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Question

The maximum work of expansion of 2 moles of an ideal gas at 300 K, occupying a volume of 20 dm3,isothermally and reversibly until the volume becomes 40 dm3 is:
(R=8.314 JK−1 mol−1)

A
34.57 J
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B
3.457 J
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C
37.92 kJ
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D
34.57 kJ
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Solution

The correct option is D 34.57 kJ
As we know that work done in an isothermally reversible process is given as-
W=2.303nRTlogVfVi.....(1)
Given that:-
n=2 moles
T=300K
Vf=40dm3
Vi=20dm3
R=8.314J/molK
Substituting all these values in eqn(1), we have
W=2.303×2×8.314×300×log4020
W=3.458kJ
Hence the work done for the given process is 3.458kJ.

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