Question

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

$T_v^{\gamma }=constant$

$T_2=T_1(\frac{v_1}{v_2}{)}^{\gamma -1}$

$=150$

Since,$(C_v=\frac{R}{\gamma -1})$

Hence,

$w=-n{C}_{v}dT$

$=-3\times 2(150-300)=900Cal$

$\gamma =\frac{8}{6}=1.33$

$T_2=300\times {\frac{v}{8v}}^{\frac{8}{6}-1}$

$\frac{300}{2}=150K$