Chemistry - Real Gas Question with Solution | TestHub

Given the equation $P(V_m-b)=RT$.

The compressibility factor $Z=\frac{P{V}_m}{RT}$.

From the given equation, $V_m-b=\frac{RT}{P}$, so $V_m=\frac{RT}{P}+b$.

Substitute $V_m$ into the expression for $Z$:

$Z=\frac{P(\frac{RT}{P}+b)}{RT}=\frac{RT+Pb}{RT}=1+\frac{Pb}{RT}$.

Now, differentiate $Z$ with respect to $P$ at constant $T$:

$(\frac{\partial Z}{\partial P}{)}_T=\frac{\partial }{\partial P}(1+\frac{Pb}{RT}{)}_T=\frac{b}{RT}$.

Comparing this to the given form $\frac{xb}{RT}$, we find $x=1$.