If a real gas is adiabatically expanded against constant pressure, then which of the following will definitely increase?

Compressibility factor (Z) or Entropy (S)

Correct answer: Entropy

My answer: Compressibility

My Reasoning :


In adiabatic $TV^{γ-1}=\text{constant}$

Therefore, $T$$\frac{1}{V^{γ-1}}$

$γ-1$ is greater than zero always. So we can say is volume increases, temp decreases. Therefore Z should increase as its numerator increases and denominator decreases. Entropy for isobaric adiabatic has $\ln\frac{T_2}{T_1}$. Since $T_2$ is less than $T_1$, entropy decreases. What am I doing wrong ?

  • $\begingroup$ First of all, that TV equation is for perfect gas only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change. $\endgroup$
    – TheLearner
    Commented Aug 8, 2020 at 5:26
  • $\begingroup$ @The99sLearner so how do i proceed to solve it then ? $\endgroup$ Commented Aug 8, 2020 at 6:51
  • $\begingroup$ The compressibility vs pressure shows a decrease then increase so depending on what the initial pressure is the change could be positive or negative. The entropy change in a reversible adiabatic expansion is zero for an ideal gas and positive for a non-reversible adiabatic change (positive because less work is done so the temperature is lowered less than the reversible case and no longer balances the volume change). The entropy in the non-ideal gas case will increase as work always has to be done against intermolecular interactions. $\endgroup$
    – porphyrin
    Commented Aug 8, 2020 at 8:08

1 Answer 1


First note that the process is an irreversible adiabatic expansion. If it were a reversible one, the gas has to expand against a varying pressure because pressure of gas decreases during expansion and this varying gas pressure must be matched with external pressure in all infinitesimal steps of expansion. "The gas expanded against constant pressure" ⇒ the gas must have a pressure which is always higher than the constant external pressure during the expansion ⇒ irreversible.

Will entropy definitely increase?

ΔSsur = 0 as q = 0. Irreversible ⇒ ΔStot > 0. ΔStot = ΔSsys + ΔSsur = 0 + ΔSsys = ΔSsys. So ΔSsys > 0, so entropy of system will definitely increase.

Will compressibility factor definitely increase?

In adiabatic irreversible expansion all p, V, T changes. If Z is to increase, the increment in pV must be greater than the increment in T, or the magnitude of decrement in T must be greater than that in pV, which cannot be sure will happen as all p, V, T change.

What did you do wrong?

First of all, TV equation is for perfect gas doing reversible adiabatic expansion only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change. All p, V, T will change. You said as volume increases, temperature decreases, but pressure of gas will decrease too.

  • 1
    $\begingroup$ You are correct in saying this process is irreversible. But, I think you should add a sentence about why the process is irreversible. $\endgroup$ Commented Aug 8, 2020 at 12:23
  • 1
    $\begingroup$ Note that the TV equation for adiabatic processes has two restrictions. First, as you mentioned, it applies only to ideal gases. Second (and it's important to state this explicitly), it applies only to reversible processes. Those are the two fundamental ways the OP went wrong in applying that equation. $\endgroup$
    – theorist
    Commented Aug 8, 2020 at 19:25

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