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Question

The entropy change when two moles of ideal monoatomic gas is heated from 200C to 300C reversibly and isochorically?

A
32R ln(300200)
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B
52R ln(573273)
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C
3 R ln(573473)
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D
32R ln(573473)
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Solution

The correct option is C 3 R ln(573473)
Entropy is given as:
S=nCv lnT2T1+nR lnV2V1

So, entropy change for an isochoric process (constant volume) is given as:

S=nCv lnT2T1
where
T2=573 K
T1=473 K
n=2
for ideal monoatomic gas we know
Cv=32R

Putting the values we get:
S=2×32R×ln573473=3R ln(573473)

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