Heat capacity and its relation with internal energy

[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? please help with the derivation and clarification...

advanced regards.

 

[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

 

[Studiot]

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

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

As regards your query about C_v,

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

Since your version of the first law probably reads something like

dU = δQ - pdV and your definition of C_v 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 = C_vdT

Substituting into the first law

dU = C_vdT - pdV

But at constant volume dV = 0

dU = C_vdT

rearranging


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

 

[cooper607]

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

regards

 

[cooper607]

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

regards

 

[Studiot]

Glad you got it sorted