Answer to Question #303801 in Physical Chemistry for Chemistry

Question

Derive an expression for Entropy Change accompanying an isothermal reversible

expansion of an ideal gas.

Answer 1

Let us consider n moles of ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The work of expansion for a small change of volume dV against the external pressure P is given by-
dW=−PdV
∴ Total work done when the gas expands from initial volume V1​ to the final volume V2​, will be-
W=−∫V1​V2​​PdV
For an ideal gas,
PV=nRT
⇒P=V
nRT​
Therefore,
W=−∫V1​V2​​V
nRT​dV
For isothermal expansion, T is constant.Therefore,
W=−nRT∫V1​V2​​V
dV​
⇒W=−nRT[lnV]V1​V2​​
⇒W=−nRT(lnV2​−lnV1​)
⇒W=−nRTln(V1​
V2​​)
⇒W=−2.303nRTlog(V1​
V2​​).....(1)
At constant temperature,
P1​V1​=P2​V2​
⇒V1​
V2​​=P2​
P1​​
Therefore,
W=−2.303nRTlog(P2​
P1​​).....(2)
Equation (1)&(2) are the expression for the work obtained in an isothermal reversible expansion of an ideal gas.