# Heat capacities definition

1. Jan 27, 2012

### dapias09

Hi all,

I'm working with the heat capacities definition and I have got a confusion. I don't understand why we can express them like

Cp = T(∂S/∂T)p
Cv = T(∂S/∂T)v

I know that Cp=(dQ/dT)p = (∂H/∂T)p with H equal to TdS + VdP and Cv=(dQ/dT)v = (∂U/∂T)v with U equal to TdS + PdV,

My guess: If I begin for instance with the enthalpy, H, and I constrain it to constant pressure I get just the TdS term, that is the thermodynamic definition of "Heat (Q)" . So I get the definition of the heat capacity if I derive Q respect to T, or TdS respect to T (is the same thing). Doing it, I get:

dQ/dT = (TdS)/dT = Td^2S/d^2T + dS/dT.

An expression very different of the definition.

Can anyone help me?

2. Jan 27, 2012

### Andrew Mason

You cannot equate TdS to dQ unless dQ = dU + PdV, and that is true only if the process is reversible. Generally, you have to use the first law: dQ = dU + ∂W.

I think you mean dU = TdS - PdV. Again, that is only true if it is a reversible process.

I don't follow what you are doing. Why would it not just be T(dS/dT)? Again, this is true only in a reversible process.

AM

3. Jan 27, 2012

### juanrga

( ∂Q/∂T )p = ( T∂S/∂T )p = T (∂S/∂T)p

Cp = T (∂S/∂T)p = (∂Q/∂T)p
Cv = T (∂S/∂T)v = (∂Q/∂T)v

4. Jan 27, 2012

### pabloenigma

The problem you have encountered in the italicized step is mathematical.
(TdS)/dT is T(dS/dT).But in the mentioned step,you have computed the derivative of (T(dS/dT)).

5. Jan 28, 2012

### dapias09

Thanks everybody,