Question

1 mole of $N{H}_3$ gas at ${27}^{o}C$ is expanded in reversible adiabatic condition to make volume 8 times $(\gamma =1.33).$ Final temperature and magnitude of work done respectively are

• A. 150 K, 900 cal
• B. 150 K, 400 cal
• C. 250 K, 1000 cal
• D. 200 K, 800 cal

Answer 1

The correct option is A 150 K, 900 cal

For a reversible adiabatic process :
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
Given, $T_1=300K$
$V_2=8V_1$
putting the values we get,
$\Rightarrow 300\times V_1^{1/3}=T_2(8V_1{)}^{1/3}{T}_2=150K$
So, final temperature is $150K$

and work done will be ,
$W=n{C}_v(T_2-T_1)$
we know $C_v=\frac{R}{\gamma -1}$
so,
$W=1\times \frac{R}{\gamma -1}(150-300)=\frac{2}{1.33-1}(-150)=-900cal$

so, option (a) is correct.