Question

A cylindrical vessel is divided into two parts by a fixed partition which is perfectly heat conducting. The wall and the piston are thermally insulated from the surroundings. The left side contains $0.5$ moles of gas with $C_V=2R$ at temperature of $300K$ . The right side contains $4$ moles of mixture of gas with $C_V=1.75R$ at the same temperature at $300K$. The piston compresses slowly the right side from volume $V_0$ to $\frac{V_0}{4}$. If total change in internal energy of gases is $n\times {10}^4\text{J}$, the value of $n$ is ($R=25/3SI$ units)

Answer 1

$dQ=0$
From the first law of thermodynamics,
$\Rightarrow n_1{C}_{V_1}dT+n_2{C}_{V_2}dT+PdV=0$
$\Rightarrow \frac{1}{2}\left(2R\right)dT+4\left(\frac{7R}{4}\right)dT+\left(\frac{4RT}{V}\right)dV=0$
$\Rightarrow 8RdT+\left(\frac{4RT}{V}\right)dV=0$
$\Rightarrow 2\frac{dT}{T}+\frac{dV}{V}=0$
Integrating the equation from initial volume and temperature to final volume and temperature.
$\Rightarrow 2{\int }_{300}^{T_f}\frac{dT}{T}+{\int }_{V_0}^{\frac{V_0}{4}}\left(\frac{dV}{V}\right)=0$
$\Rightarrow 2\ln \left(\frac{Tf}{300}\right)+\ln \left(\frac{1}{4}\right)=0$
$T_f=600K$
$dU=n_1{C}_{1}dT+n_2{C}_{2}dT$
$=\frac{1}{2}\left(2R\right)(600-300)+4\times \frac{7R}{4}(600-300)$
$=2400RJ$
$\Rightarrow dU=2400\left(\frac{25}{3}\right)=20000J=2\times {10}^{4}J$