Deriving an expression for internal energy.

[L-x]

Homework Statement

Show that the change in internal energy of a simple system between states (V1, T1)
and (V2, T2) is given by

$$∆U = \int^{T1}_{T2} C_v\ dT + \int^{V1}_{V2} T.\frac{\partial p}{\partial T}|_V - p \ dV$$

Homework Equations

dU=dQ-pdV

The Attempt at a Solution

As U is a function of state i wrote down $$dU =\frac{\partial U}{\partial T}|_V dT + \frac{\partial U}{\partial V}|_T dV$$

$$\frac{\partial U}{\partial T}|_V$$ is clearly just Cv but i can't get the other part into the correct form, my manipulation is just going around in circles.

 

[danago]

Keep in mind that, in terms of its fundamental variables, a differential in internal energy is also given by:

$$dU = T dS - P dV$$

You can then write:

$$<br /> \left( {\frac{{\partial U}}{{\partial V}}} \right)_T = T\left( {\frac{{\partial S}}{{\partial V}}} \right)_T - P<br />$$

Can you figure out what to do from there?

 

[L-x]

Keep in mind that, in terms of its fundamental variables, a differential in internal energy is also given by:

$$dU = T dS - P dV$$

Doesn't this only hold for a reversible process?

 

[danago]

The equation is derived for a reversible process, however internal energy is a state function so it can be applied to non-reversible processes.

 

[L-x]

Ah of course! Thanks very much for your help.