# Heat capacity and its relation with internal energy

1. Jun 25, 2012

### cooper607

hi everyone,
in thermodynamics, when we calculate the heat capacity in constant volume, we assume Cv=dQ/dT..

well, but at isothermal condition suddenly they came up with Cv=dU/dT...
so i am getting stuck with this concept how they replace dQ with dU?

i know U= internal energy is only a function of T, but what is the explanation behind it? plz help with the derivation and clarification...

2. Jun 25, 2012

### Rap

You have to be careful with that "dQ" - there is no such thing as "Q" so there is no such thing as "dQ". Usually it is written "$\delta Q$", which is an abbreviation for "T dS" to emphasize this fact. You only use the "d" in front of state variables, and there is no "Q" state variable. A state variable X obeys $\oint dX=0$. In other words the integral around a closed path of a state variable is zero, and the integral between two points is not dependent on the path. $\oint \delta Q$ is really $\oint T dS$ and that's not necessarily zero.

The correct way to define the C's are $TdS=C_V dT$ at constant volume and $TdS=C_P dT$ at constant pressure.

You can write the fundamental equation (at constant number of particles) as $$dU=TdS-PdV$$ and since, by the chain rule, $$dU=\left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$ you can see that at constant volume, $$TdS= \left(\frac{\partial U}{\partial T}\right)_V dT$$ so that $$C_V=\left(\frac{\partial U}{\partial T}\right)_V$$

You can also write the fundamental equation (at constant number of particles) as $$dH=TdS+VdP$$ where H is the enthalpy (H=U+PV) and since, by the chain rule, $$dH=\left(\frac{\partial H}{\partial T}\right)_P dT + \left(\frac{\partial H}{\partial P}\right)_T dP$$ you can see that at constant pressure, $$TdS= \left(\frac{\partial H}{\partial T}\right)_P dT$$ so that $$C_P=\left(\frac{\partial H}{\partial T}\right)_P$$

3. Jun 25, 2012

### Studiot

I think you are mixing up several things you have been told.
Don't worry this is not uncommon there is a lot to get hold of in thermodynamics.

So firstly your statement about U is only true for ideal gasses. It is one of the possible definitions of an ideal gas and was proved experimentally by Joule. However it is not relevant here.

$${\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0$$

It is not necessary to invoke entropy and considering your definition of Cv you may not have met entropy anyway.

dU = δQ - pdV and your definition of Cv is

$${C_v} = {\left( {\frac{{\delta Q}}{{\partial T}}} \right)_V}$$

(RAP is correct it is not a good idea to use dQ)

So δQ = CvdT

Substituting into the first law

dU = CvdT - pdV

But at constant volume dV = 0

dU = CvdT

rearranging

$${C_v} = {\left( {\frac{{dU}}{{dT}}} \right)_V}$$

4. Jun 25, 2012

### cooper607

oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

regards

5. Jun 25, 2012

### cooper607

oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

regards

6. Jun 25, 2012