Question

20 litres of a monoatomic ideal gas at $0^{o}C$ and 20 $atm$ pressure is suddenly exposed to 1 $atm$ pressure and the gas expands adiabatically against this constant pressure to maximum possible volume. The final temperature and volume of the gas respectively are

• A. T = 169 K, V = 247.5 L
• B. T = 165 K, V = 247.5 L
• C. T = 169 K, V = 257.5 L
• D. none of these

Answer 1

The correct option is A T = 169 K, V = 247.5 L

This is adiabatic irreversible process,
$W=\left(\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}\right)$....(i)
And work done against pressure is given as:
$W=-P_{ext}(V_2-V_1)$....(ii)

$P_1=20atm$, $P_2=P_{ext}=1atm$
$T_1=273K$, $V_1=20L$
$\gamma =5/3$ (for monoatomic gas)

Also we know:
$P_1{V}_1=nR{T}_{1}20\times 20=n\times 0.082\times 273n=17.86mol$

Equating eq. (i) and (ii):

$-P_{ext}(V_2-V_1)=\left(\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}\right)\Rightarrow -1\times (V_2-20)=\frac{1\times V_2-20\times 20}{1.67-1}$
$\Rightarrow (20-V_2)=\frac{V_2-400}{0.67}\Rightarrow 13.4-0.67V_2=V_2-400$
$\Rightarrow 413.4=1.67V_2\Rightarrow V_2=247.5L$

now using;
$P_2{V}_2=nR{T}_{2}1\times 247.5=17.86\times 0.082\times T_2{T}_2=168.997K$
Hence option (a) is correct.