Compressibility Factor (Z) - Thermodynamics

At constant temperature, as pressure increases for most real gases, the compressibility factor Z typically:
Answer: First decreases below 1, reaches a minimum, then increases above 1 at very high pressures 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.
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?
Answer: $Z = 0.95$ to $1.05$ (approaches ideal gas behavior at higher temperature) 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.
A gas mixture contains $60\%$ methane and $40\%$ ethane by mole fraction. At $400$ K and $80$ bar, the pure component compressibility factors are $Z_{CH_4} = 0.91$ and $Z_{C_2H_6} = 0.73$ . Using Kay's rule for pseudocritical properties, the compressibility factor of the mixture is estimated as $Z_{mix} = 0.85$ . What does this suggest about the mixture behavior?
Answer: The mixture shows positive deviations due to weaker cross-interactions between unlike molecules Simple mole-fraction weighted average: $Z_{weighted} = 0.6(0.91) + 0.4(0.73) = 0.838$ . The Kay's rule prediction (0.85) is slightly higher than this weighted average, suggesting the mixture is less compressed (closer to ideal) than expected from individual components. This indicates positive deviations, typically caused by weaker interactions between unlike molecules compared to like molecules. Choice A is wrong because ideal mixing would give the weighted average. Choice C has the wrong sign. Choice D is incorrect because the difference isn't extreme enough to invalidate Kay's rule.
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?
Answer: Higher temperature reduces the relative importance of intermolecular attractive forces, making gas behavior more ideal 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.
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?
Answer: 1.80 L/mol 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.
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)
Answer: 0.329 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).
For most real gases at moderate temperatures, which pressure range typically shows the minimum value of compressibility factor Z?
Answer: Intermediate pressures around 10-100 atm where attractive forces dominate over repulsive effects 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.
At what condition would you expect the compressibility factor Z to be closest to 1.00 for a real gas?
Answer: High temperature and low pressure, where intermolecular forces become negligible and molecular volume is insignificant 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.
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)
Answer: 0.656 L/mol 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: $V = \frac{(0.91)(1 \text{ mol})(0.08206 \text{ L}\cdot\text{atm/mol}\cdot\text{K})(350 \text{ K})}{40 \text{ atm}} = 0.656 \text{ L/mol}$ 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.
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?
Answer: 136.4 atm 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.
Two engineers measure the compressibility factor of the same gas sample. Engineer A uses the virial equation truncated after the second coefficient: $Z = 1 + BP$ . Engineer B uses the van der Waals equation. At moderate pressures, Engineer A consistently reports higher Z values than Engineer B. This systematic difference most likely indicates:
Answer: The van der Waals equation accounts for both attractive and repulsive forces while the virial form captures only one effect The van der Waals equation explicitly includes both attractive forces ( $a$ parameter) and repulsive forces ( $b$ parameter), providing a more complete description of molecular interactions. The truncated virial equation $Z = 1 + BP$ captures the leading-order deviation from ideality but may not fully account for the balance between attractive and repulsive forces that van der Waals includes. At moderate pressures, this can lead to systematic differences. Choice A is incorrect about the sign relationship. Choice B assumes error rather than systematic physical differences. Choice C incorrectly describes the virial equation's behavior.
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?
Answer: The gas exhibits repulsive intermolecular forces that decrease in importance as pressure decreases 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.
A van der Waals gas has $Z = 0.88$ at certain conditions. When the pressure is doubled while temperature remains constant, the compressibility factor changes to $Z = 0.71$ . Based on this information, which statement best describes the relative magnitudes of the van der Waals constants?
Answer: The $a$ parameter dominates over $b$ parameter effects at both pressure conditions At both conditions, $Z < 1$ , indicating that attractive forces ( $a$ parameter) dominate over excluded volume effects ( $b$ parameter). When pressure doubles, Z decreases further from 0.88 to 0.71, showing that attractive forces become even more significant relative to repulsive forces. If $b$ were dominant, we would expect $Z > 1$ . Choice B is incorrect because $Z > 1$ would be observed. Choice C is wrong because $Z$ decreases (doesn't increase) with pressure. Choice D is incorrect because the values don't approach unity.
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?
Answer: Finite molecular volume becomes significant, causing molecules to occupy more space than point particles 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.
Using the generalized compressibility chart, a student determines that $Z = 0.82$ for a gas at $T_r = 1.1$ and $P_r = 2.8$ . However, the actual measured value is $Z = 0.76$ . Which factor most likely explains this discrepancy?
Answer: The gas has a significantly non-spherical molecular structure affecting intermolecular interactions The generalized compressibility chart is based on the principle of corresponding states, which assumes molecules are spherical and have similar intermolecular potential functions. Non-spherical molecules (like long hydrocarbons) or molecules with strong dipole moments deviate from this assumption. The measured Z is lower than predicted, suggesting stronger attractive forces than expected for a 'typical' fluid. Choice B is unlikely to cause such systematic deviation. Choice C is wrong because $T_r = 1.1$ is reasonably far from critical. Choice D would affect the chart reading but wouldn't consistently give lower Z values.
Two identical vessels contain the same gas at identical reduced temperatures ( $T_r = 0.9$ ). Vessel A contains the gas at $P_r = 0.5$ with $Z_A = 0.92$ , while vessel B contains the gas at $P_r = 3.0$ with $Z_B = 0.78$ . If the contents of both vessels are mixed adiabatically in a third vessel with volume equal to the sum of the original volumes, the compressibility factor of the mixture will be closest to:
Answer: $Z = 0.82$ (weighted toward the higher pressure gas properties) The final mixture pressure will be between the original pressures but closer to the high-pressure vessel's conditions due to the significant pressure difference. The high-pressure gas (vessel B) contains much more mass than the low-pressure gas (vessel A) at the same temperature. When mixed, the final state will be dominated by the properties of the denser, higher-pressure gas, so Z will be closer to 0.78 than to 0.92. Choice A ignores the mass difference. Choice C incorrectly weights toward low pressure. Choice D incorrectly assumes mixing improves ideality.
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?
Answer: 1.63 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.
A laboratory reports that nitrogen gas at $126$ K and $25$ bar has a compressibility factor of $Z = 0.86$ . A student calculates that these conditions correspond to $T_r = 1.0$ and $P_r = 0.74$ and expects $Z \approx 0.92$ from the generalized chart. The discrepancy is most likely explained by:
Answer: Nitrogen exhibits significant quantum effects at these low temperatures that classical equations cannot predict At $T_r = 1.0$ (near critical temperature) and relatively high pressure, nitrogen exhibits significant quantum effects because it's a light molecule at low absolute temperature (126 K). Classical corresponding states theory, which underlies the generalized compressibility chart, assumes classical molecular behavior. Quantum effects make the gas more compressible than classical theory predicts, explaining why measured Z (0.86) is lower than chart prediction (0.92). Choice B is unlikely given standard critical constants. Choice C is wrong because triple point affects solid-liquid equilibrium, not gas compressibility. Choice D mischaracterizes the chart's basis.
Which statement correctly describes the relationship between compressibility factor Z and the extent of deviation from ideal gas behavior?
Answer: The magnitude of |Z - 1| indicates the extent of deviation, with larger values representing greater non-ideality 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.
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?
Answer: 71.4 atm 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.

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