Question

The equation of state of a gas is $P(V-nb)=nRT$, where $b$ and $R$ are constants. If the pressure and the temperature are such that $V_m=10b,$ what is the value of compressibility factor?

• A. $\frac{10}{9}$
• B. $\frac{9}{11}$
• C. $\frac{11}{10}$
• D. $\frac{10}{11}$

Answer 1

The correct option is A $\frac{10}{9}$

Given,
$P(V-nb)=nRT$
for 1 mole $P(V_m-b)=RT$
$P{V}_m-Pb=RT$
$P{V}_m=Pb+RT$
$\frac{P{V}_m}{RT}=\frac{Pb}{RT}+1$

$Z=\frac{Pb}{RT}+1$
$Z$ is the compressibility factor.

Also,
$P(V_m-b)=nRT$ for 1 mole
$P{V}_m-Pb=RT$
Given $V_m=10b$
$P\left(10b\right)-Pb=RT$
$9Pb=RT$
Hence, $\frac{Pb}{RT}=\frac{1}{9}$
Therefore compressibility factor,
$Z=\frac{1}{9}+1=\frac{10}{9}$