Question

1 mole of ${NH}_3$ at ${27}^{o}C$ is expanded in reversible adiabatic condition to make volm 8 times ($\gamma =1.33$). What is the work done?

Answer 1

Given, $T_1={27}^{o}C=300K,r=1.33$

Let $V_1=V$

Then $V_2=8V$

By using,

$T_1{V}_1^{r-1}=T_2{V}_2^{r-1}$

$\Rightarrow \frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{r-1}$

$\Rightarrow \frac{300K}{T_2}={\left(\frac{8V}{V}\right)}^{1.33-1}$

$\Rightarrow T_2=151.04K$

For adiabatic condition, $dq=0$

Thus, $dW=dU$ (by First law of Thermodynamics)

$dW=n{C}_{v}dT⟶\left(1\right)$

as $C_P-C_V=nR$

$\Rightarrow r=\frac{C_P}{C_V}$

$\therefore r-1=\frac{C_P}{C_V}-1=\frac{C_P-C_V}{C_V}=\frac{nR}{C_V}$

$\Rightarrow 1.33-1=\frac{n_R}{C_V}\Rightarrow C_V=\frac{n_R}{0.33}$

Putting this value of $C_v$ in equation (1), we get

$W=1\times \frac{nR}{0.33}\times (151.04-300)K$

$W=1\times \frac{1\times R}{0.33}\times (151.04-300)K$

$W=1mole\times \frac{8.314}{0.33}Jmo{l}^{-1}{K}^{-1}(-148.96)K$

$W=-3.752kJ$

This is the work done.