Demonstrate that Cv depends only on temperature

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mwa1
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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 ?
 
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but I thought it's value was determined by the nature of the gas only (3/2 for monoatomic and 5/2 for diatomic).
The 5/2 is not valid in the whole temperature range, as it does not take vibrations into account.
 
Ok, but do you have any idea of how I could prove that it depends only on temperature ? (other than experimentally)
I can't seem to find anything about this on the internet or my textbook.
 
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 ?
 
mwa1 said:
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 ?
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
 
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