For a real gas that follows the van der Waals equation of state, how does the difference in specific heats (Cp − Cv) compare with the ideal-gas value? Explain on the basis of thermodynamic identities and stability criteria.

Difficulty: Medium

Correct Answer: > R

Explanation:


Introduction / Context:
The difference between the constant-pressure and constant-volume specific heats, Cp − Cv, is a fundamental quantity that links caloric and thermal equations of state. For an ideal gas, Cp − Cv equals the universal gas constant R on a molar basis. For real gases described by the van der Waals model, non-ideality changes thermal expansion and compressibility, which in turn modifies Cp − Cv. This question checks your understanding of the general identity relating Cp − Cv to measurable volumetric properties and why real-gas behavior typically yields a value larger than R under stable conditions.


Given Data / Assumptions:

  • Real gas modeled by the van der Waals equation.
  • Stable single-phase states (no crossing into spinodal region).
  • Calorically simple behavior (Cp and Cv defined in the usual way).


Concept / Approach:
The general identity is Cp − Cv = T * V * α^2 / κ_T (molar basis: replace V with molar volume). Here α is the volumetric thermal expansion coefficient and κ_T is the isothermal compressibility. For an ideal gas, α = 1/T and κ_T = 1/P, giving Cp − Cv = R. For a van der Waals gas at the same T and P, κ_T is usually smaller (the fluid is less compressible) and α does not decrease proportionally, so the ratio α^2/κ_T is larger, making Cp − Cv greater than R in stable regions.


Step-by-Step Solution:

Use identity: Cp − Cv = T V α^2 / κ_T.Insert ideal-gas limits: α_id = 1/T, κ_T,id = 1/P → Cp − Cv = R.For van der Waals fluids, κ_T < κ_T,id at comparable T, P (less compressible), while α often remains of similar order.Therefore Cp − Cv (real) > Cp − Cv (ideal) = R, for stable states.


Verification / Alternative check:
Another exact form is Cp − Cv = −T * (∂P/∂T)_V^2 / (∂P/∂V)_T. For stable fluids, (∂P/∂V)_T < 0, making the fraction positive. Real-gas deviations generally make the magnitude larger than the ideal case at the same conditions.


Why Other Options Are Wrong:

  • R: true only for ideal gases.
  • < R or 0.5 R: contradict typical real-gas trends in stable regions.
  • Negative: impossible for a mechanically stable single phase because (∂P/∂V)_T is negative.


Common Pitfalls:
Confusing per-mole vs per-mass bases; forgetting that α and κ_T govern Cp − Cv in all fluids; applying the ideal-gas result blindly to real fluids at high pressures.


Final Answer:
> R

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