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Question

Calculate q for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure 1 bar to a final pressure 0.1 bar at constant temperature
of 273 K.

A
-5227 J
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B
5227 J
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C
2240 J
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D
None of these
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Solution

The correct option is B 5227 J
For isothermal reversible expansion

W=2.303 nRT logP1P2

=2.303×1×8.314×273log1.00.1=5227.2J

At constant temperature, for expansion ΔT=0

Therefore , ΔE=0 Also, ΔU=n×Cv×ΔT=0

And ΔU=q+W (for 1 mole)

q=W=5227.2J

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