Question

A certain gas obeys $P(V_m-b)=RT$ . The value of ${\left(\frac{\partial Z}{\partial P}\right)}_T$ is $\frac{xb}{RT}.$ The value of $x$ is ______.

Answer 1

$P(V_m-b)=RTP{V}_m–Pb=RT\frac{P{V}_m}{RT}=1+\frac{Pb}{RT}Z=1+\frac{Pb}{RT}\frac{\partial Z}{\partial P}=0+\frac{b}{RT}\because \frac{b}{RT}=\frac{xb}{RT}\therefore x=1$