Question

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

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

Answer 1

Answer: (A)

$150K,900cal$

Given ,

reversible adiabatic expansion $(q=0),V_f=8V_1,n=1,T=300K,\gamma =1.33$

Using, $T{V}^{\gamma -1}=constant$

$\frac{T_1}{T_2}=(\frac{V_2}{V_1}{)}^{(\gamma -1)}$

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

$\Rightarrow T_2=150K\left(approx\right)$

Using, $\gamma =\frac{C_P}{C_V}$

$1.33=\frac{4R}{3R}=\frac{C_P}{C_V}$

So, by comparing both sides we can say that, $C_V=3R$

Using, $△E=q+E$

As, $q=0,△E=W$

$W=n{C}_{V}deltaT$

$W=1\left(3R\right)(T_2-T_1)$

$W=3R(150-300)$

$W=-150R$

So, work done will be $900cal$.