Thermodynamics Quiz: Compressibility Factor Z

20 questions · exam conditions

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Compressibility Factor ZQuestion 1 of 20

A compressed gas cylinder shows a pressure reading of 120 atm at 300 K. If the gas has Z = 0.88 under these conditions, what pressure would be measured if the same amount of gas behaved ideally in the same volume and temperature?

136.4 atm

120.0 atm

105.6 atm

88.0 atm

150.0 atm

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Thermodynamics Quiz

Thermodynamics Quiz: Compressibility Factor Z

Practice Compressibility Factor Z 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 Compressibility Factor Z, 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

136.4 atm (correct answer)
120.0 atm
105.6 atm
88.0 atm
150.0 atm

Explanation: When you encounter problems involving real gases and compressibility factors, you're dealing with deviations from ideal gas behavior. The compressibility factor Z relates real gas behavior to ideal gas behavior through the equation

PV = ZnRT

, where Z = 1 for ideal gases and Z ≠ 1 for real gases. Given that the real gas shows 120 atm at Z = 0.88, you can find what pressure an ideal gas would show under identical conditions (same n, V, T). Since

PV = ZnRT

for the real gas and

PV = nRT

for the ideal gas, you can write: For real gas:

P_{real} \cdot V = Z \cdot nRT = 0.88 \cdot nRT

For ideal gas:

P_{ideal} \cdot V = nRT

Therefore:

P_{ideal} = \frac{nRT}{V} = \frac{P_{real}}{Z} = \frac{120 \text{ atm}}{0.88} = 136.4 \text{ atm}

Looking at the wrong answers: B (120.0 atm) incorrectly assumes the pressure would remain the same, ignoring the compressibility factor entirely. C (105.6 atm) represents the common error of multiplying rather than dividing by Z, giving 120 × 0.88. D (88.0 atm) appears to be an arbitrary calculation with no thermodynamic basis. Remember this key relationship: when Z < 1 (as with most real gases at high pressure), the gas is more compressible than ideal, meaning an ideal gas would require higher pressure to occupy the same volume. Always divide the real pressure by Z to find the equivalent ideal pressure.

Question 2

A gas sample at 400 K and 50 atm has a compressibility factor Z = 0.85. If the gas were to behave ideally under these same conditions, what would be the ratio of the actual molar volume to the ideal molar volume?

0.85 (correct answer)
1.18
1.00
0.72
1.85

Explanation: The compressibility factor Z measures how much a real gas deviates from ideal gas behavior. When you see Z in a problem, remember that it directly relates actual and ideal molar volumes through the equation

Z = \frac{V_{actual}}{V_{ideal}}

. Since Z = 0.85 for this gas, you can directly determine that

\frac{V_{actual}}{V_{ideal}} = 0.85

. The compressibility factor is defined precisely as this ratio, so no additional calculations are needed. A Z value less than 1 indicates the gas is more compressible than an ideal gas, meaning intermolecular attractive forces dominate, pulling molecules closer together and reducing the actual volume below what an ideal gas would occupy. Looking at the wrong answers: Choice B (1.18) represents the reciprocal calculation

\frac{1}{0.85}

, which would give you

\frac{V_{ideal}}{V_{actual}}

instead of what the question asks for. Choice C (1.00) would only be correct if the gas behaved ideally (Z = 1), which contradicts the given information. Choice D (0.72) might result from incorrectly manipulating the ideal gas equation or confusing Z with other thermodynamic factors. The correct answer is A) 0.85. Study tip: Always remember that Z is defined as the ratio of actual to ideal molar volume. When Z < 1, attractive forces dominate and the gas occupies less space than predicted by ideal behavior. When Z > 1, repulsive forces dominate and the gas takes up more space. This direct relationship makes compressibility factor problems straightforward once you recognize the definition.

Question 3

For a real gas with Z < 1 at a given temperature and pressure, which statement best explains the dominant intermolecular forces present?

Attractive forces dominate, causing molecules to occupy less space than predicted by ideal gas law (correct answer)
Repulsive forces dominate, causing molecules to occupy more space than predicted by ideal gas law
No intermolecular forces are present, indicating ideal gas behavior throughout the system
Both attractive and repulsive forces are equally balanced, resulting in perfect gas behavior
Attractive forces dominate, causing molecules to occupy more space than predicted by ideal gas law

Explanation: When you encounter questions about real gas behavior and the compressibility factor Z, focus on what deviations from ideal gas behavior tell you about intermolecular forces. The compressibility factor

Z = \frac{PV}{nRT}

compares real gas behavior to ideal gas predictions. Since Z < 1 in this scenario, the real gas occupies less volume than an ideal gas would at the same temperature and pressure. This volume reduction occurs because attractive intermolecular forces (like van der Waals forces) pull molecules closer together, causing the gas to be more compressed than the ideal gas law predicts. Looking at the wrong answers: Option B incorrectly associates Z < 1 with repulsive forces. Repulsive forces would actually cause molecules to spread out more, increasing volume and making Z > 1. Option C is wrong because Z ≠ 1 clearly indicates non-ideal behavior with significant intermolecular forces present. Option D misunderstands the situation entirely—balanced forces would give Z ≈ 1, not Z < 1, and wouldn't result in "perfect" gas behavior anyway. The correct answer is A because attractive forces dominate when Z < 1, pulling molecules closer and reducing the volume below ideal gas predictions. Remember this pattern: Z < 1 means attractive forces dominate (molecules pulled together), while Z > 1 indicates repulsive forces dominate (molecules pushed apart). At high pressures, most real gases show Z < 1 due to attractive forces becoming significant.

Question 4

At constant temperature, as pressure increases for most real gases, the compressibility factor Z typically:

First decreases below 1, reaches a minimum, then increases above 1 at very high pressures (correct answer)
Remains constant at 1 throughout all pressure ranges, indicating perfect ideal gas behavior
Continuously increases linearly with pressure, always remaining above 1 for all real gases
Continuously decreases with pressure, approaching zero at extremely high pressures for all gases
First increases above 1, reaches a maximum, then decreases below 1 at very high pressures

Explanation: When you encounter questions about the compressibility factor Z, you're dealing with how real gases deviate from ideal gas behavior. The compressibility factor is defined as

Z = \frac{PV}{nRT}

, where Z = 1 for an ideal gas, Z < 1 indicates the gas is more compressible than ideal, and Z > 1 means it's less compressible. At constant temperature, real gases exhibit a characteristic behavior as pressure increases. Initially, intermolecular attractive forces (van der Waals forces) dominate, making molecules pull together more than in an ideal gas. This creates greater compressibility, driving Z below 1. As pressure continues increasing, Z reaches a minimum value, then begins rising as repulsive forces between molecules start dominating due to decreased intermolecular distances. At very high pressures, these repulsive forces make the gas less compressible than ideal, pushing Z above 1. Choice B is wrong because no real gas behaves ideally across all pressure ranges – intermolecular forces always cause deviations. Choice C incorrectly suggests Z always stays above 1 and increases linearly, missing the initial attractive-force-dominated region where Z drops below 1. Choice D falsely claims Z continuously decreases toward zero, ignoring the repulsive forces that become dominant at high pressures and cause Z to increase. Remember this pattern: attractive forces dominate at moderate pressures (Z < 1), while repulsive forces dominate at high pressures (Z > 1). The characteristic "dip and rise" behavior of Z versus pressure is fundamental to understanding real gas behavior in thermodynamics.

Question 5

Two different gases at the same temperature and pressure have compressibility factors Z₁ = 0.88 and Z₂ = 1.15. Which statement correctly compares their molecular behavior?

Gas 1 experiences stronger attractive forces while Gas 2 experiences stronger repulsive forces relative to ideal behavior (correct answer)
Gas 1 experiences stronger repulsive forces while Gas 2 experiences stronger attractive forces relative to ideal behavior
Both gases experience identical intermolecular forces since they are at the same temperature and pressure conditions
Gas 1 has larger molecular size while Gas 2 has smaller molecular size compared to point particles
Both gases exhibit ideal behavior since their Z values are close to 1 within experimental uncertainty

Explanation: When you encounter compressibility factor problems, you're analyzing how real gases deviate from ideal gas behavior due to intermolecular forces and molecular size effects. The compressibility factor

Z = \frac{PV}{nRT}

tells you how a real gas compares to an ideal gas. For an ideal gas, Z = 1.0. When Z < 1, the gas occupies less volume than predicted by the ideal gas law, indicating that attractive intermolecular forces are pulling molecules together and dominating the behavior. When Z > 1, the gas occupies more volume than ideal, showing that repulsive forces (often from molecular size effects) are pushing molecules apart and dominating. Gas 1 has Z₁ = 0.88 < 1, meaning attractive forces dominate its behavior. Gas 2 has Z₂ = 1.15 > 1, indicating repulsive forces dominate. This makes choice A correct. Choice B reverses the relationship between Z values and force types - this is a common misconception. Remember: Z < 1 means attraction dominates, Z > 1 means repulsion dominates. Choice C incorrectly assumes that identical temperature and pressure conditions produce identical intermolecular forces. Different molecules have different force characteristics regardless of external conditions. Choice D focuses only on molecular size effects while ignoring attractive forces. While molecular size contributes to repulsive behavior, the complete picture requires considering both attractive and repulsive contributions to determine which dominates. Remember this pattern: Z < 1 = attraction wins, Z > 1 = repulsion wins. This relationship is fundamental for analyzing real gas behavior in thermodynamics problems.

Question 6

At what condition would you expect the compressibility factor Z to be closest to 1.00 for a real gas?

High temperature and low pressure, where intermolecular forces become negligible and molecular volume is insignificant (correct answer)
Low temperature and high pressure, where intermolecular forces are maximized and molecular interactions dominate
High temperature and high pressure, where both attractive and repulsive forces are equally maximized
Low temperature and low pressure, where molecular motion is minimized and gas density approaches liquid density
Moderate temperature and pressure, where attractive forces exactly balance repulsive forces at all molecular distances

Explanation: When you encounter questions about the compressibility factor Z, remember that this measures how much a real gas deviates from ideal gas behavior, where

Z = \frac{PV}{nRT}

. For an ideal gas, Z = 1.00, so you're looking for conditions where real gases behave most like ideal gases. Real gases deviate from ideal behavior due to two main factors: intermolecular attractive forces (which become significant at high pressures and low temperatures) and the finite volume occupied by gas molecules themselves (which matters at high pressures). The closer we can get to eliminating both effects, the closer Z approaches 1.00. Option A is correct because high temperature gives molecules enough kinetic energy to overcome intermolecular attractions, while low pressure means molecules are far apart, making both attractive forces and molecular volume negligible compared to the container volume. Option B is wrong because low temperature and high pressure maximize deviations from ideality—molecules move slowly and are close together, emphasizing both attractive forces and finite molecular size. Option C is incorrect because high pressure alone causes significant deviation from ideal behavior due to molecular crowding, regardless of temperature effects on intermolecular forces. Option D represents conditions approaching condensation, where gas behavior becomes highly non-ideal as molecules pack closely together and attractive forces dominate. Study tip: Remember "Hot and Sparse" for ideal gas conditions. High temperature overcomes intermolecular attractions, while low pressure (sparse molecules) minimizes both crowding effects and attractive interactions. This combination pushes Z closest to 1.00.

Question 7

A gas cylinder contains 5.0 moles of gas at 450 K and 80 atm with Z = 1.12. What would be the pressure if this same amount of gas behaved ideally at the same temperature and volume?

71.4 atm (correct answer)
80.0 atm
89.6 atm
112.0 atm
62.5 atm

Explanation: When you encounter a problem involving the compressibility factor (Z), you're dealing with real gas behavior versus ideal gas behavior. The compressibility factor tells you how much a real gas deviates from ideality, where Z = 1 represents perfect ideal behavior. The key relationship here is

Z = \frac{P_{real}V}{nRT}

, which can be rearranged to

PV = ZnRT

for real gases, compared to

PV = nRT

for ideal gases. Since the problem asks what pressure an ideal gas would have at the same temperature and volume, you need to find the volume first using the real gas data, then calculate the ideal pressure. From the real gas:

V = \frac{ZnRT}{P_{real}} = \frac{(1.12)(5.0)(0.08206)(450)}{80} = 2.59 \text{ L}

For an ideal gas in this same volume:

P_{ideal} = \frac{nRT}{V} = \frac{(5.0)(0.08206)(450)}{2.59} = 71.4 \text{ atm}

Choice A (71.4 atm) is correct—this lower pressure reflects that ideal gases exert less pressure than this particular real gas under these conditions. Choice B (80.0 atm) incorrectly assumes the pressure stays the same regardless of gas behavior. Choice C (89.6 atm) mistakenly multiplies the real pressure by Z instead of dividing. Choice D (112.0 atm) appears to multiply 80 atm by 1.4, showing confusion about how Z relates to pressure calculations. Remember: when Z > 1, the real gas is less compressible than ideal, meaning an ideal gas would occupy more volume at the same pressure, or exert less pressure at the same volume.

Question 8

The compressibility factor for hydrogen gas at 273 K varies with pressure as follows: Z = 1.00 at 1 atm, Z = 0.98 at 50 atm, and Z = 1.05 at 200 atm. What physical phenomenon explains the increase in Z at very high pressure?

Finite molecular volume becomes significant, causing molecules to occupy more space than point particles (correct answer)
Increased kinetic energy at high pressure overcomes all intermolecular attractive forces completely
Temperature-dependent attractive forces become stronger at higher pressures, expanding the gas volume
Molecular collisions become more frequent, increasing the effective gas volume through collision dynamics
Gas molecules begin to dissociate into atoms, increasing the total number of particles in the system

Explanation: When you encounter compressibility factor questions, focus on how real gases deviate from ideal behavior due to two competing molecular effects: intermolecular attractions and finite molecular volume. The compressibility factor

Z = \frac{PV}{nRT}

tells us how real gases differ from ideal gases. At low pressures, hydrogen behaves nearly ideally (Z ≈ 1.00). The dip to Z = 0.98 at 50 atm shows intermolecular attractions dominating—molecules are pulled together, reducing volume below ideal predictions. However, the jump to Z = 1.05 at 200 atm reveals a different effect taking over. At very high pressures, molecules are compressed so tightly that their finite size becomes significant. Real molecules aren't point particles—they have volume. When packed closely, this molecular volume becomes "excluded volume" that can't be compressed further, forcing the gas to occupy more space than an ideal gas would. This makes Z > 1, confirming answer A. Answer B is wrong because kinetic energy depends on temperature, not pressure, and attractive forces aren't "completely overcome." Answer C incorrectly suggests attractive forces expand volume—they actually contract it, and they don't become stronger with pressure alone. Answer D misunderstands collision dynamics; frequent collisions don't increase effective volume. Study tip: Remember the pressure trend for real gases: low pressure (Z ≈ 1, nearly ideal) → moderate pressure (Z < 1, attractions dominate) → high pressure (Z > 1, molecular volume dominates). This pattern appears frequently in thermodynamics problems.

Question 9

A real gas at 500 K and 150 atm has a molar volume that is 15% less than predicted by the ideal gas law. If the temperature is increased to 750 K at constant pressure, and the new molar volume is only 8% less than the ideal prediction, what can be concluded about temperature effects on gas behavior?

Higher temperature reduces the relative importance of intermolecular attractive forces, making gas behavior more ideal (correct answer)
Higher temperature increases the relative importance of intermolecular attractive forces, making gas behavior less ideal
Temperature has no significant effect on gas ideality since both cases show deviation from ideal behavior
Higher temperature increases molecular repulsions, causing greater deviation from ideal behavior at all pressures
Higher temperature causes gas molecules to associate, reducing the effective number of particles in the system

Explanation: When analyzing real gas behavior, focus on how deviations from ideality change with temperature and pressure. Real gases deviate from ideal behavior due to two main factors: intermolecular attractive forces (which make the gas more compressible) and molecular volume/repulsive forces (which make the gas less compressible). The key insight here is in the numbers: at 500 K, the molar volume is 15% less than ideal, but at 750 K, it's only 8% less than ideal. Since both measurements show volumes smaller than predicted by the ideal gas law, intermolecular attractions are the dominant cause of deviation. More importantly, the deviation decreases as temperature increases (from 15% to 8%). At higher temperatures, molecules move faster and spend less time near each other, weakening the relative effect of intermolecular attractions. This makes the gas behave more ideally, confirming answer A. Answer B incorrectly suggests attractions become more important at higher temperature, which contradicts both theory and the observed decrease in deviation. Answer C misses the crucial trend - while both cases show deviation, the magnitude of deviation significantly decreases with temperature. Answer D focuses on repulsions, but repulsions would cause volumes larger than ideal, not smaller as observed here. Study tip: Remember that attractive forces dominate at high pressures (causing V < V_ideal), while repulsive forces dominate at very high pressures or low temperatures (causing V > V_ideal). Higher temperatures always reduce the relative importance of intermolecular forces.

Question 10

Which statement correctly describes the relationship between compressibility factor Z and the extent of deviation from ideal gas behavior?

The magnitude of |Z - 1| indicates the extent of deviation, with larger values representing greater non-ideality (correct answer)
Only Z values greater than 1 indicate deviation from ideal behavior, while Z < 1 represents perfect ideality
The compressibility factor Z is always positive and values closer to zero indicate more ideal behavior
Z values between 0.5 and 1.5 indicate ideal behavior, while values outside this range show significant deviation
The sign of (Z - 1) is more important than its magnitude in determining the degree of non-ideal behavior

Explanation: When analyzing real gas behavior, the compressibility factor Z serves as your key indicator of how much a gas deviates from ideal gas law predictions. For an ideal gas, Z equals exactly 1, so any departure from this value signals non-ideal behavior. The correct approach is to examine |Z - 1| — the absolute difference between Z and 1. Answer A correctly identifies that larger magnitudes of |Z - 1| indicate greater deviation from ideality. Whether Z is 0.5 or 1.5, both represent the same degree of non-ideality since |0.5 - 1| = |1.5 - 1| = 0.5. Answer B incorrectly suggests that only Z > 1 indicates deviation. This misses that Z < 1 also represents significant non-ideality — it typically occurs when intermolecular attractions dominate, making the gas more compressible than predicted by ideal gas law. Answer C falsely claims that Z values closer to zero indicate more ideal behavior. This is backwards; Z approaching zero represents extreme non-ideality where the gas is highly compressible due to strong intermolecular forces or phase transitions. Answer D arbitrarily defines a range (0.5 to 1.5) as "ideal," which contradicts the fundamental definition. Real gases can show significant deviation even within this range, and the boundaries have no theoretical basis. Remember this pattern: for any property comparing real vs. ideal behavior, focus on the absolute deviation from the ideal reference point. The direction of deviation tells you about the dominant molecular interactions, but the magnitude tells you about the extent of non-ideality.

Question 11

A gas storage tank contains gas at Z = 1.08 and 25°C. If the gas expands isothermally until the pressure drops by half, and the new compressibility factor becomes Z = 1.03, what does this behavior suggest about the gas properties?

The gas exhibits repulsive intermolecular forces that decrease in importance as pressure decreases (correct answer)
The gas exhibits attractive intermolecular forces that increase in importance as pressure decreases
The gas behavior becomes less ideal at lower pressure due to increased molecular volume effects
The gas undergoes a phase transition during expansion, changing its intermolecular force characteristics
The temperature change during expansion causes the observed change in compressibility factor values

Explanation: When you encounter compressibility factor problems, focus on how Z values relate to intermolecular forces and how they change with pressure conditions. The compressibility factor Z measures deviation from ideal gas behavior, where Z = 1 represents perfect ideality. Values above 1 indicate repulsive forces dominate, while values below 1 suggest attractive forces are stronger. Here, Z decreases from 1.08 to 1.03 as pressure drops by half, showing the gas becomes more ideal-like (closer to Z = 1) at lower pressure. Since both Z values exceed 1, repulsive forces dominate throughout the expansion. The decrease in Z indicates these repulsive forces become less significant as molecules spread out at lower pressure, allowing the gas to behave more ideally. This confirms answer A. Answer B incorrectly suggests attractive forces, but Z > 1 throughout the process indicates repulsive forces dominate. Answer C misinterprets the data - the gas actually becomes MORE ideal (not less) at lower pressure, as shown by Z moving closer to 1. Answer D assumes a phase transition, but the smooth, continuous change in Z values suggests normal gas expansion without phase changes. Study tip: Remember that Z > 1 always indicates repulsive forces dominate, Z < 1 means attractive forces dominate, and movement toward Z = 1 represents behavior becoming more ideal. For most real gases at moderate to high pressures, you'll see Z > 1 decreasing toward 1 as pressure drops.

Question 12

Using the compressibility factor approach, what would be the molar volume of a real gas at 350 K and 40 atm if Z = 0.91? (R = 0.08206 L·atm/mol·K)

0.656 L/mol (correct answer)
0.721 L/mol
0.792 L/mol
0.871 L/mol
0.598 L/mol

Explanation: When you encounter real gas problems, remember that real gases deviate from ideal behavior due to intermolecular forces and molecular volume. The compressibility factor Z accounts for these deviations and modifies the ideal gas law. The equation for real gases using the compressibility factor is

PV = ZnRT

, which rearranges to

V = \frac{ZnRT}{P}

for molar volume (n = 1 mol). Substituting the given values:

V = \frac{(0.91)(1 \text{ mol})(0.08206 \text{ L·atm/mol·K})(350 \text{ K})}{40 \text{ atm}} = 0.656 \text{ L/mol}

This confirms answer A (0.656 L/mol) is correct. Answer B (0.721 L/mol) represents a common error where students use Z = 1.0 instead of the given Z = 0.91, essentially treating this as an ideal gas problem. Answer C (0.792 L/mol) occurs when students incorrectly multiply by Z instead of using it as a correction factor in the numerator. Answer D (0.871 L/mol) results from calculation errors, often involving incorrect unit conversions or arithmetic mistakes with the temperature or pressure values. Remember that Z < 1 indicates the real gas occupies less volume than predicted by ideal gas behavior, typically due to attractive intermolecular forces. Always use the given Z value directly in your calculation, and double-check that your final volume is smaller than the ideal gas prediction when Z < 1.

Question 13

For most real gases at moderate temperatures, which pressure range typically shows the minimum value of compressibility factor Z?

Intermediate pressures around 10-100 atm where attractive forces dominate over repulsive effects (correct answer)
Very low pressures near 1 atm where ideal gas behavior is most closely approached by all gases
Very high pressures above 500 atm where molecular volume effects become completely dominant over attractions
The minimum Z occurs at constant pressure regardless of temperature for any given gas composition
Atmospheric pressure conditions where intermolecular forces are perfectly balanced by thermal energy effects

Explanation: When you encounter questions about compressibility factor Z, remember that Z measures how much a real gas deviates from ideal behavior, with Z = PV/nRT. For ideal gases, Z = 1, but real gases show interesting patterns due to competing molecular forces. The correct answer is A because at intermediate pressures (10-100 atm), attractive intermolecular forces (van der Waals forces) dominate over repulsive forces from molecular volume. These attractions cause gas molecules to pull together more than an ideal gas would predict, reducing the actual volume below the ideal volume. This makes Z < 1, and this pressure range typically contains the minimum Z value for most real gases at moderate temperatures. Option B is wrong because at very low pressures near 1 atm, real gases actually behave most like ideal gases, so Z approaches 1, not its minimum value. Option C is incorrect because at very high pressures above 500 atm, repulsive forces from finite molecular volume dominate, causing Z to rise well above 1 as molecules resist compression. Option D is false because the pressure at which minimum Z occurs definitely depends on temperature - higher temperatures shift the minimum to higher pressures. Study tip: Remember the Z-curve pattern for real gases - it starts near 1 at low pressure, dips below 1 at intermediate pressures due to attractions (minimum Z), then rises above 1 at high pressures due to molecular volume effects. This creates a characteristic "dip and rise" shape that's fundamental to understanding real gas behavior.

Question 14

Consider two gas samples at identical temperature and pressure conditions. Gas A has Z = 0.76 and Gas B has Z = 1.24. If equal numbers of moles of each gas are present, what is the ratio of molar volume of Gas B to Gas A?

1.63 (correct answer)
0.61
1.00
1.48
0.76

Explanation: When you encounter compressibility factor (Z) problems, you're dealing with real gas behavior and how it deviates from ideal gas conditions. The compressibility factor relates real gas behavior to the ideal gas law through the equation

PV = nZRT

, where Z accounts for intermolecular forces and molecular volume. Since both gases are at identical temperature and pressure with equal moles, you can set up a ratio. For Gas A:

PV_A = nZRT

where

Z_A = 0.76

. For Gas B:

PV_B = nZRT

where

Z_B = 1.24

. Taking the ratio of molar volumes:

\frac{V_B}{V_A} = \frac{Z_B}{Z_A} = \frac{1.24}{0.76} = 1.63

This makes physical sense: Gas A (Z < 1) experiences strong attractive forces that compress it below ideal behavior, while Gas B (Z > 1) has repulsive forces or significant molecular volume causing expansion beyond ideal behavior. Answer A (1.63) is correct as shown by the direct calculation above. Answer B (0.61) represents the inverse ratio - you'd get this if you mistakenly calculated

\frac{Z_A}{Z_B}

. Answer C (1.00) would only be true if both gases were ideal (Z = 1 for both) or if their Z values were equal. Answer D (1.48) has no clear mathematical relationship to the given Z values and likely represents a calculation error. Remember: when comparing real gases at identical conditions, their volume ratios directly equal their compressibility factor ratios. Always check that your ratio makes physical sense given whether Z values are above or below 1.

Question 15

If a real gas has a compressibility factor Z = 0.75 at certain conditions, and the ideal gas law predicts a molar volume of 2.40 L/mol at those same conditions, what is the actual molar volume of the real gas?

1.80 L/mol (correct answer)
2.40 L/mol
3.20 L/mol
0.75 L/mol
1.65 L/mol

Explanation: When you encounter compressibility factor problems, you're dealing with how real gases deviate from ideal behavior. The compressibility factor Z is defined as the ratio of actual molar volume to ideal molar volume:

Z = \frac{V_{actual}}{V_{ideal}}

. To find the actual molar volume, rearrange this equation:

V_{actual} = Z \times V_{ideal}

. With Z = 0.75 and the ideal gas prediction of 2.40 L/mol, you get:

V_{actual} = 0.75 \times 2.40 = 1.80 \text{ L/mol}

. This confirms answer A is correct. Looking at the wrong answers: Answer B (2.40 L/mol) represents the ideal gas volume, which ignores the real gas behavior entirely—a common trap when students forget to apply the compressibility factor. Answer C (3.20 L/mol) comes from incorrectly dividing the ideal volume by Z instead of multiplying (

\frac{2.40}{0.75} = 3.20

), which flips the relationship and suggests the gas is more compressible than ideal when Z < 1 actually means it's less compressible. Answer D (0.75 L/mol) simply restates the compressibility factor value, showing confusion about what the question is asking for. Remember that Z < 1 means the real gas occupies less volume than predicted by ideal gas law (attractive forces dominate), while Z > 1 means it occupies more volume (repulsive forces dominate). Always multiply the ideal volume by Z to get the actual volume—never divide.

Question 16

A real gas has a compressibility factor Z = 1.25 at 300 K and 100 atm. If 2.0 moles of this gas occupy a volume of 0.615 L under these conditions, what volume would the same amount of gas occupy if it behaved ideally at the same temperature and pressure?

0.492 L (correct answer)
0.615 L
0.769 L
1.230 L
0.328 L

Explanation: When you encounter compressibility factor problems, you're dealing with how real gases deviate from ideal behavior. The compressibility factor Z relates real gas volume to what the volume would be if the gas were ideal:

Z = \frac{V_{real}}{V_{ideal}}

. Since Z = 1.25 and the real gas occupies 0.615 L, you can find the ideal volume by rearranging:

V_{ideal} = \frac{V_{real}}{Z} = \frac{0.615 \text{ L}}{1.25} = 0.492 \text{ L}

. This means the real gas takes up more space than an ideal gas would under the same conditions, which makes sense since Z > 1 indicates the gas molecules have significant volume themselves or weak intermolecular attractions. Looking at the wrong answers: (A) 0.492 L is actually correct. (B) 0.615 L represents the real gas volume you started with - this would only be correct if Z = 1 (perfect ideal behavior). (C) 0.769 L and (D) 1.230 L both come from incorrectly multiplying instead of dividing; (C) might result from using Z = 1.25 incorrectly in calculations, while (D) equals 0.615 × 2, suggesting confusion about the relationship entirely. Remember this key relationship: when Z > 1, the real gas occupies more volume than ideal, so divide the real volume by Z to get the smaller ideal volume. When Z < 1, the real gas is more compressed than ideal. Always check whether your final answer makes physical sense given the Z value.

Question 17

For a gas with molar volume 0.180 L/mol at 400 K and 60 atm, calculate the compressibility factor Z. (R = 0.08206 L·atm/mol·K)

0.329 (correct answer)
0.548
1.000
1.824
3.040

Explanation: When you encounter compressibility factor problems, you're dealing with how real gases deviate from ideal gas behavior. The compressibility factor Z compares a real gas's molar volume to what it would be if it behaved ideally. To find Z, use the equation

Z = \frac{V_{real}}{V_{ideal}}

. You're given the real molar volume (0.180 L/mol), so you need to calculate what the ideal molar volume would be under the same conditions using the ideal gas law:

PV = nRT

. For one mole of gas:

V_{ideal} = \frac{RT}{P} = \frac{(0.08206)(400)}{60} = \frac{32.824}{60} = 0.547 \text{ L/mol}

Now calculate Z:

Z = \frac{0.180}{0.547} = 0.329

This confirms answer A) 0.329 is correct. The Z value less than 1 indicates the real gas occupies less volume than predicted by ideal gas law, typical at high pressures where intermolecular attractions dominate. Answer B) 0.548 represents a common error of confusing the ideal molar volume with the compressibility factor itself. Answer C) 1.000 would indicate perfect ideal gas behavior, which doesn't occur at 60 atm pressure. Answer D) 1.824 results from incorrectly inverting the Z formula (using

V_{ideal}/V_{real}

), which would give Z > 1. Remember: Z = 1 for ideal gases, Z < 1 when real gases are more compressed than ideal (high pressure/low temperature), and Z > 1 when real gases are less compressed than ideal (moderate pressures).

Question 18

A gas mixture has Z = 0.92 at 350 K and 75 atm. If the pressure is reduced to 25 atm at the same temperature and Z becomes 0.97, what can be concluded about the effect of pressure on gas non-ideality?

Lower pressure reduces the deviation from ideal behavior, with attractive forces becoming less significant (correct answer)
Lower pressure increases the deviation from ideal behavior, with repulsive forces becoming more significant
Pressure has no effect on gas behavior since both Z values are less than 1
Lower pressure maintains constant deviation from ideal behavior regardless of intermolecular force changes
Lower pressure causes the gas to become more non-ideal due to increased molecular collisions

Explanation: When analyzing gas non-ideality, the compressibility factor Z tells you how much a real gas deviates from ideal behavior. For an ideal gas, Z = 1. Values less than 1 indicate that attractive intermolecular forces dominate, while values greater than 1 suggest repulsive forces are more significant. In this problem, Z increases from 0.92 to 0.97 as pressure decreases from 75 atm to 25 atm at constant temperature. Since both Z values are below 1, attractive forces dominate at both conditions. However, the key insight is that Z is moving closer to 1 (ideal behavior) as pressure decreases. This means the gas is becoming more ideal-like at lower pressure. At high pressures, molecules are forced closer together, making attractive intermolecular forces more significant and causing greater deviation from ideality (lower Z). As pressure decreases, molecules spread out more, weakening these attractive interactions and reducing the deviation from ideal behavior. Answer A correctly identifies that lower pressure reduces deviation from ideal behavior by making attractive forces less significant. Answer B incorrectly suggests deviation increases and repulsive forces become dominant - but Z values below 1 indicate attractive forces still dominate. Answer C wrongly claims pressure has no effect, ignoring the clear change in Z values. Answer D incorrectly states deviation remains constant, contradicting the observed Z increase. Study tip: Remember that Z < 1 means attractive forces dominate, Z > 1 means repulsive forces dominate, and changes in Z reveal how intermolecular forces respond to pressure and temperature changes.

Question 19

In the equation of state PV = ZnRT, if a gas sample has P = 90 atm, V = 8.2 L, n = 3.0 mol, T = 400 K, what is the value of Z? (R = 0.08206 L·atm/mol·K)

0.75 (correct answer)
1.33
1.00
0.56
1.78

Explanation: When you encounter the equation of state

PV = ZnRT

, you're working with the compressibility factor Z, which measures how much a real gas deviates from ideal gas behavior. This equation is a modification of the ideal gas law where Z accounts for intermolecular forces and molecular volume effects. To find Z, rearrange the equation:

Z = \frac{PV}{nRT}

. Substituting the given values:

Z = \frac{(90 \text{ atm})(8.2 \text{ L})}{(3.0 \text{ mol})(0.08206 \text{ L·atm/mol·K})(400 \text{ K})} = \frac{738}{98.472} = 0.75

This confirms answer A) 0.75 is correct. A Z value less than 1.0 indicates the gas is more compressible than an ideal gas, typically due to attractive intermolecular forces dominating at high pressure. Answer B) 1.33 would suggest the gas occupies more volume than predicted by ideal gas law, which contradicts the high pressure conditions. Answer C) 1.00 would mean perfect ideal gas behavior, which is unlikely at 90 atm where intermolecular forces become significant. Answer D) 0.56 results from calculation errors, possibly using incorrect unit conversions or arithmetic mistakes. Remember that Z values tell a story about molecular behavior: Z < 1 means attractive forces dominate (gases are more compressible), while Z > 1 means repulsive forces or molecular size effects dominate. At high pressures like 90 atm, expect Z < 1 due to intermolecular attractions.

Question 20

A gas at $T_r = 1.2$ and $P_r = 2.5$ has a compressibility factor of $Z = 0.85$ . If the temperature is increased to $T_r = 1.8$ while maintaining the same molar density, what is the most likely range for the new compressibility factor?

$Z = 0.75$ to $0.80$ (decreased due to higher intermolecular forces)
$Z = 0.85$ to $0.90$ (slightly increased due to reduced relative pressure effects)
$Z = 0.95$ to $1.05$ (approaches ideal gas behavior at higher temperature) (correct answer)
$Z = 1.10$ to $1.20$ (increased due to enhanced molecular repulsion at constant density)

Explanation: At constant molar density, increasing temperature reduces the reduced pressure (

P_r \propto T_r

), but the net effect is movement toward ideal gas behavior. Higher temperature reduces the relative importance of intermolecular attractions, and the gas moves closer to

Z = 1

. The original conditions show significant deviation from ideality (

Z = 0.85

), but at higher

T_r

with correspondingly higher

P_r

, the gas approaches ideal behavior. Choice A incorrectly assumes attractions increase. Choice B underestimates the temperature effect. Choice D incorrectly suggests strong repulsive behavior dominates.