Deriving Entropy Formula: Thermo Solution

[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 can't 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

 

[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.

 

[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.

 

[Mapes]

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.

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.