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

Options

A. Zero
B. R In T
C. R In `(V_1)/(V_2)`
D. R In `(V_1)/(V_1)`

Answer

D

Explanation

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))`