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The change in the entropy of a 1 mole of...

The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process form an initial state `(P_1,V_1,T)` to the final state `(P_2,V_2 , T)` is equal to

A

Zero

B

R In T

C

R In `(V_1)/(V_2)`

D

R In `(V_1)/(V_1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Change in entropy of an ideal gas -
`DeltaS = (DeltaQ)/(T)`
For isothermal process `DeltaU = 0` Now, `DeltaQ=DeltaU +DeltaW implies DeltaQ = DeltaW `
`DeltaQ=DeltaW = n R T In ((V_2)/(V_1))`
for n = 1 (one mole) `DeltaS =(DeltaQ)/(T) = R In ((V_2)/(V_1))`
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