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Question

Two moles of an ideal gas initially at 270C and 50 L volume are compressed isothermally and reversibly till the final volume of the gas is 5 L. Calculate q for the process.

A
11.4 kJ
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B
25.4 kJ
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C
30.1 kJ
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D
21.4 kJ
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Solution

The correct option is A 11.4 kJ
work done in an isothermal reversible process is given as:

w=2.303nRT(logV2V1)
Here, n=2 mol
T=27+273=300 K
V1=50 L, V2=5 atm
R=8.314 JK1mol1

Using the formula we get,
w=2.303×2×8.314×300×log550
w=11488 J

For isothermal process, U=0 (as temperature is constant)
so, from first law
U=q+w
q=w=11488 J
or
q=11.4 kJ

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