Question

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

Answer 1

Given : $n=3$

The equation $P{V}^{19/13}$ $⟹\gamma =\frac{19}{13}$

From $P{V}^{\gamma }=constant$ we get $P{V}^{\gamma }=P^{\prime }(\frac{V}{5}{)}^{\gamma }$

$⟹P^{\prime }=5^{\gamma }P$

Compressibility $K=\frac{1}{B}=\frac{1}{\gamma P}$

where B is bulk modulus

$\therefore$ $\Delta K=K^{\prime }-K=\frac{1}{\gamma }[\frac{1}{P^{\prime }}-\frac{1}{P}]$

OR $\Delta K=\frac{1}{\gamma P}[\frac{1}{5^{\gamma }}-\frac{1}{1}]$

Using ideal gas equation, $PV=3RT⟹P=\frac{3RT}{V}$

$⟹$ $\Delta K=\frac{V}{3RT\gamma }[\frac{1}{5^{\gamma }}-\frac{1}{1}]$ where $\gamma =\frac{19}{13}=1.46$

$\therefore$ $\Delta K=\frac{V}{3\times 8.3T\times 1.46}[\frac{1}{5^{1.46}}-\frac{1}{1}]=\frac{-0.0248V}{T}$