Thermo Derivation

1. Jan 5, 2009

olechka722

Hello,

I am trying to derive a formula for entropy. I have:

dS= Cv/T dT + R/(V-b) dV

and want to get:

S= Cv*ln(T) + R*ln(V-b) + constant.

Math rules seem to say i cant just integrate this up even though it looks obvious since i have two different d's on the right hand side. Maybe something using Maxwell relations? Not sure.

Thank you for any help!

olechka

2. Jan 5, 2009

Mapes

The first term on the right side contains only a constant, a function of T, and dT. The second term contains only constants, a function of V, and dV. It's OK to integrate the terms separately in this case. You can verify this by differentiating the second expression.

3. Jan 5, 2009

Renge Ishyo

You can just integrate it up I believe. The derivatives in this instance are actually partial derivates if I am not mistaken (the evidence is that Cv is the heat capacity at constant V...which implies that the second term is the change due to volume at constant T). So the total change in S is the sum of the integration of the two partial derivatives.

4. Jan 5, 2009

Mapes

Not quite. It's true that the terms represent partial derivatives; that is, we're looking at

$$dS=\left(\frac{\partial S}{\partial T}\right)_V dT+\left(\frac{\partial S}{\partial V}\right)_T dV$$

However, this observation isn't sufficient to conclude that the integral of the right hand side equals the sum of the integrals of the individual terms. We must also require that $C_V(T,V)=T\left(\frac{\partial S(T,V)}{\partial T}\right)_V$, which is generally a function of both T and V, is idealized as a constant, as I stated above.