Compressibility factor Z for an ideal gas:\nSelect the correct statement about Z for an ideal gas over all temperatures and pressures.

Difficulty: Easy

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.

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