$5$ moles of an ideal gas expand isothermally and reversibly from a pressure of $10$ atm to $2$ atm at $300 K$ . What is the mass which can be lifted through a height of $1$ metre in this expansion?

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Solution

Work done by the system(w)
= $- n R T {log}_{e} \frac{P_{1}}{P_{2}} = 2.303 n R T log \frac{P_{1}}{P_{2}}$
$= - 2.303 \times 5 \times 8.314 \times 300 log \frac{10}{2} = - 20.075 \times 10^{3} J$
Let $M$ be the mass which can be lifted through a height of $1 m$ .
Work done in lifting the mass
$= M g h = M \times 9.8 \times 1 J$
So, $M \times 9.8 = 20.075 \times 10^{3}$
$M = 2048.469 k g$

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5 moles of an ideal gas expand isothermally and reversibly from a pressure of 10atm to 2atm at 300K. What is the mass which can be lifted through a height of 1 metre in this expansion?

Work done by the system(w)= −nRTlogeP1P2=2.303nRTlogP1P2=−2.303×5×8.314×300log102=−20.075×103JLet M be the mass which can be lifted through a height of 1 m.Work done in lifting the mass=Mgh=M×9.8×1 JSo, M×9.8=20.075×103M=2048.469 kg

$5$ moles of an ideal gas expand isothermally and reversibly from a pressure of $10$ atm to $2$ atm at $300 K$ . What is the mass which can be lifted through a height of $1$ metre in this expansion?