Question

20 litres of a monoatomic ideal gas at $0^{o}C$ and 10 atm pressure is suddenly released to 1 atm pressure and the gas expands adiabatically against this constant pressure. The volume of the gas respectively are

• A. $V=164L$
• B. $V=57L$
• C. $V=123.7L$
• D. $V=68.3L$

Answer 1

The correct option is C $V=123.7L$

This is an adiabatic irreversible process, so for this process
$P{V}^{\gamma }=$ Constant, is not applicable
$W=-P_{ext}(V_2-V_1)$
$V_2\text{is final volume},V_1\text{is initial volume},\text{W is work done}$
$T_2\text{is final temperature,}{T}_1\text{is initial temperature}$
But for adiabatic process,

$W=dU=\left(\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}\right)$
Also, we know;
$PV=nRT\Rightarrow 10\times 20=n\times 0.082\times 273\Rightarrow n=\mathrm{8.92.}moles$
Also, Comparing the work done, We get
$-P_{ext}(V_2-V_1)=\left(\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}\right)$
$\Rightarrow -1\times (V_2-10)=\frac{1\times V_2-10\times 20}{1.67-1}$
$\Rightarrow (10-V_2)=\frac{V_2-200}{0.67}$
$\Rightarrow 6.7-0.67V_2=V_2-200$
$\Rightarrow 206.7=1.67V_2$
$\Rightarrow V_2=123.75L$