Homework Help: Internal Energy of virial expansion

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1. May 25, 2017

Mikhail_MR

Hello,
I have some trouble understanding the virial expansion of the ideal gas.

1. The problem statement, all variables and given/known data

I have given the state equation:
$$pV = N k_b T \left(1+\frac{A\left(T\right)}{V}\right)$$

2. Relevant equations

and a hint how to calculate the caloric equation of state $$\left(\frac{\partial U}{\partial V}\right)_T = T^2 \left(\frac{\partial}{\partial T}\left(\frac{p}{T}\right)\right)_V$$

3. The attempt at a solution

I calculated $\frac{\partial}{\partial T} \left(\frac{p}{T}\right)_V = \frac{N k_b}{V^2} \left(\frac{\partial A\left(T\right)}{\partial T}\right)$ first. So using the hint $$U = \int_0^V \frac{N k_b}{V^2} T^2 \frac{\partial A\left(T\right)}{\partial T} dV = -N k_b T^2 \frac{\partial A\left(T\right)}{\partial T} \frac{1}{V}$$
If my calculations are correct, why is my internal energy negative? Unfortenately, I don't know which sign does $\frac{\partial A\left(T\right)}{\partial T}$ have. My next task is to calculate $c_V$ as a function of $T$ and $V$ and determine when it is volume independent.

$$c_V = \left(\frac{\partial U}{\partial T}\right)_V = - \frac{1}{V} N k_B \left(2 T \frac{\partial A\left(T\right)}{\partial T} + T^2 \frac{\partial^2 A\left(T\right)}{\partial^2 T}\right)$$

If I look at this equation I cannot say when it does not depend on $V$. Where have I made a mistake?

Any help would be greatly appreciated

2. May 25, 2017

kuruman

Your lower limit of integration should not be zero. The gas is expanding, but not from zero initial volume. When you integrate, you get $-\frac{1}{V}$. When the gas expands, from initial volume $V_i$ to final volume $V$, $V_i~<V$ and $- \left[ \frac{1}{V} \right]_{V_i}^{V} >0$.

3. May 25, 2017

Mikhail_MR

Thank you, kuruman. I should have noticed that I can't use zero as the lower limit. If I use $V_i$ then I get
$$U = - \left[\frac{1}{V}-\frac{1}{V_i}\right] N k_B T^2 \frac{\partial A}{\partial T} > 0$$
How do I determine now $A$, so that my $c_V$ does not depend on $V$

4. May 25, 2017

Staff: Mentor

Your limits of integration should go from the current volume to infinite volume (where the ideal gas law applies). At the infinite volume limit, $U=U^{IG}(T)$

5. May 25, 2017

Mikhail_MR

The internal energy is a total differential. In case of an ideal gas it depends only on the specific heat capacity by constant volume. In this case I calculate using the hint something like specific heat capacity by constant temperature. I mean $c_T$ in $dU=c_V dT + c_T dV$. But I have no information about my first term. I must calculate the heat capacity by constant volume in the next part of the task.

Last edited: May 25, 2017
6. May 25, 2017

Staff: Mentor

The internal energy in your problem is $$U(T,V)=C_vT-\frac{N k_b T^2}{V} \frac{d A}{dT}$$

7. May 25, 2017

Mikhail_MR

Thank you, Chestermiller. Maybe you think $$U(T,V)=C_vT +\frac{N k_b T^2}{V} \frac{d A}{dT}$$ since $\int_V^{\infty} 1 / V^2 = - \left[1 / V\right]_V^{\infty} = - 0 + 1 / V$

8. May 25, 2017

Staff: Mentor

No. Differentiate it with respect to V and see what you get.

9. May 25, 2017

Mikhail_MR

It is strange: $$\frac{d\left( 1/V\right)}{dV} = -1 / V^2,$$ but it is also true, that $$\int_V^{\infty} 1 / V^2 = - \left[1 / V\right]_V^{\infty} = - 0 + 1 / V$$

10. May 25, 2017

Staff: Mentor

Flip the limits of integration.

11. May 25, 2017

Mikhail_MR

Thank you! To the second part of the task. I can define new $C_V\left(T, V\right)$, so that if I let $V$ to be infinite (ideal gas), I will get the internal energy of the ideal gas. So it is just
$$C_V\left(V, T\right) = 3/2 N k_B - N k_B / V \frac{d}{dt}\left(T^2 \frac{d}{dt} A\right)$$
It means, when $dA/dT$ or $\left(2 T \frac{\partial A\left(T\right)}{\partial T} + T^2 \frac{\partial^2 A\left(T\right)}{\partial^2 T}\right)$ are zero, I have no volume dependency.