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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 lt T)` is equal to

A

zero

B

R ln T

C

R ln `(V_1)/(V_2)`

D

`Rln(V_2)/(V_1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaQ=DeltaW=nRTln((V_2)/(V_1))=RTln((V_2)/(V_1))`
`DeltaS=(DeltaQ)/(T)=Rln((V_2)/(V_1))`
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CP SINGH-LAWS OF THERMODYNAMICS-EXERCISE
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  19. The change in the entropy of a 1 mole of an ideal gas which went throu...

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