Change in Entropy for Isothermal Expansion

[ragingbuddha]

Homework Statement

Derive change in entropy for Van der Waal gas in isothermal expansion

TdS = ∫(dP/dV)dV + ∫(Cv/T)dT

Homework Equations

P = NRT/(V-bN) + a(N/V)^2

The Attempt at a Solution

TdS = ∫dP + 0 = Pf - Pi

TΔS = ∫dP = ∫(Pf - Pi)

TΔS = [ ∫ (NRT / Vf-bN) - a(N/Vf)^2] - [ ∫ (NRT / Vi-bN) - a (N/Vi)^2]

= [ NRT ln (Vf-bN) + aN^2/Vf ] - [ NRT ln (Vi-bN) + aN^2/Vi ]

= NRT ln (Vf-bN / Vi-bN) + a(N^2/Vf) - a (N^2/Vi)

 

[Andrew Mason]

Homework Statement

Derive change in entropy for Van der Waal gas in isothermal expansion

TdS = ∫(dP/dV)dV + ∫(Cv/T)dT

Where did you get this? The left side has dimensions of heat flow and the first integral on the right side has dimensions of pressure. Perhaps you meant (using the first law dQ = dU + PdV) :

$$\Delta S = \int dS = \int dQ/T = \int (nC_vdT + PdV)/T = n\int C_v\frac{dT}{T} + \int \frac{P}{T}dV$$

Use your expression for P for a Van der Waal gas and see if you can work it out from there.

AM

 

[ragingbuddha]

Where did you get this? The left side has dimensions of heat flow and the first integral on the right side has dimensions of pressure. Perhaps you meant (using the first law dQ = dU + PdV) :

This equation is given by the problem. Frankly, I don't know where that equation comes from either because I cannot derive it from anything I have learned so far.

 

[Andrew Mason]

This equation is given by the problem. Frankly, I don't know where that equation comes from either because I cannot derive it from anything I have learned so far.

It is wrong. Use the expression I have given you. You can multiply both sides by T since T is constant in the reversible, isothermal process.

AM