Thermodynamics: find internal energy given the equation of state

AI Thread Summary
The discussion focuses on deriving the internal energy equation from a given equation of state, specifically p(V-b) = RT, using the relationship between specific heats. The participants derive the differential form of internal energy, dU, and relate it to the specific heat at constant volume, C_V, which is determined to be 2R based on the given C_p of 3R. The final expression for internal energy change is ΔU = 2RΔT, and the conversation touches on the implications of not determining absolute internal energy but rather a change in internal energy. Additionally, the classical interpretation of internal energy at absolute zero is debated, noting that while U=0 at T=0K is standard, quantum mechanics introduces the concept of zero-point energy.
gerardpc
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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!
 
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Have you tried applying what you know about "internal energy" from your coursework so far?
Same with specific heats.
 
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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?
 
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?
 
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.
 
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.
 
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