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Question

Work done by 0.1 mole of a gas at 27C to double its volume at constant pressure is
[Take R=2 cal mol1K1]

A
54 cal
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B
600 cal
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C
60 cal
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D
546 cal
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Solution

The correct option is C 60 cal
Given
Number of moles of gas = 0.1 mole
Initial temperature (T1)=27C=300 K

From ideal gas equation,
PV=nRT
P=nRTV
Pressure is constant
nRTV=constantTV=constant
Hence, T1V1=T2V2
From the data given in the question,
V2=2V1
T2=2T1
ΔT=T2T1=2T1T1=300 K
Work done in an isobaric process is given by
W=PΔV
Using the ideal gas equation, we can write it as,
W=nRΔT {using ideal gas equation}
=0.1×2×300
=60 cal
Thus, option (c) is the correct answer.

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