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Question

When 2 moles of an ideal monoatomic gas are heated from 300 K to 600 K at constant pressure. The change in entropy of gas (S) is:

A
3Rln2
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B
32Rln2
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C
5Rln2
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D
52Rln2
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Solution

The correct option is C 5Rln2
The entropy change for an isobaric process is given as:
S=nCp,m lnT2T1
where
T2=600 K
T1=300 K
n=2
for ideal monoatomic gas we know
Cp,m=52R

Putting the values we get:
S=2×52R×ln600300
= 5R ln2

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