# Entropy and Heat Capacity Relation

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• cwill53

#### cwill53

I have a simple question sort of about exact differentials and deciding which variables matter and when.

I know we can write entropy ##S## as ##S(P,T)## and ##S(V,T)## to derive different relations between heat capacities ##C_V## and ##C_P##. I was wondering if it is technically correct to write

$$dS(P,V,T)=\left ( \frac{\partial S}{\partial P} \right )_{V,T}dP+\left ( \frac{\partial S}{\partial V} \right )_{P,T}dV+\left ( \frac{\partial S}{\partial T} \right )_{P,V}dT$$

$$\delta Q_{rev}=TdS(P,V,T)=T\left ( \frac{\partial S}{\partial P} \right )_{V,T}dP+T\left ( \frac{\partial S}{\partial V} \right )_{P,T}dV+T\left ( \frac{\partial S}{\partial T} \right )_{P,V}dT$$

Then how would we define heat capacities ##C,C_V,C_P## in terms of these partial derivatives?

I know we can define them as (at constant volume and constant pressure, respectively)

$$C_V=T\left ( \frac{\partial S}{\partial T} \right )_{V}; C_P=T\left ( \frac{\partial S}{\partial T} \right )_{P}$$

But I can't understand how to reconcile the gap between the second and third equations I typed.

There's a couple points of confusion here.

First, the first equation you wrote is not correct. Those partial derivatives are not independent. A given thermodynamic system has state variables P,V,T, and N. For a closed system, N is fixed. That means you're down to 3 remaining variables, as in your equation. However, there's always another constraint: the equation of state (e.g., ideal gas law or van der waals gas law), which reflects the unique physics of the system. This brings you down to two independent variables. In other words, any set of three state variables (P,V,T) or (P,S,T) or what have you, is over-defined.

Second, what is this third heat capacity C (as opposed to ##C_v## or ##C_p##) that you mention? The other two contain everything you need. Remember, any reversible process can be broken down into and isobaric or isochoric part plus an isothermal part. Irreversible processes that are internally reversible can be handled by the same approach if the rate of irreversibility is known.

• cwill53
There's a couple points of confusion here.

First, the first equation you wrote is not correct. Those partial derivatives are not independent. A given thermodynamic system has state variables P,V,T, and N. For a closed system, N is fixed. That means you're down to 3 remaining variables, as in your equation. However, there's always another constraint: the equation of state (e.g., ideal gas law or van der waals gas law), which reflects the unique physics of the system. This brings you down to two independent variables. In other words, any set of three state variables (P,V,T) or (P,S,T) or what have you, is over-defined.

Second, what is this third heat capacity C (as opposed to ##C_v## or ##C_p##) that you mention? The other two contain everything you need. Remember, any reversible process can be broken down into and isobaric or isochoric part plus an isothermal part. Irreversible processes that are internally reversible can be handled by the same approach if the rate of irreversibility is known.
I ended up finding out that the phase space isn’t three dimensional; and also that ##S(P,V,T)## is illegal anyway as the third variable is already determined by the other two, and also S is not a thermodynamic potential. I can derive the heat capacities using ##S(P,T)## and ##S(V,T)## easily, I was just confused on how ##S(P,V,T)## was ill-defined. I understand now though.

• Twigg