Question

$3$ moles of a gas mixture having volume $V$ and temperature $T$ is compressed to ${\frac{1}{5}}^{th}$ of the initial volume. Find the change in its adiabatic compressibility, if the gas obeys $P{V}^{19/13}=\text{constant}$ $[R=8.3\text{J/mol-K}]$

• A. $\frac{-0.0248V}{7T}$
• B. $\frac{-0.0248V}{T}$
• C. $\frac{-0.0248V}{13T}$
• D. $\frac{-0.0248V}{19T}$

Answer 1

The correct option is B $\frac{-0.0248V}{T}$

As mixture is compressed to ${\frac{1}{5}}^{th}$ of the initial volume
$P{V}^{\gamma }=P^{\prime }{\left(\frac{V}{5}\right)}^{\gamma }$
$\text{we get}{P}^{\prime }=5^{\gamma }P,$
Given $\gamma =\frac{19}{13}$
Bulk modulus $B=\gamma P$
compressibility $C=\left(\frac{1}{B}\right)=\frac{1}{\gamma P}$
and $\Delta C=C-C$
$\Delta C=\frac{1}{\gamma }[\frac{1}{P^{\prime }}-\frac{1}{P}]$
$\Delta C=\frac{1}{\gamma P}[\frac{1}{5^{\gamma }}-\frac{1}{1}]=\frac{-13\times 0.905}{19P}$
But $PV=nRT\text{or}P=\frac{nRT}{V}$
$\Delta C=\frac{-13\left(.905\right)V}{19\times 3\times 8.3T}=\frac{-0.0248V}{T}$