Compressibility factor Z for an ideal gas: Select the correct statement about Z for an ideal gas over all temperatures and pressures.
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AZ decreases with pressure for an ideal gas.
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BZ equals exactly 1 at all temperatures and pressures.
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CZ equals 1 only at Boyle’s temperature.
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DZ is zero for an ideal gas.
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EZ exceeds 1 only at high temperatures.
Answer
Correct Answer: Z equals exactly 1 at all temperatures and pressures.
Explanation
Introduction / Context:The compressibility factor Z = PV/(nRT) measures deviation from ideal-gas behavior. For real gases, Z departs from unity, especially near condensation conditions or at high pressures/low temperatures. Understanding the exact behavior for an ideal gas avoids confusion in state estimation and property calculations.
Given Data / Assumptions:
- Ideal-gas equation of state, PV = nRT.
- No interactions or excluded volume.
- All temperatures and pressures are mathematically admissible under the model.
Concept / Approach:By definition, an ideal gas is one whose P–V–T relationship follows PV = nRT. Substituting into Z = PV/(nRT) yields Z = 1 identically. Boyle’s temperature is a real-gas concept where first-order deviations vanish (B2 ≈ 0), making Z ≈ 1 over a range, but that pertains to real gases only—not the ideal-gas postulate.
Step-by-Step Solution:
Start with the ideal-gas equation PV = nRT.Compute Z = PV/(nRT) = 1.Conclude Z is exactly 1 for all T and P in the ideal model.Verification / Alternative check:Virial EOS for real gases: Z = 1 + B2P/RT + …; setting all virial coefficients to zero recovers the ideal gas where Z = 1 identically.
Why Other Options Are Wrong:
- (a), (d), (e) contradict the identity Z ≡ 1.
- (c) confuses real-gas Boyle’s temperature with the ideal-gas definition.
Common Pitfalls:Applying real-gas intuition (e.g., Z trends) to the ideal gas; misusing Boyle’s temperature as a universal condition.
Final Answer:Z equals exactly 1 at all temperatures and pressures.