10 litres of a monoatomic ideal gas at $0^{o} C$ and 10 $a t m$ pressure is suddenly exposed to 1 $a t m$ pressure and the gas expands adiabatically against this constant pressure to maximum possible volume. The final temperature and volume of the gas respectively are

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Solution

This is adiabatic irreversible process,
$W = (\frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1})$ ....(i)
And work done against pressure is given as:
$W = - P_{e x t} (V_{2} - V_{1})$ ....(ii)

$P_{1} = 10 a t m$ , $P_{2} = P_{e x t} = 1 a t m$
$T_{1} = 273 K$ , $V_{1} = 10 L$
$γ = 5 / 3$ (for monoatomic gas)

Also we know:
$P_{1} V_{1} = n R T_{1} 10 \times 10 = n \times 0.082 \times 273 n = 4.47 m o l$

Equating eq. (i) and (ii):

$- P_{e x t} (V_{2} - V_{1}) = (\frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}) \Rightarrow - 1 \times (V_{2} - 10) = \frac{1 \times V_{2} - 10 \times 10}{1.67 - 1}$
$\Rightarrow (10 - V_{2}) = \frac{V_{2} - 100}{0.67} \Rightarrow 6.7 - 0.67 V_{2} = V_{2} - 100$
$\Rightarrow 106.7 = 1.67 V_{2} \Rightarrow V_{2} = 64 L$

now using;
$P_{2} V_{2} = n R T_{2} 1 \times 64 = 4.47 \times 0.082 \times T_{2} T_{2} = 174.6 K$
Thus, the closest option is (a).
Hence, option (a) is correct.

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