Thermodynamics: find internal energy given the equation of state

[gerardpc]

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

<br /> p(V-b) = RT<br />

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

 

[Simon Bridge]

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

 

[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?

 

[Simon Bridge]

I don't know - 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?

 

[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.

 

[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.