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Question

Calculate q (heat transfered) for an isothermal reversible expansion of 1 mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K.

A
7 kJ
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B
5.2 kJ
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C
4.3 kJ
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D
6 kJ
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Solution

The correct option is B 5.2 kJ
As in isothermal process as temperature remains constant so U is zero (as it is a function of temperature only)
U = 0

According to first law of thermodynamics,
U=W+q
0=W+q
q=W
as W=2.303nRT(logP1P2)
q=W=2.303nRT(logP1P2)
Given : P1=1 bar
P2=0.1 bar
T=273 K
n=1 mol

q=2.303×1×8.314×273(log10.1)
=5227.1 Jq=5.2 kJ


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