# Thermodynamics: find internal energy given the equation of state

1. Jun 25, 2013

### gerardpc

Given the following equation of state and $C_p = 3R$, find the equation of the internal energy in terms of T and V

$p(V-b) = RT$

Any clues? I can use also the specific heat at constant pressure or volume. Thanks!

Last edited: Jun 26, 2013
2. Jun 25, 2013

### Simon Bridge

Have you tried applying what you know about "internal energy" from your coursework so far?
Same with specific heats.

3. Jun 26, 2013

### gerardpc

Observation: I changed a little the statement of the problem, as I noticed I had additional constraints.

Now to the problem: I have found the equation (which I didn't remember...)

$dU =C_{V}dT +\left[T\left(\frac{\partial p}{\partial T}\right)_{V} - p\right]dV$

So with the equation of state it gives:

$\dfrac{\partial p}{\partial T} = \dfrac{\partial}{\partial T}\left(\dfrac{R}{V-b}T\right) = \dfrac{R}{V-b}$

And including it in the previous equation:

$dU = C_{V}dT +\left[T\dfrac{R}{V-b} - p\right]dV = C_{V}dT +\left[T\dfrac{R}{V-b} - \dfrac{R}{V-b}T \right]dV = C_{V}dT$

Now, to put $C_{V}$ in terms of $C_{P}$ we have the relation:

$C_p - C_V = T \left(\frac{\partial p}{\partial T}\right)_{V} \left(\frac{\partial V}{\partial T}\right)_{p}$

The terms of which we can find with the equation of state:

$\frac{\partial p}{\partial T} = \dfrac{R}{V-b} \qquad \frac{\partial V}{\partial T} = \dfrac{R}{p}$

So

$C_p - C_V = \cdots = R \Rightarrow C_V = C_p - R = 2R$

And finally $dU = C_{V}dT= 2R dT$ so $\Delta U = 2R \Delta T$

Is it alright? Any problem with not determining $U$ but $\Delta U$?

4. Jun 26, 2013

### Simon Bridge

I dunno - its your course.

Note: you had $$\frac{dU}{dT}=2R$$ ... so why not treat it like an initial value problem? What is U when T=0K?

5. Jun 26, 2013

### gerardpc

Well, I'm not sure of the initial condition, that's why I asked ;)
I guess it must be zero internal energy when temperature is zero, so then we can drop de deltas.

6. Jun 26, 2013

### Simon Bridge

There you go.

All you need for an "initial" condition is any known value.
The problem statement will not always include all the information you are expected to use.

Note: U=0 @ T=0 would be the classical interpretation - IRL you still have zero-point energy corresponding to the QM ground state.