# A gas cylinder of $5L$ capacity containing $4\; kg$ of He gas at $27^{\circ}C$ developed leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is $1 \;atm$, calculate the work done by the gas assuming ideal behavior.

$\begin {array} {1 1} (A)\; 2.494 \times 10^3\: kJ & \quad (B)\;-24.94 \times 10^5\: kJ \\ (C)\; -24.94 \times 10^3\: kJ & \quad (D)\;-2.494 \times 10^3\: kJ \end {array}$

Number of moles of gas $n = \large\frac{4 \times 10^3}{4}$$= 10^3 moles Initial volume, V_1 = 5L;\: T = 27 + 273 = 300\: K Final volume, V_2 = \large\frac{nRT}{P}$$= 10^3 \times 0.0821 \times 300/1 = 24630\: L$
$\Delta V = V_2 – V_1 = 24630 – 5 = 24625\: L$
$W_{PV} = -P \Delta V = -1 \times 24625\: L-atm$
$= -24625 \times 101.3\: J = -24625 \times 101.3 \times10^{-3}\: kJ$
$= -2.494 \times 10^3 \: kJ$
Ans : (D)