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? plz 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 "[itex]\delta Q[/itex]", 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 [itex]\oint dX=0[/itex]. 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. [itex]\oint \delta Q[/itex] is really [itex]\oint T dS[/itex] and that's not necessarily zero.

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

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

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

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.


[tex]{\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0[/tex]

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


[tex]{C_v} = {\left( {\frac{{\delta Q}}{{\partial T}}} \right)_V}[/tex]

(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


[tex]{C_v} = {\left( {\frac{{dU}}{{dT}}} \right)_V}[/tex]

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