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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

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.

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

1 Answer

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)

 

answered Mar 15, 2014 by thanvigandhi_1
 

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