Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at 127∘C from a volume of 10 dm3 to 20 dm3 Take ln2=0.3 and R=8.3 J K−1 mol−1
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=0⇒q=−W
Again, for isothermal process, W=−2.303nRTlogV2V1=−2.303×1×8.3×400log2010=−2.303×1×8.3×400×0.3J=−2293.78J