Question

$32$ gram of oxygen gas at temperature ${27}^{o}C$ is compressed adiabatically to $\frac{1}{3}$ of its initial volume. Calculate the change in internal energy. ($\gamma =1.5$ for oxygen)

Answer 1

For , adiabatically ,

$dU=nCvdT=\frac{nR}{(\gamma -1)}dT$

now,

$P{V}^{\gamma }=constant$

so, $V^{\gamma }\times \frac{nRT}{V}=constant$

$T{V}^{(\gamma -1)}=constant$

so, $T_1{V}_1^{(\gamma -1)}=T_2{V}_2^{(\gamma -1)}$

$300\times V^{(\gamma -1)}=T_2\times {\left(\frac{V}{3}\right)}^{(\gamma -1)}$

$\frac{300}{T_2}={\left(\frac{1}{3}\right)}^{(1.5-1)}$

$T_2=\frac{300}{\left(1/\sqrt{3}\right)}=300\sqrt{3}K$

n = no of mole = $\frac{32}{32}=1$

now,

$dU=1\times \frac{R}{(1.5-1)}\times (300\sqrt{3}-300)$

$⟹=2R\times 300(\sqrt{3}-1)$

$⟹=2\times \frac{25}{3}\times 300(1.732-1)$

$=50\times 100\times 0.732J=3660J$