# How to solve constants out of the internal energy equation?

Imagine we deal with a new kind of matter, whose state is described by:

$$PV = AT^3$$

Its internal energy is given by

$$U = BT^n \ln\left(\frac{V}{V_0}\right) + f(T)$$

where $$A$$, $$B$$ and $$V_0$$ is a constant and $$f(T)$$ is a polynomial function.

Find $$B$$ and $$n$$.

### What I know

The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.

$$\Delta U = -W$$

And work is:

$$W = -P\Delta V$$

### Some thoughts on how to solve the problem

We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:

$$\mathrm{d}U = \left[nBT^{n-1}\ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T$$

$$\left[nBT^{n-1} \ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T = \frac{AT^3}{V}\mathrm{d}V$$

I do not see a solution to this differential equation. There has to be a easier way to get both $$B$$ and $$n$$, but how?

• Your differential equation doesn't make sense to me: you have infinitesimals ($\mathrm dU$ and $\mathrm df$) and finite quantities ($nBT^{n-1}\ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout? – orthocresol Apr 27 at 16:16
• @orthocresol that is a typo let me fix it. – JD_PM Apr 27 at 16:18
• Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$) – JD_PM Apr 27 at 16:25
• By using the equation $\big( \frac{\partial U}{\partial V}\big)_T = T\big(\frac{\partial P}{\partial T}\big)_V - P$, you will get $B$ as $2A$ and $n=3$. – Soumik Das Apr 27 at 16:50

## 2 Answers

You need to use the equation $$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$

• Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + \mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term. – JD_PM Apr 27 at 17:48
• The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV. – Chet Miller Apr 27 at 17:55
• I use to study from two books I really like: Physics for Scientists and Engineers (Serway and Jewett) and Thermal Physics (Schroeder). The former is more general and has a good part on Thermodynamics but it doesn't deal with thermodynamic identities. The later does (pages 156 and 157) but it is not enough for me. May you recommend a source to study thermodynamic identities deeper (where all the formulas are derived and so on)? Please let me know if I should ask this in a separate post. Thanks – JD_PM Apr 30 at 8:15
• Fundamentals of Engineering Thermodynamics, Moran et al. Introduction to Chemical Engineering Thermodynamics, Smith and van Ness. Principles of Chemical Equilibrium, Denbigh. – Chet Miller Apr 30 at 11:43

I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.

Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP