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Work Done in Reversible Adiabatic Process

One mole of a...

Question

One mole of an ideal gas expands isothermally from initial state ( 1 atm, 22.4 L, 273 K ) to the final state ( $0.25 a t m, 89.6 L$ ) in reversible manner. One mole of the same gas with the same initial state in a different cylinder expands against a constant external pressure of $0.25 a t m$ to final volume $89.6 L$ . The difference in work done in $J$ is (Round the answer to nearest integer)
$(T a k e 1 L a t m = 101 J, R = 8.3 J K^{- 1} m o l^{- 1}, l o g 2 = 0.3)$

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Solution

We know, for reversible isothermal process
$W = - 2.303 nRT log \frac{V_{2}}{V_{1}}$
Here, n = number of moles of the gas
R = gas constant
T = temperature
$V_{1}, V_{2}$ are initial and final volumes respectively.

$W = - 2.303 \times 8.3 \times 273 \times l o g \frac{89.6}{22.4}$
$= - 3132.02 J$
Work done in irrversible process $= - P_{e x t} (V_{f} - V_{i}) \times 101 J$
$= - 0.25 (89.6 - 22.4) \times 101$
$= - 1696.8 J$
Difference in work done = $3131.02 - 1696.80 = 1434.22 J \approx 1434 J$

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