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Question

Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at 127C from a volume of 10 dm3 to 20 dm3 Take ln2=0.3 and R=8.3 J K1 mol1

A
2293.78 J
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B
2480.78 J
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C
-2293.78 J
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D
-2480.78 J
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Solution

The correct option is A 2293.78 J
Since the process is isothermal
E=H=0
according to first law of thermodynamics,
E=q+w=0q=W
Again, for isothermal process,
W=2.303nRTlogV2V1=2.303×1×8.3×400log2010=2.303×1×8.3×400×0.3 J=2293.78 J

Again, q=W=2293.78 J

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