Question

1 mol of ammonia gas at ${27}^{o}C$ is expanded in adiabatic reversible condition to make volume 8 times ($\gamma =\frac{4}{3}$). Final temperature and work done respectively are:

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

Answer 1

Answer: (A)

150K, 900cal

Solution:- (A) $150K,900cal$

As we know that, for an adiabatic expansion,

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

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

Given:-

$T_1=27℃=(27+273)K=300K$

$T_2=T\left(\text{say}\right)=?$

$V_1=V\left(\text{say}\right)$

$V_2=8V$

$\gamma =\frac{4}{3}$

$\therefore \frac{T}{300}={\left(\frac{V}{8V}\right)}^{\frac{4}{3}-1}$

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

Now, using first law of thermodynamics,

$\Delta E=q+w$

As we know that, for adiabatic expansion, $q=0$

$\therefore \Delta E=w$

If gas expands, the internal energy decreases.

$\therefore w=\Delta E=-n{C}_{V}dT$

$\Rightarrow w=-[1\times \frac{R}{\gamma -1}\times (150-300)]$

$\Rightarrow w=-[-150\times \frac{2}{\left(\frac{4}{3}\right)-1}]$

$w=900cal$

Hence the value of final temperature and work done are $150K$ and $900cal$ respectively.