationship: T2T1=(ρ2ρ1)γ−1\frac{T_2}{T_1} = \left(\frac{\rho_2}{\rho_1}\right)^{\gamma-1} Since density increases by a factor of 3.2, we have ρ2ρ1=3.2\frac{\rho_2}{\rho_1} = 3.2. With γ=1.25\gamma = 1.25, the exponent becomes γ−1=0.25\gamma - 1 = 0.25. Calculating: T2T1=(3.2)0.25=1.78\frac{T_2}{T_1} = (3.2)^{0.25} = 1.78 Therefore: T2=T1×1.78=350 K×1.78=623 KT_2 = T_1 \times 1.78 = 350 \text{ K} \times 1.78 = 623 \text{ K} This confirms answer A is correct. Answer B (1120 K) likely comes from using the wrong exponent, perhaps γ\gamma instead of γ−1\gamma - 1, which would give (3.2)1.25=3.2(3.2)^{1.25} = 3.2, yielding an unrealistically high temperature. Answer C (545 K) might result from calculation errors or using an incorrect form of the isentropic relationship, possibly confusing it with isothermal or other process equations. Answer D (700 K) could stem from using a linear relationship or incorrect manipulation of the density-temperature correlation, missing the exponential nature of the isentropic process. Remember: for isentropic processes, always use (γ−1)(\gamma - 1) as the exponent when relating density and temperature ratios. The key insight is that compression during adiabatic processes increases both density and temperature, with the specific relationship governed by the heat capacity ratio.

Question 9

During an isentropic expansion of air (γ=1.4\gamma = 1.4), the pressure drops from 1000 kPa to 300 kPa. If the initial specific volume is v1=0.8v_1 = 0.8 m³/kg, what is the final specific volume?

  1. v2=2.03v_2 = 2.03 m³/kg (correct answer)
  2. v2=2.67v_2 = 2.67 m³/kg
  3. v2=1.85v_2 = 1.85 m³/kg
  4. v2=3.20v_2 = 3.20 m³/kg
  5. v2=1.52v_2 = 1.52 m³/kg
Explanation: When you encounter isentropic processes in thermodynamics, you're dealing with reversible adiabatic processes where entropy remains constant. For an ideal gas undergoing an isentropic process, pressure and specific volume are related by the equation pvγ=constantpv^{\gamma} = \text{constant}, where γ\gamma is the specific heat ratio. To find the final specific volume, you can write p1v1γ=p2v2γp_1v_1^{\gamma} = p_2v_2^{\gamma}. Rearranging to solve for v2v_2: v2=v1(p1p2)1/γv_2 = v_1 \left(\frac{p_1}{p_2}\right)^{1/\gamma} Substituting the given values: v2=0.8(1000300)1/1.4=0.8×(3.333)0.714=0.8×2.54=2.03 m³/kgv_2 = 0.8 \left(\frac{1000}{300}\right)^{1/1.4} = 0.8 \times (3.333)^{0.714} = 0.8 \times 2.54 = 2.03 \text{ m³/kg} This confirms answer A is correct. Answer B (2.67 m³/kg) likely results from using γ\gamma instead of 1/γ1/\gamma in the exponent, which would give (3.333)1.4(3.333)^{1.4}. Answer C (1.85 m³/kg) might come from incorrectly using the isothermal relation pv=constantpv = \text{constant} instead of the isentropic relation. Answer D (3.20 m³/kg) appears to result from simply multiplying the initial volume by the pressure ratio without applying any exponent. Remember that isentropic processes always use the pvγ=constantpv^{\gamma} = \text{constant} relationship, and the exponent in your final calculation should be 1/γ1/\gamma, not γ\gamma itself. This distinction is crucial for getting the correct answer.

Question 10

An ideal gas with γ=1.25\gamma = 1.25 undergoes isentropic compression. If the work done per unit mass is w=85w = 85 kJ/kg and the initial temperature is T1=320T_1 = 320 K, what is the final temperature assuming cv=0.8c_v = 0.8 kJ/kg⋅K?

  1. T2=426T_2 = 426 K (correct answer)
  2. T2=456T_2 = 456 K
  3. T2=405T_2 = 405 K
  4. T2=491T_2 = 491 K
  5. T2=378T_2 = 378 K
Explanation: When you encounter isentropic processes with work calculations, you need to connect the work done to temperature changes through the gas properties. For isentropic compression of an ideal gas, the work done per unit mass relates directly to the change in internal energy since no heat transfer occurs. For an isentropic process, the work done per unit mass equals the change in internal energy: w=cv(T2−T1)w = c_v(T_2 - T_1). This relationship comes from the first law of thermodynamics with q=0q = 0 for isentropic processes. Substituting the given values: 85=0.8(T2−320)85 = 0.8(T_2 - 320) Solving for T2T_2: T2=320+850.8=320+106.25=426.25T_2 = 320 + \frac{85}{0.8} = 320 + 106.25 = 426.25 K This confirms answer A) T2=426T_2 = 426 K is correct. Answer B) T2=456T_2 = 456 K represents a common error where students might use cpc_p instead of cvc_v in the work equation, or make calculation mistakes with the division. Answer C) T2=405T_2 = 405 K suggests using an incorrect sign convention, perhaps treating compression work as negative when the problem already accounts for the sign in stating work as positive. Answer D) T2=491T_2 = 491 K likely results from confusing isentropic work formulas or incorrectly applying the γ\gamma value in an alternative work expression without proper unit conversions. Remember: for isentropic processes, always start with w=cvΔTw = c_v \Delta T when work per unit mass is given directly. The γ\gamma value becomes relevant when using pressure or volume ratios, but isn't needed for this direct approach.

Question 11

During an isentropic expansion of an ideal gas, the temperature decreases from 500 K to 350 K. If γ=1.4\gamma = 1.4, by what factor does the pressure decrease?

  1. Pressure decreases by factor of 3.5
  2. Pressure decreases by factor of 2.8
  3. Pressure decreases by factor of 4.2 (correct answer)
  4. Pressure decreases by factor of 1.43
  5. Pressure decreases by factor of 2.4
Explanation: When you encounter isentropic processes with ideal gases, you're dealing with adiabatic changes where entropy remains constant. The key relationship here connects pressure and temperature through the equation P1P2=(T1T2)γγ−1\frac{P_1}{P_2} = \left(\frac{T_1}{T_2}\right)^{\frac{\gamma}{\gamma-1}}. Let's solve this step by step. You have T1=500 KT_1 = 500 \text{ K}, T2=350 KT_2 = 350 \text{ K}, and γ=1.4\gamma = 1.4. First, calculate the exponent: γγ−1=1.41.4−1=1.40.4=3.5\frac{\gamma}{\gamma-1} = \frac{1.4}{1.4-1} = \frac{1.4}{0.4} = 3.5. Now substitute: P1P2=(500350)3.5=(1.429)3.5=4.2\frac{P_1}{P_2} = \left(\frac{500}{350}\right)^{3.5} = (1.429)^{3.5} = 4.2. This means pressure decreases by a factor of 4.2, confirming answer C. Looking at the wrong answers: A gives 3.5, which is exactly the exponent value—a common trap where students confuse the exponent with the final answer. B shows 2.8, which might result from calculation errors or using an incorrect relationship. D gives 1.43, which is simply the temperature ratio 500350\frac{500}{350}—this ignores the exponential relationship entirely and treats the problem as if pressure and temperature change proportionally. The key study tip: For isentropic processes, remember that the exponent γγ−1\frac{\gamma}{\gamma-1} amplifies the temperature ratio effect on pressure. Small temperature changes create much larger pressure changes due to this exponential relationship. Always calculate this exponent first, then apply it to your temperature ratio.

Question 12

An ideal gas undergoes an isentropic process in which the volume changes by a factor of nn and the pressure changes by a factor of mm. If γ=1.3\gamma = 1.3, what is the relationship between mm and nn?

  1. m=n−1.3m = n^{-1.3} (correct answer)
  2. m=n−0.77m = n^{-0.77}
  3. m=n1.3m = n^{1.3}
  4. m=n−1m = n^{-1}
  5. m=n0.3m = n^{0.3}
Explanation: When you encounter an isentropic process problem, you're dealing with an adiabatic process where entropy remains constant. The key relationship to remember is the isentropic equation for an ideal gas: PVγ=constantPV^{\gamma} = \text{constant}. Let's work through this step by step. If the initial state is (P1,V1)(P_1, V_1) and the final state is (P2,V2)(P_2, V_2), then: P1V1γ=P2V2γP_1 V_1^{\gamma} = P_2 V_2^{\gamma} Given that volume changes by a factor of nn (so V2=nV1V_2 = nV_1) and pressure changes by a factor of mm (so P2=mP1P_2 = mP_1), we can substitute: P1V1γ=(mP1)(nV1)γP_1 V_1^{\gamma} = (mP_1)(nV_1)^{\gamma} Simplifying by canceling P1P_1 and V1γV_1^{\gamma}: 1=m⋅nγ1 = m \cdot n^{\gamma} Therefore: m=n−γ=n−1.3m = n^{-\gamma} = n^{-1.3} Looking at the wrong answers: B gives m=n−0.77m = n^{-0.77}, which would result from incorrectly using 1/γ1/\gamma instead of −γ-\gamma. C shows m=n1.3m = n^{1.3}, representing the common error of forgetting the negative sign in the exponent relationship. D gives m=n−1m = n^{-1}, which would apply to an isothermal process (where PV=constantPV = \text{constant}) rather than an isentropic one. Answer A is correct. Study tip: For isentropic processes, always remember that pressure and volume are inversely related with the exponent γ\gamma. When volume increases by factor nn, pressure decreases by factor nγn^{\gamma}, giving the relationship m=n−γm = n^{-\gamma}.

Question 13

An ideal gas undergoes an isentropic compression from state 1 to state 2. If the initial temperature is 300 K300 \text{ K} and the pressure ratio P2/P1=8P_2/P_1 = 8, what is the final temperature for a diatomic gas (γ=1.4\gamma = 1.4)?

  1. 548 K548 \text{ K}
  2. 692 K692 \text{ K} (correct answer)
  3. 2400 K2400 \text{ K}
  4. 1714 K1714 \text{ K}
Explanation: For an isentropic process of an ideal gas, the temperature-pressure relation is T2/T1=(P2/P1)(γ−1)/γT_2/T_1 = (P_2/P_1)^{(\gamma-1)/\gamma}. Substituting values: T2=300×8(1.4−1)/1.4=300×80.4/1.4=300×82/7=300×2.307=692 KT_2 = 300 \times 8^{(1.4-1)/1.4} = 300 \times 8^{0.4/1.4} = 300 \times 8^{2/7} = 300 \times 2.307 = 692 \text{ K}. Choice A uses the wrong exponent (0.4 instead of 2/7). Choice C incorrectly multiplies by the pressure ratio directly. Choice D uses γ\gamma instead of (γ−1)/γ(\gamma-1)/\gamma as the exponent.

Question 14

Two identical masses of air (γ=1.4\gamma = 1.4) undergo different isentropic processes starting from the same initial conditions. Gas A is compressed to half its initial volume, while Gas B is expanded to twice its initial volume. What is the ratio of final pressures PA2/PB2P_{A2}/P_{B2}?

  1. 10.610.6
  2. 21.121.1 (correct answer)
  3. 5.285.28
  4. 42.242.2
Explanation: For isentropic processes: PA2/P1=(V1/VA2)γ=(1/0.5)1.4=21.4=2.639P_{A2}/P_1 = (V_1/V_{A2})^\gamma = (1/0.5)^{1.4} = 2^{1.4} = 2.639 and PB2/P1=(V1/VB2)γ=(1/2)1.4=0.51.4=0.125P_{B2}/P_1 = (V_1/V_{B2})^\gamma = (1/2)^{1.4} = 0.5^{1.4} = 0.125. Therefore, PA2/PB2=2.639/0.125=21.1P_{A2}/P_{B2} = 2.639/0.125 = 21.1. Choice A uses only the volume ratio without proper exponentiation. Choice C incorrectly applies the square of the volume ratio. Choice D doubles the correct answer, representing a common computational error.

Question 15

A piston-cylinder device contains helium (γ=5/3\gamma = 5/3, R=2077 J/(kg\cdotpK)R = 2077 \text{ J/(kg·K)}) that undergoes an isentropic compression from 100 kPa100 \text{ kPa} and 300 K300 \text{ K} to 800 kPa800 \text{ kPa}. What is the work done per unit mass during this process?

  1. −243 kJ/kg-243 \text{ kJ/kg}
  2. −728 kJ/kg-728 \text{ kJ/kg}
  3. −364 kJ/kg-364 \text{ kJ/kg}
  4. −485 kJ/kg-485 \text{ kJ/kg} (correct answer)
Explanation: For an isentropic process, w=R(T1−T2)γ−1w = \frac{R(T_1 - T_2)}{\gamma - 1}. First find T2T_2: T2=T1(P2/P1)(γ−1)/γ=300(8)(2/3)/(5/3)=300(8)2/5=300×2.297=689 KT_2 = T_1(P_2/P_1)^{(\gamma-1)/\gamma} = 300(8)^{(2/3)/(5/3)} = 300(8)^{2/5} = 300 \times 2.297 = 689 \text{ K}. Then w=2077(300−689)5/3−1=2077(−389)2/3=−8079532/3=−485 kJ/kgw = \frac{2077(300 - 689)}{5/3 - 1} = \frac{2077(-389)}{2/3} = \frac{-807953}{2/3} = -485 \text{ kJ/kg}. The negative sign indicates work done on the gas during compression. Choice B uses incorrect temperature calculation. Choice C uses wrong denominator. Choice D applies isothermal work formula.

Question 16

An isentropic nozzle accelerates air from negligible velocity to 400 m/s400 \text{ m/s}. The inlet temperature is 500 K500 \text{ K} and cp=1005 J/(kg\cdotpK)c_p = 1005 \text{ J/(kg·K)}. Assuming the process is isentropic and using isentropic relations, what is the exit temperature?

  1. 420 K420 \text{ K} (correct answer)
  2. 460 K460 \text{ K}
  3. 540 K540 \text{ K}
  4. 580 K580 \text{ K}
Explanation: For an isentropic process with kinetic energy changes, the stagnation temperature remains constant: T01=T1+V12/(2cp)=T02=T2+V22/(2cp)T_{01} = T_1 + V_1^2/(2c_p) = T_{02} = T_2 + V_2^2/(2c_p). Since V1≈0V_1 \approx 0: 500=T2+(400)2/(2×1005)=T2+79.6500 = T_2 + (400)^2/(2 \times 1005) = T_2 + 79.6. Therefore, T2=500−79.6=420.4 K≈420 KT_2 = 500 - 79.6 = 420.4 \text{ K} \approx 420 \text{ K}. Choice B uses an incorrect kinetic energy calculation. Choice C adds instead of subtracts the kinetic energy term. Choice D uses the wrong specific heat value.

Question 17

Consider an isentropic process where an ideal gas expands from state 1 to state 2. The pressure decreases by a factor of 16, and the gas has γ=1.25\gamma = 1.25. If the initial density is ρ1=2.5 kg/m3\rho_1 = 2.5 \text{ kg/m}^3, what is the final density?

  1. 0.313 kg/m30.313 \text{ kg/m}^3
  2. 0.156 kg/m30.156 \text{ kg/m}^3
  3. 0.625 kg/m30.625 \text{ kg/m}^3
  4. 0.391 kg/m30.391 \text{ kg/m}^3 (correct answer)
Explanation: For an isentropic process, P1/ρ1γ=P2/ρ2γP_1/\rho_1^\gamma = P_2/\rho_2^\gamma. Since P2/P1=1/16P_2/P_1 = 1/16, we have ρ2γ/ρ1γ=P2/P1=1/16\rho_2^\gamma/\rho_1^\gamma = P_2/P_1 = 1/16. Therefore, ρ2/ρ1=(1/16)1/γ=(1/16)1/1.25=(1/16)0.8=0.156\rho_2/\rho_1 = (1/16)^{1/\gamma} = (1/16)^{1/1.25} = (1/16)^{0.8} = 0.156. So ρ2=2.5×0.156=0.391 kg/m3\rho_2 = 2.5 \times 0.156 = 0.391 \text{ kg/m}^3. Choice A uses γ\gamma instead of 1/γ1/\gamma. Choice B shows the ratio calculation but not the final multiplication. Choice C uses the square root instead of the proper fractional exponent.