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 '19 at 16:16
• @orthocresol that is a typo let me fix it. – JD_PM Apr 27 '19 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 '19 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 '19 at 16:50

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 '19 at 17:48