Question

1 mole of $N{H}_3$ gas at ${27}^{o}C$ is expanded in reversible adiabatic condition to make volume 64 times $(\gamma =1.33).$ Final temperature and magnitude of work done respectively are

• A. 150 K, 900.00 cal
• B. 75 K, 1363.63 cal
• C. 75 K, 1263.63 cal
• D. 150 K, 1363.63 cal

Answer 1

The correct option is B 75 K, 1363.63 cal

For a reversible adiabatic process :
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
Given, $T_1=300K$
$V_2=64V_1$
putting the values we get,
$\Rightarrow 300\times V_1^{1/3}=T_2(64V_1{)}^{1/3}{T}_2=75K$
So, final temperature is $75K$

and work done will be ,
$W=n{C}_v(T_2-T_1)$
we know $C_v=\frac{R}{\gamma -1}$
so,
$W=1\times \frac{R}{\gamma -1}(75-300)=\frac{2}{1.33-1}(-225)=-1363.63cal$

so, option (b) is correct.