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

Thermodynamics

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Thermodynamics

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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)

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