# Demonstrate that Cv depends only on temperature

1. Sep 13, 2014

### mwa1

Hello,

I stumbled upon this question and I don't know how to answer it...

I know that Cv is defined as Cv = δQv/dT = (∂U/∂T)v but I thought it's value was determined by the nature of the gas only (3/2 for monoatomic and 5/2 for diatomic).

Can someone help me figure this out ?

2. Sep 13, 2014

### Staff: Mentor

The 5/2 is not valid in the whole temperature range, as it does not take vibrations into account.

3. Sep 13, 2014

### mwa1

Ok, but do you have any idea of how I could prove that it depends only on temperature ? (other than experimentally)

4. Sep 13, 2014

### AlephZero

That's strange. I typed "Cv depends only on temperature" into Google and got lots of explanations.

5. Sep 13, 2014

### mwa1

Well I haven't. I wouldn't bother posting and waiting for an answer if I had found something convincing...

As I understand it, Internal Energy is defined as (for monatomic gases) the mean Kinetic Energy of all molecules and Temperature is a measurement of it :

Nm<v2>/2 = U = 3NkBT/2

and Cv is just 3NkB/2

Cv can always be written in terms of U and T but then how do I get rid of the U ?

6. Sep 13, 2014

### Staff: Mentor

Cv does not depend only on temperature. It also depends on pressure, as does U. If you are focusing exclusively on gases, then Cv depends only on temperature in the limit of very low pressures. This is what we call the ideal gas region. At higher pressures, Cv depends on pressure.

Chet

7. Sep 13, 2014

### Andrew Mason

It is an empirical fact that for all gases at low pressures Cv depends only on temperature. But in order to show mathematically the relationship between Cv and temperature for a particular gas you would need to know the equation of state for the gas. For an ideal monatomic gas, Cv is constant:

PV=nRT

dQ/dT = d/dT(U + PdV)

(dQ/dT)V = (dU/dT)V = Cv

From Kinetic Theory, U = 3nRT/2. So (dU/dT)V = 3nR/2 = Cv = constant

AM

Last edited: Sep 13, 2014