Question

Two moles of an ideal monoatomic gas at $5$ bar and $300K$ are expanded adiabatically, irreversibly up to an external pressure of $\text{1}$ bar. The work is done by the gas is $-xR$ (in bar.Litre). The value of $x$ is [ $R$ is gas constant]

Answer 1

Given: $p_{f i n a l}=p_{e x t}$ so $p_{1}=5$ , $p_{2}=1$ and $p_{e x t}=1$ , $T_{1}=300,T_{2}=?$
for the calculation of $T_{2}$
$nC_{v}\left(T_{2} - T_{1}\right)=-p_{e x t}\left(V_{2} - V_{1}\right)=-p_{e x t}\left[\frac{n R T_{2}}{P_{2}} - \frac{n R T_{1}}{P_{1}}\right]$
$nC_{v}\left(T_{2} - T_{1}\right)=-p_{e x t}\left[\frac{n R T_{2}}{P_{2}} - \frac{n R T_{1}}{P_{1}}\right]$
$nC_{v}\left(T_{2} - T_{1}\right)=-p_{e x t}\left[\frac{n R T_{2}}{P_{2}} - \frac{n R T_{1}}{P_{1}}\right]$
$C_{v}\left(T_{2} - T_{1}\right)=p_{e x t}\left[\frac{R T_{2}}{P_{2}} - \frac{R T_{1}}{P_{1}}\right]$
$\left(\frac{3 R}{2}\right)\left(\right.T_{2} -300\left.\right)=-1\left[\frac{R \times T_{2}}{1} - \frac{R \times 300}{5}\right]$
$\text{T}_{\text{2}} \, \text{=} \, \text{204} \, \text{K}$
$W_{i r r , a d i a}=-1\left[\frac{2 \times R \times 204}{1} - \frac{2 \times R \times 300}{5}\right]$
$=-288 \, R$