Question

At low pressure, compressibility factors of gases are less than 1. In an experiment at a constant temperature, certain moles of gas ‘X’ exhibits two different compressibility factors 0.4 and 0.8 respectively at two different volumes $V_1$ & $V_2$. What is the value of $V_1$:$V_2$?

• A. 1:2
• B. 2:1
• C. 1:3
• D. 3:1

Answer 1

The correct option is C 1:3

Real gas follows the eqaution - $(P+\frac{a{n}^2}{V^2})(V-nb)$ = nRT
At lower Pressure the volume correction term can be neglected -
$\Rightarrow$ $(P+\frac{a{n}^2}{V^2})V$ = nRT
$\Rightarrow$ P(V - nb) = nRT
$\Rightarrow$ $\frac{PV}{nRT}$ = 1 - $\frac{an}{VRT}$
$\Rightarrow$ Z = 1 - $\frac{an}{VRT}$
$\Rightarrow$ $\frac{V_2}{V_1}$ = $\frac{1-Z_1}{1-Z_2}=\frac{1-0.4}{1-0.8}$ = 3 : 1
$\Rightarrow$ $\frac{V_1}{V_2}$ = 1 : 3
Hence correct answer is option (c)