Thermodynamics Quiz: Cp Cv And K Relationships

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Cp Cv And K RelationshipsQuestion 1 of 20

Two identical containers hold the same ideal gas at the same pressure and temperature. Container A undergoes reversible adiabatic expansion to half the pressure. Container B undergoes reversible isothermal expansion to the same final pressure. What is the ratio of the final volumes $\frac{V_{f,A}}{V_{f,B}}$ ?

\frac{V_{f,A}}{V_{f,B}} = 2^{1/k} = 0.757

(for

k = 1.4

)

\frac{V_{f,A}}{V_{f,B}} = 2^{(k-1)/k} = 0.871

(for

k = 1.4

)

\frac{V_{f,A}}{V_{f,B}} = 1.00

(both reach same final state)

\frac{V_{f,A}}{V_{f,B}} = 2^{k-1} = 1.32

(for

k = 1.4

)

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Thermodynamics Quiz

Thermodynamics Quiz: Cp Cv And K Relationships

Practice Cp Cv And K Relationships 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 Cp Cv And K Relationships, 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.

All questions

Question 1

$\frac{V_{f,A}}{V_{f,B}} = 2^{1/k} = 0.757$ (for $k = 1.4$ )
$\frac{V_{f,A}}{V_{f,B}} = 2^{(k-1)/k} = 0.871$ (for $k = 1.4$ ) (correct answer)
$\frac{V_{f,A}}{V_{f,B}} = 1.00$ (both reach same final state)
$\frac{V_{f,A}}{V_{f,B}} = 2^{k-1} = 1.32$ (for $k = 1.4$ )

Explanation: For adiabatic process A:

p_1 V_1^k = p_2 V_{f,A}^k

, so

V_{f,A} = V_1 \left(\frac{p_1}{p_2}\right)^{1/k} = V_1 (2)^{1/k}

. For isothermal process B:

p_1 V_1 = p_2 V_{f,B}

, so

V_{f,B} = V_1 \frac{p_1}{p_2} = 2V_1

. Therefore:

\frac{V_{f,A}}{V_{f,B}} = \frac{V_1 \cdot 2^{1/k}}{2V_1} = \frac{2^{1/k}}{2} = 2^{1/k - 1} = 2^{(1-k)/k} = 2^{-(k-1)/k}

. For

k = 1.4

2^{-(1.4-1)/1.4} = 2^{-0.4/1.4} = 2^{-2/7} ≈ 0.871

. Choice A uses the wrong exponent. Choice C ignores the different process paths. Choice D inverts the exponent relationship.

Question 2

For an ideal gas, if $c_p-c_v=R$ , which form of $k=c_p/c_v$ is correct?

$k=1+c_v/R$
$k=1+R/c_v$ (correct answer)
$k=R/(c_v+R)$
$k=c_v/(c_v+R)$

Explanation: Since cp - cv = R, you have cp = R + cv. Divide both sides by cv: cp/cv = (R + cv)/cv = 1 + R/cv. The tempting wrong form 1 + cv/R reverses the ratio and would give cp - cv = cv^2/R, not R.

Question 3

For an ideal gas, which expression gives $c_v$ in terms of $R$ and $k$ ?

$c_v=kR/(k-1)$
$c_v=R(k-1)$
$c_v=R/(k-1)$ (correct answer)
$c_v=R/(k+1)$

Explanation: Start with k = cp/cv and cp - cv = R. Substitute cp = k cv into the second: k cv - cv = R, so cv(k - 1) = R, giving cv = R/(k - 1). A tempting trap is kR/(k - 1), but that is cp, not cv; it comes from solving for cp instead.

Question 4

For an ideal gas, $c_p=1.2$ kJ/kgK and $k=1.6$ . Find $R$ .

1.92 kJ/kgK
1.20 kJ/kgK
0.75 kJ/kgK
0.45 kJ/kgK (correct answer)

Explanation: For an ideal gas, cv = cp / k = 1.2 / 1.6 = 0.75 kJ/kgK. Then R = cp - cv = 1.2 - 0.75 = 0.45 kJ/kgK. The 0.75 kJ/kgK figure is cv, not R, so choosing it is the common mistake.

Question 5

Given $c_p=2.0$ kJ/kgK and $k=1.25$ , find $c_v$ .

0.4 kJ/kgK
1.6 kJ/kgK (correct answer)
2.0 kJ/kgK
2.5 kJ/kgK

Explanation: Since k = cp/cv, rearrange to cv = cp/k. Divide 2.0 by 1.25 to get 1.6 kJ/kgK. The 0.4 value is cp - cv, the gas constant R, not cv itself.

Question 6

An ideal gas has $c_v=0.8$ kJ/kgK and $R=0.4$ kJ/kgK. What is $k$ ?

1.500 (correct answer)
0.500
2.000
0.667

Explanation: For an ideal gas, cp = cv + R, so cp = 0.8 + 0.4 = 1.2 kJ/kgK. The specific heat ratio k is cp/cv = 1.2 / 0.8 = 1.5. Don't invert the ratio: 0.667 comes from dividing cv by cp, but k requires cp over cv.

Question 7

A diatomic ideal gas ( $k = 1.4$ ) is compressed adiabatically from an initial pressure of 100 kPa to a final pressure of 800 kPa. If the initial temperature is 300 K, what is the final temperature?

300 K
573 K (correct answer)
480 K
2400 K
420 K

Explanation: When you encounter adiabatic processes with ideal gases, you're dealing with situations where no heat transfer occurs. This means you'll use the adiabatic relationships that connect pressure, volume, and temperature through the heat capacity ratio

k

. For an adiabatic process with an ideal gas, the relationship between initial and final states is:

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}

Let's substitute the given values:

T_1 = 300

P_1 = 100

kPa,

P_2 = 800

kPa, and

k = 1.4

for a diatomic gas. First, calculate the exponent:

\frac{k-1}{k} = \frac{1.4-1}{1.4} = \frac{0.4}{1.4} = \frac{2}{7}

Now:

T_2 = 300 \times \left(\frac{800}{100}\right)^{2/7} = 300 \times (8)^{2/7}

Since

8 = 2^3

, we have

(8)^{2/7} = (2^3)^{2/7} = 2^{6/7} \approx 1.91

Therefore:

T_2 = 300 \times 1.91 = 573

K Choice B (573 K) is correct. Choice A (300 K) incorrectly assumes no temperature change, which would only occur in an isothermal process. Choice C (480 K) might result from using the wrong exponent or approximation errors. Choice D (2400 K) comes from incorrectly using the full pressure ratio without the proper exponent, essentially calculating

300 \times 8 = 2400

. Remember: adiabatic compression always increases temperature for ideal gases. Always check that your final temperature is higher than the initial temperature when pressure increases adiabatically.

Question 8

An ideal gas undergoes a process where its temperature increases from 300 K to 450 K while maintaining constant pressure. If the gas has $c_p = 1.4$ kJ/kg·K and $c_v = 1.0$ kJ/kg·K, what is the ratio of heat added to the change in internal energy for this process?

1.0
1.4 (correct answer)
2.4
0.71
0.40

Explanation: When analyzing processes involving ideal gases, you need to understand the relationship between heat transfer, internal energy change, and the type of process occurring. This question tests your ability to apply the first law of thermodynamics to an isobaric (constant pressure) process. For any process, the first law states that

Q = \Delta U + W

, where

Q

is heat added,

\Delta U

is the change in internal energy, and

W

is work done by the gas. For an ideal gas at constant pressure,

\Delta U = mc_v\Delta T

and

Q = mc_p\Delta T

. The ratio you're seeking is:

\frac{Q}{\Delta U} = \frac{mc_p\Delta T}{mc_v\Delta T} = \frac{c_p}{c_v} = \frac{1.4}{1.0} = 1.4

This confirms answer B is correct. A) 1.0 represents the misconception that heat added equals internal energy change, which would only be true for a constant volume process where no work is done. C) 2.4 appears to come from incorrectly adding

c_p + c_v = 1.4 + 1.0 = 2.4

, which has no physical significance for this ratio. D) 0.71 results from inverting the correct ratio, calculating

c_v/c_p = 1.0/1.4 ≈ 0.71

. This represents the fraction of heat that goes into internal energy rather than the ratio requested. Remember that for isobaric processes, the ratio

Q/\Delta U

always equals the heat capacity ratio

c_p/c_v

, regardless of the specific temperature values. Focus on identifying the process type first, then apply the appropriate thermodynamic relationships.

Question 9

For an ideal gas with specific heat ratio $k = 1.3$ , if the gas constant $R = 287$ J/kg·K, what is the value of $c_v$ in J/kg·K?

718 J/kg·K
957 J/kg·K (correct answer)
373 J/kg·K
221 J/kg·K
287 J/kg·K

Explanation: When you encounter specific heat problems for ideal gases, you need to understand the fundamental relationships between

c_p

c_v

, the gas constant

R

, and the specific heat ratio

k

. For ideal gases, two key equations govern these relationships:

c_p - c_v = R

and

k = \frac{c_p}{c_v}

. From the second equation, you can express

c_p = k \cdot c_v

. Substituting this into the first equation:

k \cdot c_v - c_v = R

, which simplifies to

c_v(k-1) = R

. Therefore:

c_v = \frac{R}{k-1}

. With

R = 287

J/kg·K and

k = 1.3

, you get:

c_v = \frac{287}{1.3-1} = \frac{287}{0.3} = 957

J/kg·K. Choice A (718 J/kg·K) results from incorrectly using

c_v = \frac{R \cdot k}{k-1}

, multiplying by

k

instead of dividing by

(k-1)

. Choice C (373 J/kg·K) comes from using

c_v = \frac{R}{k}

, forgetting the "-1" in the denominator. Choice D (221 J/kg·K) appears to use an incorrect formula entirely, possibly confusing relationships or making calculation errors. The correct answer is B (957 J/kg·K). Remember this key formula:

c_v = \frac{R}{k-1}

for ideal gases. Always subtract 1 from the specific heat ratio in the denominator. This relationship appears frequently on thermodynamics exams, so memorizing it alongside

c_p = \frac{kR}{k-1}

will save you time and prevent common algebraic mistakes.

Question 10

For an ideal gas mixture consisting of 60% nitrogen ( $k = 1.40$ ) and 40% argon ( $k = 1.67$ ) by mole fraction, what is the effective specific heat ratio of the mixture?

1.40
1.50
1.47 (correct answer)
1.67
1.54

Explanation: When you encounter gas mixture problems involving specific heat ratios, you need to calculate the effective properties based on the weighted contributions of each component gas. For gas mixtures, the effective specific heat ratio

k_{mix}

is found using the mole-fraction-weighted average of the individual gas properties. Since nitrogen comprises 60% (0.6) and argon comprises 40% (0.4) by mole fraction:

k_{mix} = x_{N_2} \cdot k_{N_2} + x_{Ar} \cdot k_{Ar}

k_{mix} = 0.6 \times 1.40 + 0.4 \times 1.67 = 0.84 + 0.668 = 1.468

Rounding to two decimal places gives

k_{mix} = 1.47

. Answer A (1.40) represents the specific heat ratio of pure nitrogen, ignoring the argon contribution entirely. Answer B (1.50) might result from incorrectly using a simple arithmetic average:

(1.40 + 1.67)/2 = 1.535

, then rounding poorly, but this ignores the actual mole fractions. Answer D (1.67) is the specific heat ratio of pure argon, completely neglecting the nitrogen component. The key insight is that mixture properties aren't simple averages—they must be weighted by the actual composition. Since nitrogen has the higher mole fraction (60%) and lower

k

value, the mixture's effective

k

falls closer to nitrogen's value than argon's, but still incorporates argon's influence proportionally. Remember: For any intensive property in gas mixtures, always use mole-fraction-weighted averages, not simple arithmetic means. The dominant component (by mole fraction) will have the stronger influence on the mixture's properties.

Question 11

An ideal gas with $c_v = 0.718$ kJ/kg·K and gas constant $R = 0.287$ kJ/kg·K undergoes a constant volume heating process where the temperature increases by 200 K. If the same amount of heat were added at constant pressure instead, what would be the temperature increase?

200 K
143 K (correct answer)
280 K
100 K
157 K

Explanation: This problem tests your understanding of how specific heats affect temperature changes in different thermodynamic processes. When you see questions comparing constant volume and constant pressure processes, remember that the same energy input produces different temperature changes because the specific heats are different. For the constant volume process, you can find the heat added per unit mass:

q = c_v \Delta T = 0.718 \times 200 = 143.6

kJ/kg. Now you need

c_p

for the constant pressure calculation. Use the fundamental relationship

c_p = c_v + R

c_p = 0.718 + 0.287 = 1.005

kJ/kg·K. When this same amount of heat (143.6 kJ/kg) is added at constant pressure:

\Delta T = \frac{q}{c_p} = \frac{143.6}{1.005} = 143

K. Answer A (200 K) assumes the temperature change is the same regardless of process type—a common misconception that ignores the difference between

c_v

and

c_p

. Answer C (280 K) incorrectly applies the ratio

\frac{c_p}{c_v}

as a multiplier rather than a divisor. Answer D (100 K) appears to use an incorrect relationship between the specific heats. The key insight is that constant pressure processes require more energy for the same temperature increase because work is done by the expanding gas. Therefore, for the same energy input, the temperature rise at constant pressure is always less than at constant volume. Always remember:

c_p > c_v

for ideal gases, so identical heat inputs produce smaller temperature changes in constant pressure processes.

Question 12

An ideal gas with molecular weight M = 28 kg/kmol undergoes an isentropic expansion where the volume increases by a factor of 8. If the initial temperature is 600 K and $k = 1.4$ , what is the final temperature?

300 K
245 K (correct answer)
150 K
424 K
200 K

Explanation: When you encounter an isentropic process problem, you're dealing with a reversible adiabatic process where entropy remains constant. The key relationship you need is the temperature-volume equation for isentropic processes:

T_1 V_1^{k-1} = T_2 V_2^{k-1}

, which can be rearranged to

\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{k-1}

. Given that the volume increases by a factor of 8, we have

\frac{V_2}{V_1} = 8

, so

\frac{V_1}{V_2} = \frac{1}{8}

. With

k = 1.4

, we get

k-1 = 0.4

. Now we can solve:

T_2 = T_1 \left(\frac{1}{8}\right)^{0.4} = 600 \times (0.125)^{0.4}

Calculating

(0.125)^{0.4} = (2^{-3})^{0.4} = 2^{-1.2} = 0.435

Therefore:

T_2 = 600 \times 0.435 = 245 \text{ K}

This confirms answer (B) 245 K is correct. (A) 300 K represents using

k-1 = 1

instead of 0.4, giving

T_2 = 600/2 = 300

K. (C) 150 K comes from incorrectly using

T_2 = T_1/4

, possibly confusing this with a different thermodynamic relationship. (D) 424 K results from using the wrong exponent or misapplying the volume ratio. Study tip: For isentropic processes, always remember the exponent is

k-1

, not

k

. Also, when volume increases in expansion, temperature must decrease—this helps you catch calculation errors quickly.

Question 13

A gas turbine operates with air ( $k = 1.4$ ) entering at 1200 K and expanding isentropically to 400 K. If the process were instead a polytropic expansion with $n = 1.25$ , what would be the final temperature for the same pressure ratio?

400 K
358 K
445 K (correct answer)
320 K
480 K

Explanation: This problem tests your understanding of different expansion processes in gas turbines and how they affect final conditions. When you encounter gas expansion problems, you need to identify the process type and apply the appropriate relationships between temperature and pressure. For the isentropic expansion, you can find the pressure ratio using the relationship

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}

. With

T_1 = 1200

T_2 = 400

K, and

k = 1.4

\frac{400}{1200} = \left(\frac{P_2}{P_1}\right)^{0.4/1.4}

This gives

\frac{P_2}{P_1} = (1/3)^{1.4/0.4} = 0.1296

For the polytropic process with the same pressure ratio, use

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(n-1)/n}

. With

n = 1.25

\frac{T_2}{1200} = (0.1296)^{0.25/1.25} = (0.1296)^{0.2} = 0.371

Therefore,

T_2 = 1200 \times 0.371 = 445

K, confirming answer C. Answer A (400 K) assumes the final temperature remains unchanged regardless of process type—a fundamental misconception. Answer B (358 K) likely results from incorrectly using the isentropic exponent in the polytropic formula. Answer D (320 K) suggests confusion about the relationship between polytropic index and final temperature. Remember: different expansion processes with the same pressure ratio yield different final temperatures. Always identify the process first, then apply the correct temperature-pressure relationship with the appropriate exponent.

Question 14

An ideal gas undergoes a reversible process where $P = aV^2$ where a is constant. If the gas has $k = 1.25$ and the volume changes from $V_1$ to $2V_1$ , what is the work done per unit mass?

$\frac{7aV_1^3}{3m}$ (correct answer)
$\frac{3aV_1^3}{m}$
$\frac{aV_1^3}{m}$
$\frac{2aV_1^3}{m}$
$\frac{5aV_1^3}{2m}$

Explanation: When you encounter a thermodynamics problem with a given pressure-volume relationship, you need to integrate to find the work done. For any reversible process, work is calculated as

W = \int P \, dV

. Given

P = aV^2

, you can substitute and integrate:

W = \int_{V_1}^{2V_1} aV^2 \, dV = a \int_{V_1}^{2V_1} V^2 \, dV = a \left[\frac{V^3}{3}\right]_{V_1}^{2V_1}

Evaluating the limits:

W = \frac{a}{3}[(2V_1)^3 - V_1^3] = \frac{a}{3}[8V_1^3 - V_1^3] = \frac{7aV_1^3}{3}

For work per unit mass, divide by mass

m

\frac{W}{m} = \frac{7aV_1^3}{3m}

This confirms answer A is correct. Answer B (

\frac{3aV_1^3}{m}

) results from incorrectly computing

(2V_1)^3 - V_1^3 = 9V_1^3

instead of

7V_1^3

, likely confusing

2^3 = 8

with some other calculation. Answer C (

\frac{aV_1^3}{m}

) comes from forgetting to evaluate the definite integral properly, possibly just using

V_1^3

without considering the volume change. Answer D (

\frac{2aV_1^3}{m}

) suggests using an incorrect formula, perhaps mistakenly thinking the volume doubled so the work should involve a factor of 2, ignoring the cubic relationship in the integration. Study tip: Always set up the integral

W = \int P \, dV

carefully and evaluate the definite integral completely. The volume relationship affects the final result significantly due to the power in the pressure equation.

Question 15

For an ideal gas undergoing a constant pressure process, the ratio of work done to heat added is 2/7. What is the specific heat ratio k for this gas?

1.40 (correct answer)
1.29
1.67
1.25
1.33

Explanation: When analyzing isentropic or constant pressure processes for ideal gases, you'll often need to connect work, heat, and the specific heat ratio through fundamental thermodynamic relationships. For a constant pressure process, the work done is

W = nR\Delta T

and the heat added is

Q = nC_p\Delta T

, where

C_p

is the molar heat capacity at constant pressure. The given ratio becomes:

\frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} = \frac{2}{7}

This means

C_p = \frac{7R}{2}

. For an ideal gas, we know that

C_p - C_v = R

, so:

C_v = C_p - R = \frac{7R}{2} - R = \frac{5R}{2}

The specific heat ratio is:

k = \frac{C_p}{C_v} = \frac{\frac{7R}{2}}{\frac{5R}{2}} = \frac{7}{5} = 1.40

Therefore, A) 1.40 is correct. B) 1.29 would result from incorrectly calculating the heat capacities or making algebraic errors in the ratio manipulation. C) 1.67 corresponds to a monatomic gas (

k = \frac{5}{3}

), which would give a different work-to-heat ratio than the given 2/7. D) 1.25 (or

k = \frac{5}{4}

) would arise from reversing the work-to-heat ratio or other computational mistakes. Remember: For constant pressure processes, always start with the fundamental definitions of work and heat, then use the relationship

C_p - C_v = R

to connect everything to the specific heat ratio.

Question 16

An ideal gas undergoes an isothermal expansion followed by an adiabatic compression back to its original volume. If the gas has $k = 1.25$ , and the pressure after isothermal expansion is half the initial pressure, what is the final pressure relative to the initial pressure?

0.50
0.84
1.19 (correct answer)
2.00
1.00

Explanation: This problem tests your understanding of how pressure changes through combined thermodynamic processes. When you see questions involving sequential processes, you need to track the state variables step by step. Let's work through each process systematically. During the isothermal expansion, temperature stays constant, so

PV = \text{constant}

. If the pressure drops to half the initial value (

P_2 = 0.5P_1

), then the volume must double to maintain this relationship:

V_2 = 2V_1

. For the adiabatic compression back to the original volume, no heat is exchanged, so

PV^k = \text{constant}

. Starting from state 2 with pressure

0.5P_1

and volume

2V_1

, we compress back to

V_1

. Using the adiabatic relationship:

P_2V_2^k = P_3V_3^k

0.5P_1(2V_1)^{1.25} = P_3V_1^{1.25}

0.5P_1 \times 2^{1.25} = P_3

P_3 = 0.5P_1 \times 2.38 = 1.19P_1

So the final pressure is 1.19 times the initial pressure, making C correct. Option A (0.50) incorrectly assumes the pressure remains at the post-expansion value, ignoring the adiabatic compression. Option B (0.84) likely results from using the wrong exponent or calculation error. Option D (2.00) might come from incorrectly doubling the pressure during compression without proper consideration of the adiabatic relationship. Study tip: For multi-step thermodynamic problems, always identify each process type first, then apply the appropriate equation sequentially. Keep track of which variables stay constant in each step.

Question 17

An ideal gas undergoes a cycle consisting of: (1) isothermal compression from state 1 to 2, (2) constant volume heating from state 2 to 3, and (3) adiabatic expansion from state 3 back to 1. If $V_2/V_1 = 0.25$ and $T_3/T_1 = 2$ , what is the pressure ratio $P_3/P_1$ for a gas with $k = 1.35$ ?

2.0
8.0 (correct answer)
4.0
6.0
3.0

Explanation: When you encounter a thermodynamic cycle problem, systematically apply the appropriate gas laws to each process, then connect the states using the given ratios. For the isothermal compression (1→2), temperature remains constant, so

P_1V_1 = P_2V_2

. Since

V_2/V_1 = 0.25

, we get

P_2/P_1 = V_1/V_2 = 4

. For the constant volume heating (2→3), volume stays fixed while temperature changes. Using the ideal gas law ratio:

P_3/P_2 = T_3/T_2

. Since the isothermal process kept temperature constant,

T_2 = T_1

, so

P_3/P_2 = T_3/T_1 = 2

. Combining both steps:

P_3/P_1 = (P_3/P_2) \times (P_2/P_1) = 2 \times 4 = 8

. The answer is (B) 8.0. (A) 2.0 represents only the temperature ratio

T_3/T_1

, ignoring the pressure change from isothermal compression. (C) 4.0 gives only the pressure ratio from the isothermal process, missing the additional pressure increase from heating. (D) 6.0 might result from incorrectly adding rather than multiplying the ratios, or from computational errors. The key insight is recognizing that you need to track how pressure changes through the entire path from state 1 to state 3, not just focus on individual processes. Always multiply sequential pressure ratios when working through multi-step thermodynamic processes, and remember that the given

k = 1.35

value would only be needed if you had to use the adiabatic relations—here it's extra information.

Question 18

An ideal gas with unknown properties undergoes a constant pressure heating process. If the work done by the gas is 300 J/kg and the heat supplied is 1200 J/kg, what is the value of $k$ for this gas?

$k = 1.33$ (correct answer)
$k = 1.25$
$k = 1.40$
$k = 4.00$

Explanation: For a constant pressure process:

Q = c_p \Delta T

and

W = R \Delta T = (c_p - c_v) \Delta T

. From the first law:

\Delta U = Q - W = c_v \Delta T

. Therefore:

\frac{W}{Q} = \frac{(c_p - c_v) \Delta T}{c_p \Delta T} = \frac{c_p - c_v}{c_p} = 1 - \frac{c_v}{c_p} = 1 - \frac{1}{k} = \frac{k-1}{k}

. Given

W = 300

J/kg and

Q = 1200

J/kg:

\frac{300}{1200} = \frac{1}{4} = \frac{k-1}{k}

. Solving:

k = 4(k-1)

, so

k = 4k - 4

, giving

3k = 4

, therefore

k = \frac{4}{3} = 1.33

. Choice B miscalculates the algebra. Choice C assumes standard diatomic gas. Choice D uses the reciprocal of the work-to-heat ratio.

Question 19

An ideal gas mixture consists of 60% nitrogen ( $k_1 = 1.40$ ) and 40% carbon dioxide ( $k_2 = 1.30$ ) by volume. If the mixture undergoes an isentropic compression where the volume decreases by a factor of 10, what is the ratio of final to initial temperature?

$\frac{T_f}{T_i} = 10^{0.357} = 2.28$ (using $k_{mix} = 1.357$ ) (correct answer)
$\frac{T_f}{T_i} = 10^{0.35} = 2.24$ (using $k_{mix} = 1.35$ )
$\frac{T_f}{T_i} = 10^{0.286} = 1.93$ (using $k_{mix} = 1.40$ )
$\frac{T_f}{T_i} = 10^{0.30} = 2.00$ (using $k_{mix} = 1.30$ )

Explanation: For a gas mixture by volume, the effective

k

can be approximated using:

k_{mix} = \frac{\sum x_i k_i c_{p,i}}{\sum x_i c_{p,i}}

where

x_i

is the mole fraction. For similar molecular weights, this simplifies to:

k_{mix} ≈ \sum x_i k_i = 0.6(1.40) + 0.4(1.30) = 0.84 + 0.52 = 1.36

. More precisely:

k_{mix} = 1.357

. For isentropic compression:

T_f/T_i = (V_i/V_f)^{k-1} = 10^{1.357-1} = 10^{0.357} = 2.28

. Choice B uses slight approximation. Choice C incorrectly uses pure nitrogen properties. Choice D incorrectly uses pure CO₂ properties.

Question 20

For an ideal gas with $c_p = 1.005$ kJ/kg·K and $c_v = 0.718$ kJ/kg·K, if the gas undergoes a polytropic process with index $n = 1.2$ , what is the ratio of the molar heat capacity of this process to $c_v$ ?

$\frac{c_n}{c_v} = 0.436$
$\frac{c_n}{c_v} = -0.436$ (correct answer)
$\frac{c_n}{c_v} = 2.29$
$\frac{c_n}{c_v} = 1.40$

Explanation: First, find

k = c_p/c_v = 1.005/0.718 = 1.40

. For a polytropic process,

c_n = c_v \frac{k-n}{1-n} = 0.718 \frac{1.40-1.2}{1-1.2} = 0.718 \frac{0.2}{-0.2} = -0.718

. Therefore,

c_n/c_v = -0.718/0.718 = -0.436

. The negative sign indicates heat is removed during compression or added during expansion in this process. Choice A omits the crucial negative sign. Choice C uses the reciprocal relationship incorrectly. Choice D simply gives the value of k.