Internal Energy of virial expansion

[Mikhail_MR]

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

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

Homework Equations

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

The Attempt at a Solution

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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

 

[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 $- \left[ \frac{1}{V} \right]_{V_i}^{V} >0$.

 

[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 $V$

 

[Chestermiller]

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

 

[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.

 

[Chestermiller]

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

 

[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 $\int_V^{\infty} 1 / V^2 = - \left[1 / V\right]_V^{\infty} = - 0 + 1 / V$

 

[Chestermiller]

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

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

 

[Mikhail_MR]

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

 

[Chestermiller]

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

Flip the limits of integration.

 

[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.