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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 atm,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 atm to final volume 89.6 L. The difference in work done in J is
(Take 1 L atm=101 J, R=8.3 J K1 mol1, log2=0.3)

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Solution

We know, for reversible isothermal process
W=2.303nRT logV2V1
Here, n = number of moles of the gas
R = gas constant
T = temperature
V1,V2 are initial and final volumes respectively.

W=2.303×8.3×273 ×log89.622.4
=3131.02 J
Work done in irrversible process =Pext(VfVi)×101 J
=0.25(89.622.4)×101
=1696.8 J
Difference in work done = 3131.201696.80=1434.40 J

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