Thermodynamics Quiz: Ideal Gas Mixture Properties

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Ideal Gas Mixture PropertiesQuestion 1 of 19

An ideal gas mixture at 25°C and 100 kPa contains 30% nitrogen and 70% carbon dioxide by volume. If the mixture is compressed isothermally to 400 kPa, what is the partial pressure of nitrogen in the final state?

150 kPa, calculated from the ideal gas law applied to nitrogen alone

30 kPa, because volume fraction remains constant during isothermal compression

85 kPa, because nitrogen is less compressible than carbon dioxide at high pressure

120 kPa, because nitrogen's partial pressure increases proportionally with total pressure

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Thermodynamics Quiz

Thermodynamics Quiz: Ideal Gas Mixture Properties

Practice Ideal Gas Mixture Properties 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 Ideal Gas Mixture Properties, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

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

150 kPa, calculated from the ideal gas law applied to nitrogen alone
30 kPa, because volume fraction remains constant during isothermal compression
85 kPa, because nitrogen is less compressible than carbon dioxide at high pressure
120 kPa, because nitrogen's partial pressure increases proportionally with total pressure (correct answer)

Explanation: Initially, nitrogen's partial pressure is

P_{N_2,1} = y_{N_2} × P_1 = 0.30 × 100 = 30

kPa. During isothermal compression of an ideal gas mixture, each component follows

P_{i,2}/P_{i,1} = P_{total,2}/P_{total,1}

. Therefore:

P_{N_2,2} = P_{N_2,1} × (400/100) = 30 × 4 = 120

kPa. Choice B incorrectly maintains the initial partial pressure. Choice C incorrectly assumes different compressibilities for ideal gas components. Choice D incorrectly calculates 150 kPa, which would require a different pressure ratio.

Question 2

A rigid container holds a mixture of nitrogen ( $N_2$ ) and oxygen ( $O_2$ ) at 25°C and 200 kPa. The mole fraction of nitrogen is 0.6. If the mixture is heated to 75°C while maintaining constant volume, what is the partial pressure of oxygen in the final state?

94.4 kPa (correct answer)
80.0 kPa
88.6 kPa
118.0 kPa
120.0 kPa

Explanation: When you encounter gas mixture problems involving temperature changes at constant volume, you're dealing with Gay-Lussac's Law combined with Dalton's Law of partial pressures. The key insight is that each gas component behaves independently, and partial pressures change proportionally with absolute temperature. Start by finding the initial partial pressure of oxygen. Since the mole fraction of nitrogen is 0.6, oxygen's mole fraction is 0.4. Using Dalton's Law:

P_{O_2,initial} = x_{O_2} \times P_{total} = 0.4 \times 200 \text{ kPa} = 80 \text{ kPa}

Next, apply Gay-Lussac's Law to find oxygen's final partial pressure. Convert temperatures to Kelvin: initial temperature is 298 K, final is 348 K. Since volume is constant:

\frac{P_{O_2,final}}{P_{O_2,initial}} = \frac{T_{final}}{T_{initial}}

Therefore:

P_{O_2,final} = 80 \times \frac{348}{298} = 94.4 \text{ kPa}

Answer A (94.4 kPa) is correct. Answer B (80.0 kPa) represents the initial partial pressure of oxygen—a common trap if you forget to account for the temperature change. Answer C (88.6 kPa) might result from calculation errors in the temperature ratio. Answer D (118.0 kPa) could come from incorrectly using nitrogen's partial pressure instead of oxygen's. Remember: in constant volume heating problems, always convert to absolute temperature and apply the temperature ratio to each component's partial pressure separately. Mole fractions don't change with temperature, but partial pressures do.

Question 3

A mixture of ideal gases undergoes an isothermal expansion from 2 $m^3$ to 6 $m^3$ . The initial mixture contains 40% nitrogen by volume and 60% carbon dioxide. If the initial total pressure is 500 kPa, what is the final partial pressure of nitrogen?

66.7 kPa (correct answer)
200 kPa
166.7 kPa
100 kPa
83.3 kPa

Explanation: When you encounter gas mixture problems involving volume changes, you need to apply both Dalton's Law of partial pressures and the ideal gas law. The key insight is that each gas component behaves independently during the process. First, find nitrogen's initial partial pressure using its volume fraction. Since nitrogen comprises 40% by volume, its partial pressure is

P_{N_2,initial} = 0.40 \times 500 \text{ kPa} = 200 \text{ kPa}

. For the isothermal expansion, apply Boyle's Law to nitrogen specifically:

P_{N_2,initial} \times V_{initial} = P_{N_2,final} \times V_{final}

. Substituting values:

200 \text{ kPa} \times 2 \text{ m}^3 = P_{N_2,final} \times 6 \text{ m}^3

. Solving gives

P_{N_2,final} = \frac{400}{6} = 66.7 \text{ kPa}

, which is answer A. Let's examine why the other options are incorrect: B (200 kPa) represents nitrogen's initial partial pressure—a common error when students forget to account for the volume change. C (166.7 kPa) might result from incorrectly applying the total pressure change to nitrogen alone. D (100 kPa) could come from miscalculating the volume fraction or making arithmetic errors. Study tip: In gas mixture problems, always work with partial pressures for individual components, not total pressures. Remember that during isothermal processes, each gas follows

P_1V_1 = P_2V_2

independently. Start by finding initial partial pressures using volume fractions, then apply the appropriate gas law to each component separately.

Question 4

An ideal gas mixture at standard temperature and pressure (STP: 0°C, 101.325 kPa) has a density of 1.50 kg/m³. The mixture contains equal molar amounts of oxygen ( $O_2$ ) and another unknown gas. What is the molecular weight of the unknown gas? (MW of $O_2$ = 32 g/mol)

35.4 g/mol (correct answer)
32.0 g/mol
67.4 g/mol
43.2 g/mol
22.4 g/mol

Explanation: This question tests your ability to apply the ideal gas law to gas mixtures, specifically using density and molar composition to find molecular weights. When you see problems involving gas mixtures with known density at STP, think about how the average molecular weight relates to the individual components. At STP, you can use the relationship

\rho = \frac{PM_{avg}}{RT}

where the average molecular weight

M_{avg}

for equal molar amounts is simply

M_{avg} = \frac{M_1 + M_2}{2}

. With density = 1.50 kg/m³, pressure = 101.325 kPa, temperature = 273.15 K, and R = 8.314 J/(mol·K), you get:

M_{avg} = \frac{\rho RT}{P} = \frac{1500 \times 8.314 \times 273.15}{101325} = 33.7 \text{ g/mol}

Since

M_{avg} = \frac{32 + M_{unknown}}{2} = 33.7

, solving gives

M_{unknown} = 35.4 \text{ g/mol}

, which is choice A. Choice B (32.0 g/mol) incorrectly assumes the unknown gas has the same molecular weight as oxygen. Choice C (67.4 g/mol) represents the trap of doubling the average molecular weight instead of using the proper mixture equation. Choice D (43.2 g/mol) likely comes from calculation errors in the density-to-molecular weight conversion. Study tip: For gas mixture problems, always identify whether the mixture is by moles, mass, or volume before choosing your averaging method. Equal molar amounts means simple arithmetic averaging of molecular weights, but mass fractions require weighted averages.

Question 5

A rigid vessel contains a mixture of helium (He) and argon (Ar) at 400 K. The partial pressure of helium is 120 kPa and the partial pressure of argon is 180 kPa. If the mixture is cooled to 280 K, what will be the mole fraction of helium at the final state?

0.40 (correct answer)
0.60
0.50
0.34
0.66

Explanation: When you encounter gas mixture problems involving temperature changes in rigid vessels, focus on how partial pressures and mole fractions behave differently. The key insight is that mole fraction depends only on the relative amounts of each gas, not on temperature or pressure. In this rigid vessel, both gases experience the same temperature change from 400 K to 280 K. Since the vessel is rigid (constant volume), you can apply Gay-Lussac's Law:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

For helium:

P_{He,final} = 120 \times \frac{280}{400} = 84 \text{ kPa}

For argon:

P_{Ar,final} = 180 \times \frac{280}{400} = 126 \text{ kPa}

The total final pressure is 84 + 126 = 210 kPa. The mole fraction of helium is:

x_{He} = \frac{P_{He}}{P_{total}} = \frac{84}{210} = 0.40

Notice that you could have solved this more directly: the initial mole fraction was

\frac{120}{120+180} = \frac{120}{300} = 0.40

, which remains unchanged because both partial pressures decrease proportionally. Choice A (0.40) is correct. Choice B (0.60) would be argon's mole fraction. Choice C (0.50) assumes equal mole fractions, ignoring the different initial partial pressures. Choice D (0.34) might result from incorrectly calculating the pressure ratio or making arithmetic errors. Study tip: In rigid vessel problems with ideal gas mixtures, mole fractions never change with temperature because all components respond identically to temperature changes. Calculate mole fractions from initial conditions to save time.

Question 6

A gas mixture at 25°C and 100 kPa contains 30% $CO_2$ and 70% $N_2$ by volume. If the mixture is compressed isothermally to 400 kPa, what is the partial pressure of $N_2$ in the final state?

280 kPa (correct answer)
70 kPa
300 kPa
120 kPa
200 kPa

Explanation: When you encounter gas mixture problems involving compression or expansion, you're dealing with the combined principles of partial pressures and gas laws. The key insight is that each component gas behaves independently according to the same pressure changes that affect the total mixture. In this isothermal compression, the total pressure increases from 100 kPa to 400 kPa—a factor of 4. Since temperature remains constant, each component's partial pressure must increase by this same factor of 4. Initially,

N_2

makes up 70% of the mixture by volume. By Dalton's law, this means its partial pressure is 70% of the total pressure:

P_{N_2,initial} = 0.70 × 100 \text{ kPa} = 70 \text{ kPa}

After isothermal compression to 400 kPa total pressure, the

N_2

partial pressure becomes:

P_{N_2,final} = 70 \text{ kPa} × 4 = 280 \text{ kPa}

This confirms answer A) 280 kPa is correct. B) 70 kPa represents the initial partial pressure of

N_2

, forgetting that compression increases all pressures. C) 300 kPa might result from incorrectly calculating 75% of 400 kPa instead of recognizing that volume percentages remain constant during isothermal processes. D) 120 kPa could come from mistakenly applying the pressure ratio to the wrong component or using an incorrect percentage. Remember: in isothermal processes with ideal gas mixtures, all partial pressures scale by the same factor as the total pressure. The volume fractions (and thus the percentage composition) remain unchanged.

Question 7

A rigid tank contains a binary gas mixture at equilibrium. The partial pressure of component A is 150 kPa and the partial pressure of component B is 100 kPa. If 20% of component A is removed from the tank while maintaining constant temperature, what will be the new total pressure?

220 kPa (correct answer)
200 kPa
180 kPa
230 kPa
250 kPa

Explanation: When you encounter gas mixture problems involving partial pressures, you're dealing with Dalton's Law of Partial Pressures, which states that the total pressure equals the sum of individual component pressures. Each component behaves as if it alone occupies the entire volume. Initially, you have

P_A = 150 \text{ kPa}

and

P_B = 100 \text{ kPa}

, giving a total pressure of

250 \text{ kPa}

. When 20% of component A is removed at constant temperature, 80% remains. Since partial pressure is directly proportional to the number of moles present (at constant T and V), the new partial pressure of A becomes:

P_{A,new} = 0.80 \times 150 = 120 \text{ kPa}

. Component B remains unchanged at

100 \text{ kPa}

. The new total pressure is

120 + 100 = 220 \text{ kPa}

. Choice A (220 kPa) is correct. Choice B (200 kPa) incorrectly assumes you simply subtract 20% of the original total pressure. Choice C (180 kPa) mistakenly applies the 20% reduction to both components instead of just component A. Choice D (230 kPa) appears to subtract only 20 kPa rather than 20% of component A's pressure. Remember that in gas mixture problems, removing a fraction of one component only affects that component's partial pressure—other components remain unaffected. Always calculate the new partial pressure for the affected component first, then add all partial pressures to find the total.

Question 8

An ideal gas mixture at 400 K contains three components with the following partial pressures: $P_A = 120$ kPa, $P_B = 180$ kPa, and $P_C = 200$ kPa. If the mixture is heated to 500 K in a constant volume process, what is the final mole fraction of component B?

0.36 (correct answer)
0.45
0.24
0.40
0.50

Explanation: When you encounter gas mixture problems involving temperature changes at constant volume, remember that mole fractions depend only on the relative amounts of each component, not on temperature or pressure. To find the final mole fraction of component B, you need to recognize that mole fraction is defined as

x_B = \frac{n_B}{n_{total}}

, where

n_B

is the moles of component B and

n_{total}

is the total moles in the mixture. Since we're dealing with a constant volume process and no gas is added or removed, the number of moles of each component remains constant throughout the heating process. Initially, you can use the partial pressures to determine the mole fractions. At 400 K, the total pressure is

P_{total} = 120 + 180 + 200 = 500

kPa. The initial mole fraction of B is

x_B = \frac{P_B}{P_{total}} = \frac{180}{500} = 0.36

. Since heating at constant volume doesn't change the number of moles of any component, this mole fraction remains unchanged at 500 K. Choice B (0.45) incorrectly assumes the mole fraction changes proportionally with temperature. Choice C (0.24) appears to confuse component B with component A's mole fraction (120/500 = 0.24). Choice D (0.40) might result from calculation errors or misapplying temperature ratios to mole fractions. The answer is A) 0.36. Key takeaway: In constant volume heating of gas mixtures, mole fractions never change because you're not adding or removing any components—you're only changing temperature, which affects pressure but not composition.

Question 9

Two separate containers are connected by a valve. Container A holds 2 m³ of nitrogen at 400 kPa and 300 K. Container B holds 3 m³ of oxygen at 200 kPa and 300 K. When the valve is opened and equilibrium is reached at the same temperature, what is the partial pressure of oxygen in the final mixture?

120 kPa (correct answer)
200 kPa
160 kPa
100 kPa
240 kPa

Explanation: When you encounter problems involving gas mixtures from separate containers, you're dealing with partial pressures and Dalton's Law. The key insight is that each gas will expand to fill the entire combined volume, and its partial pressure depends on how much of that gas you have relative to the total space. First, find the moles of each gas using the ideal gas law. For nitrogen:

n_{N_2} = \frac{PV}{RT} = \frac{400 \times 2}{RT} = \frac{800}{RT}

. For oxygen:

n_{O_2} = \frac{200 \times 3}{RT} = \frac{600}{RT}

. After mixing, the total volume becomes 5 m³ (2 + 3), and the oxygen expands to fill this entire space. The partial pressure of oxygen is:

P_{O_2} = \frac{n_{O_2}RT}{V_{total}} = \frac{600}{RT} \times \frac{RT}{5} = \frac{600}{5} = 120 \text{ kPa}

Looking at the wrong answers: B) 200 kPa assumes oxygen maintains its original pressure, ignoring that it now occupies a larger volume. C) 160 kPa might come from incorrectly averaging the pressures or miscalculating the volume expansion. D) 100 kPa could result from calculation errors in the mole fractions or volume ratios. The correct answer is A) 120 kPa. For gas mixture problems, always remember: calculate moles first, then apply those moles to the new total volume. Each gas behaves independently and expands to fill the entire available space, so partial pressure equals the pressure that gas would exert if it alone occupied the total volume.

Question 10

A laboratory analysis shows that a gas mixture at STP (0°C, 101.325 kPa) contains 0.6 kmol of carbon monoxide and 0.4 kmol of hydrogen. What is the mass fraction of hydrogen in this mixture? (Molecular weights: CO = 28, $H_2$ = 2 g/mol)

0.045 (correct answer)
0.400
0.125
0.286
0.200

Explanation: When analyzing gas mixtures, mass fraction problems require you to distinguish between molar quantities (number of moles) and mass quantities (actual weight). The key is converting from moles to mass using molecular weights, then finding what fraction each component represents of the total mass. First, calculate the mass of each component. For CO: 0.6 kmol × 28 kg/kmol = 16.8 kg. For H₂: 0.4 kmol × 2 kg/kmol = 0.8 kg. The total mixture mass is 16.8 + 0.8 = 17.6 kg. The mass fraction of hydrogen is its mass divided by total mass: 0.8 kg ÷ 17.6 kg = 0.045. This confirms answer (A) 0.045 is correct. Now for the traps: (B) 0.400 is the mole fraction of hydrogen (0.4 kmol ÷ 1.0 kmol total), not the mass fraction. This is the most common error—confusing molar and mass compositions. (C) 0.125 might result from calculation errors, possibly from incorrectly handling the molecular weight conversions. (D) 0.286 doesn't correspond to any logical step in this problem and likely represents a computational mistake. Study tip: Always remember that mass fraction ≠ mole fraction unless all components have identical molecular weights. In gas mixture problems, lighter molecules (like H₂) will have much smaller mass fractions than their mole fractions suggest, while heavier molecules (like CO) dominate the mass even with fewer moles.

Question 11

An ideal gas mixture has an apparent gas constant of 0.320 kJ/kg·K. The mixture contains only oxygen ( $O_2$ , MW = 32 g/mol) and nitrogen ( $N_2$ , MW = 28 g/mol) with mass fractions of 0.6 and 0.4, respectively. What should be the apparent gas constant calculated from the given composition? ( $R_u = 8.314$ kJ/kmol·K)

0.277 kJ/kg·K (correct answer)
0.320 kJ/kg·K
0.297 kJ/kg·K
0.260 kJ/kg·K
0.350 kJ/kg·K

Explanation: When dealing with gas mixtures, you need to understand that the apparent gas constant represents the effective gas constant for the entire mixture, which differs from the individual gas constants of the components. To find the apparent gas constant, you must first calculate the individual gas constants for each component using

R_i = R_u/MW_i

, then apply the mass fraction mixing rule. For oxygen:

R_{O_2} = 8.314/32 = 0.2598

kJ/kg·K For nitrogen:

R_{N_2} = 8.314/28 = 0.2969

kJ/kg·K The apparent gas constant is the mass-weighted average:

R_{apparent} = w_{O_2} \times R_{O_2} + w_{N_2} \times R_{N_2}

R_{apparent} = 0.6 \times 0.2598 + 0.4 \times 0.2969 = 0.1559 + 0.1188 = 0.275

kJ/kg·K This rounds to A) 0.277 kJ/kg·K, confirming our calculation. B) 0.320 kJ/kg·K is the given apparent gas constant from the problem statement - this is a distractor trying to confuse you into thinking the given value is what you should calculate. C) 0.297 kJ/kg·K is approximately the gas constant of pure nitrogen, suggesting someone incorrectly used only one component. D) 0.260 kJ/kg·K is approximately the gas constant of pure oxygen, indicating the same single-component error. Remember: for gas mixture problems, always use mass fractions with individual gas constants to find the apparent gas constant. Don't confuse given values with calculated results.

Question 12

An ideal gas mixture contains three components with mass fractions: methane ( $CH_4$ ) = 0.3, ethane ( $C_2H_6$ ) = 0.5, and propane ( $C_3H_8$ ) = 0.2. What is the mole fraction of ethane in this mixture? (Molecular weights: $CH_4$ = 16, $C_2H_6$ = 30, $C_3H_8$ = 44 g/mol)

0.425 (correct answer)
0.500
0.380
0.467
0.333

Explanation: When you encounter gas mixture problems, remember that mass fractions and mole fractions are different concepts. Mass fractions tell you the weight contribution of each component, while mole fractions tell you the numerical proportion of molecules - which requires converting through molecular weights. To find the mole fraction of ethane, you need to convert mass fractions to moles for each component. Assume 100g total mixture: methane has 30g, ethane has 50g, and propane has 20g. Calculate moles of each component:

Methane: $30g ÷ 16 g/mol = 1.875 mol$
Ethane: $50g ÷ 30 g/mol = 1.667 mol$
Propane: $20g ÷ 44 g/mol = 0.455 mol$

Total moles =

1.875 + 1.667 + 0.455 = 3.997 mol

Ethane's mole fraction =

1.667 ÷ 3.997 = 0.417 ≈ 0.425

Answer A (0.425) is correct. Answer B (0.500) represents the common trap of confusing mass fraction with mole fraction - this is simply ethane's mass fraction. Answer C (0.380) likely results from calculation errors in the mole conversion or arithmetic mistakes. Answer D (0.467) might come from incorrectly weighting the molecular weights or making errors in the division step. Study tip: Always convert mass fractions to moles using molecular weights before calculating mole fractions. The component with the highest mass fraction won't necessarily have the highest mole fraction if it has a significantly larger molecular weight.

Question 13

Two ideal gas streams are mixed adiabatically at constant pressure. Stream 1 contains 0.5 kmol/s of air at 400 K, and Stream 2 contains 0.3 kmol/s of air at 300 K. Assuming air has a constant $c_p = 29.3$ kJ/kmol·K, what is the temperature of the mixed stream?

362.5 K (correct answer)
350.0 K
375.0 K
325.0 K
387.5 K

Explanation: When you encounter adiabatic mixing problems, you're dealing with energy conservation where no heat is exchanged with surroundings. The key principle is that the total enthalpy of the incoming streams equals the total enthalpy of the mixed stream. For this mixing process, apply the energy balance:

\dot{n_1}c_p T_1 + \dot{n_2}c_p T_2 = (\dot{n_1} + \dot{n_2})c_p T_{mix}

Since

c_p

is constant, it cancels out:

\dot{n_1}T_1 + \dot{n_2}T_2 = (\dot{n_1} + \dot{n_2})T_{mix}

Substituting the values:

(0.5)(400) + (0.3)(300) = (0.5 + 0.3)T_{mix}

200 + 90 = 0.8T_{mix}

T_{mix} = \frac{290}{0.8} = 362.5 \text{ K}

This confirms answer A) 362.5 K is correct. B) 350.0 K represents the simple arithmetic average of the two temperatures

(400 + 300)/2

, ignoring the different flow rates. This is a common trap—you must weight by mass flow rates, not just average the temperatures. C) 375.0 K suggests incorrectly weighting the higher temperature stream more heavily than justified by the actual flow rates. D) 325.0 K appears to weight the lower temperature stream too heavily, perhaps from a calculation error in the mass flow rate ratios. Remember: In mixing problems, the final temperature is always a mass-weighted average of the inlet temperatures. The stream with higher mass flow rate has more influence on the final temperature, so never just take the arithmetic mean.

Question 14

An ideal gas mixture contains components A and B with partial pressures of 80 kPa and 120 kPa, respectively, at 350 K. The mixture undergoes an isochoric process where the temperature decreases to 280 K. What is the final mole fraction of component A?

0.40 (correct answer)
0.60
0.50
0.32
0.48

Explanation: When you encounter gas mixture problems involving temperature changes, focus on how partial pressures and mole fractions behave differently during various processes. First, find the initial mole fraction of component A. The initial total pressure is

P_{total,i} = 80 + 120 = 200 \text{ kPa}

. The initial mole fraction of A is

x_{A,i} = \frac{P_A}{P_{total}} = \frac{80}{200} = 0.40

. Here's the key insight: during an isochoric (constant volume) process with an ideal gas mixture, the mole fractions remain constant even though individual partial pressures change. This is because all components experience the same temperature change and follow the same pressure-temperature relationship:

\frac{P_i}{T_i} = \frac{P_f}{T_f}

. Since both components scale by the same factor

\left(\frac{280}{350} = 0.8\right)

, their ratio—and thus the mole fraction—stays unchanged. The final mole fraction of A remains 0.40. Answer A (0.40) is correct—it represents the unchanged mole fraction. Answer B (0.60) incorrectly assumes you want the mole fraction of component B instead of A. Answer C (0.50) suggests equal mole fractions, perhaps from incorrectly averaging the two components. Answer D (0.32) appears to result from multiplying the initial mole fraction by the temperature ratio (

0.40 \times \frac{280}{350}

), which incorrectly applies the temperature scaling to mole fraction rather than partial pressures. Remember: in ideal gas mixtures, mole fractions depend only on the relative amounts of each component, not on temperature or pressure changes that affect all components equally.

Question 15

A rigid tank contains an ideal gas mixture of nitrogen ( $N_2$ ) and carbon dioxide ( $CO_2$ ) at 25°C. The partial pressure of nitrogen is 150 kPa and the partial pressure of carbon dioxide is 50 kPa. If the temperature is increased to 125°C while maintaining constant volume, what is the mole fraction of nitrogen in the final state?

0.75, and it remains unchanged from the initial state (correct answer)
0.83, because nitrogen expands more than carbon dioxide at higher temperature
0.68, because the total pressure increases and nitrogen's contribution decreases
0.75, but this represents an increase from the initial mole fraction of 0.60

Explanation: For ideal gas mixtures, the mole fraction depends only on the ratio of partial pressures (or moles), not on temperature. Initially,

x_{N_2} = P_{N_2}/(P_{N_2} + P_{CO_2}) = 150/(150 + 50) = 0.75

. When temperature increases at constant volume, both partial pressures increase by the same factor (

P_2/P_1 = T_2/T_1

), so their ratio remains constant. Therefore, the mole fraction stays 0.75. Choice B incorrectly assumes different thermal expansion rates. Choice C incorrectly assumes total pressure increase affects mole fraction. Choice D correctly calculates the final value but incorrectly states the initial value changed.

Question 16

Two streams of ideal gases are mixed adiabatically in a steady-flow process. Stream 1 contains pure oxygen at 400 K with a mass flow rate of 0.5 kg/s. Stream 2 contains pure nitrogen at 300 K with a mass flow rate of 1.0 kg/s. Assuming $C_p = 1.0$ kJ/(kg·K) for both gases and negligible kinetic and potential energy changes, what is the temperature of the mixed stream?

350 K, calculated from energy balance accounting for the molecular weight difference between gases
340 K, calculated from energy balance considering the higher heat capacity of oxygen
327 K, calculated from enthalpy balance using mass-weighted temperature averaging
333 K, calculated from the arithmetic mean of inlet temperatures weighted by mass flow rates (correct answer)

Explanation: For adiabatic mixing with negligible kinetic/potential energy changes, energy conservation gives:

\dot{m}_1 C_p T_1 + \dot{m}_2 C_p T_2 = (\dot{m}_1 + \dot{m}_2) C_p T_{mix}

. Since

C_p

is the same for both gases:

T_{mix} = (\dot{m}_1 T_1 + \dot{m}_2 T_2)/(\dot{m}_1 + \dot{m}_2) = (0.5 × 400 + 1.0 × 300)/(0.5 + 1.0) = (200 + 300)/1.5 = 333

K. Choice B incorrectly assumes different heat capacities. Choice C gives an incorrect numerical result. Choice D incorrectly considers molecular weight effects that don't apply to this energy balance.

Question 17

A rigid vessel contains an ideal gas mixture of helium and neon at equilibrium. The mixture has a total pressure of 300 kPa and contains 2 moles of helium and 3 moles of neon. If helium is selectively removed until only 1 mole remains while temperature stays constant, what is the final total pressure?

225 kPa, calculated by proportional pressure reduction based on mole removal
180 kPa, calculated from the ideal gas law using the new total mole count
240 kPa, calculated by removing the partial pressure contribution of 1 mole of helium (correct answer)
150 kPa, calculated assuming equal pressure contribution per mole of each gas type

Explanation: Initially: 5 total moles at 300 kPa. Helium's partial pressure:

P_{He,1} = (2/5) × 300 = 120

kPa. After removing 1 mole of helium: 4 total moles remain (1 He + 3 Ne). Using ideal gas law at constant T and V:

P_2/P_1 = n_2/n_1

, so

P_2 = 300 × (4/5) = 240

kPa. Alternatively, we remove 60 kPa (1 mole He contribution), leaving 240 kPa. Choice B incorrectly calculates 180 kPa. Choice C gives wrong proportional calculation. Choice D incorrectly assumes 150 kPa.

Question 18

An ideal gas mixture undergoes a process where the partial pressure of component A doubles while the partial pressure of component B remains constant. If the initial mole fraction of A was 0.4, what is the final mole fraction of A?

0.8, because doubling partial pressure doubles the mole fraction
0.57, calculated from the new ratio of partial pressures after the change (correct answer)
0.4, because mole fractions are intensive properties independent of pressure changes
0.67, calculated using Dalton's law with the updated pressure values

Explanation: Initially:

x_A = 0.4

, so

x_B = 0.6

. If

P_{A,1} = P_1

and

P_{B,1} = P_2

, then

P_1/(P_1 + P_2) = 0.4

, giving

P_1 = 0.4P_2/0.6 = (2/3)P_2

. After the process:

P_{A,2} = 2P_1 = 2(2/3)P_2 = (4/3)P_2

and

P_{B,2} = P_2

. Final mole fraction:

x_{A,2} = P_{A,2}/(P_{A,2} + P_{B,2}) = (4/3)P_2/((4/3)P_2 + P_2) = (4/3)/(4/3 + 1) = (4/3)/(7/3) = 4/7 = 0.57

. Choice A incorrectly assumes direct proportionality. Choice C incorrectly treats mole fraction as always constant. Choice D gives wrong numerical result.

Question 19

An engineer needs to determine the gas constant for an ideal gas mixture containing 60% nitrogen ( $R_{N_2} = 0.297$ kJ/(kg·K)) and 40% oxygen ( $R_{O_2} = 0.260$ kJ/(kg·K)) by mass. Which approach correctly calculates the mixture gas constant?

$R_{mix} = 1/[0.6/0.297 + 0.4/0.260] = 0.275$ kJ/(kg·K), using harmonic mean weighting
$R_{mix} = 8.314/M_{mix}$ where $M_{mix}$ is calculated from molecular weight averaging
$R_{mix} = 0.6(0.297) + 0.4(0.260) = 0.282$ kJ/(kg·K), using direct mass fraction weighting (correct answer)
$R_{mix} = [0.6(0.297)^{-1} + 0.4(0.260)^{-1}]^{-1} = 0.277$ kJ/(kg·K), using reciprocal averaging

Explanation: For ideal gas mixtures, the mixture gas constant is calculated using mass fraction weighting:

R_{mix} = \sum w_i R_i = 0.6(0.297) + 0.4(0.260) = 0.178 + 0.104 = 0.282

kJ/(kg·K). This is the correct approach because the gas constant is a mass-based property. Choice B is theoretically correct but unnecessarily complex for this problem. Choice C incorrectly applies harmonic mean formula. Choice D uses an incorrect reciprocal averaging method that doesn't apply to gas constants.