Adiabatic expansion of gas 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 :
$Z=\frac{PV}{nRT}$
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 ?
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. &quot;The gas expanded against constant pressure&quot; ⇒ 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 &gt; 0. ΔStot = ΔSsys + ΔSsur = 0 + ΔSsys = ΔSsys. So ΔSsys &gt; 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.