Obtaining this form for molar energy under virial expansion (Callen)

[EE18]

In his Chapter 13.3 (2nd edition), Callen gives the standard form for the virial expansion for the mechanical equation of state of a fluid as an exapnsion in powers of the molar volume $v$:
$$P = \frac{RT}{v}\left(1 + \frac{B(T)}{v} + \frac{C(T)}{v^2} + \dots \right) \equiv P_{ideal} + \frac{RT}{v}\left( \frac{B(T)}{v} + \frac{C(T)}{v^2} + \dots \right),$$
where I have shown explicitly that the first term in the expansion is the pressure of an ideal gas under the same conditions.

Callen then indicates that this fact implies that the molar energy of such a gas should be expressible as
$$u = u_{ideal} + RT^2\left( \frac{1}{v}\frac{dB}{dT} + \frac{1}{2v^2}\frac{dC}{dT} + \dots \right).$$
How can I see this? I know that
$$P = - \left(\frac{\partial u}{\partial v} \right)_s,$$
but the presence of the $s$ makes this difficult (for me) to see how to proceed. In contrast, since
$$P = - \left(\frac{\partial f}{\partial v} \right)_T,$$
we can immediately integrate the expression for $P$ (which is in terms of $v$ and $T$) term-wise to obtain $f$.

 

[Chestermiller]

$$dU=TdS-PdV=T\left(\frac{\partial S}{\partial T}\right)_VdT-PdV+T\left(\frac{\partial S}{\partial V}\right)_TdV$$But $$T\left(\frac{\partial S}{\partial T}\right)_V=C_v$$and $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$So, $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$$$=C_vdT-\frac{\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]}{\left(\frac{\partial P}{\partial V}\right)_T}dP$$

 

[EE18]

$$dU=TdS-PdV=T\left(\frac{\partial S}{\partial T}\right)_VdT-PdV+T\left(\frac{\partial S}{\partial V}\right)_TdV$$But $$T\left(\frac{\partial S}{\partial T}\right)_V=C_v$$and $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$So, $$dU=C_vdT+\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$$$=C_vdT+\frac{\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]}{\left(\frac{\partial P}{\partial V}\right)_T}dP$$

I'm not sure I see how the final expression gets me there. The form for $C_v$ is not immediately clear to me, and it seems like you're tacitly saying I could use that in your final expression?

 

[Chestermiller]

I'm not sure I see how the final expression gets me there. The form for $C_v$ is not immediately clear to me, and it seems like you're tacitly saying I could use that in your final expression?

What part don't you understand? All you need to do is substitute the equation for P in post #1 into this equation, and you get the Callen result.

 

[EE18]

What part don't you understand? All you need to do is substitute the equation for P in post #1 into this equation, and you get the Callen result.

You have $c_v$ in the expression too though? I think that's where I'm not quite following.

 

[Chestermiller]

You have $c_v$ in the expression too though? I think that's where I'm not quite following.

OK. How about this: $$U=U_{IG}(T)+\int_v^{\infty}\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$where V is a dummy variable integration for v.

 

[EE18]

OK. How about this: $$U=U_{IG}(T)+\int_v^{\infty}\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$where V is a dummy variable integration for v.

I guess I'm not quite sure how you got to that form from the total differential you gave above. Did you integrate first at infinite volume (fixed) so that you had an ideal gas, and then proceed with the volume part of the integral? But then it seems like there's no denominator now?

 

[TSny]

The virial expansion $\frac{P}{T} = \frac{R}{v}\left[1+\frac{B(T)}{v} + \frac{C(T)}{v^2} + ...\right]$ is one equation of state of the gas. This equation of state alone is not sufficient information to derive the virial expansions for the other quantities $f$, $c_v$, and $u$ that Callen gives in the second edition of the text. See section 3-3 in the 2nd edition for a general discussion of the information content of equations of state.

As a simple illustration, for the classical ideal gas, we have the equation of state $\frac {P}{T} = \frac {R}{v}$. This expression does not depend on the molecular structure (monatomic, diatomic, etc.) But the expression for the molar energy $u(T)$ does depend on the molecular structure. So, it is clear that you can't derive $u(T)$ from the $\frac{P}{T}$ equation alone.

In the first edition, Callen gives the virial expansions only for $\frac{P}{T}$ and $c_v(T,v)$. After giving $\frac{P}{T}$, he states,

“The specific heat $c_v$ of a real gas also depends upon the virial coefficients. It is found by statistical mechanical calculations that $c_v = …$".

So, he’s referring to results of statistical mechanics rather than implying that $c_v$ can be derived from the $\frac{P}{T}$ equation using just thermodynamics. In the 2nd edition, he is not as explicit in indicating how the results are obtained.

In comparing the second edition with the first edition and also with another book on statistical mechanics, I believe there are some sign errors in the 2nd edition for the virial expansions of $f, c_v$ and $u$. These are equations 13.28, 13.29, and 13.30. In each of these equations, the sign in front of $RT$ is wrong.

 

[Chestermiller]

I guess I'm not quite sure how you got to that form from the total differential you gave above. Did you integrate first at infinite volume (fixed) so that you had an ideal gas, and then proceed with the volume part of the integral? But then it seems like there's no denominator now?

I start out at the ideal gas state at temperature T and very large specific volume, and then decrease the specific volume at constant temperature T until I reach the desired final state at (T,v). The latter is what the integral calculates. No denominator is needed if we are working with v.

 

[EE18]

The virial expansion $\frac{P}{T} = \frac{R}{v}\left[1+\frac{B(T)}{v} + \frac{C(T)}{v^2} + ...\right]$ is one equation of state of the gas. This equation of state alone is not sufficient information to derive the virial expansions for the other quantities $f$, $c_v$, and $u$ that Callen gives in the second edition of the text. See section 3-3 in the 2nd edition for a general discussion of the information content of equations of state.

As a simple illustration, for the classical ideal gas, we have the equation of state $\frac {P}{T} = \frac {R}{v}$. This expression does not depend on the molecular structure (monatomic, diatomic, etc.) But the expression for the molar energy $u(T)$ does depend on the molecular structure. So, it is clear that you can't derive $u(T)$ from the $\frac{P}{T}$ equation alone.

In the first edition, Callen gives the virial expansions only for $\frac{P}{T}$ and $c_v(T,v)$. After giving $\frac{P}{T}$, he states,

“The specific heat $c_v$ of a real gas also depends upon the virial coefficients. It is found by statistical mechanical calculations that $c_v = …$".

So, he’s referring to results of statistical mechanics rather than implying that $c_v$ can be derived from the $\frac{P}{T}$ equation using just thermodynamics. In the 2nd edition, he is not as explicit in indicating how the results are obtained.

In comparing the second edition with the first edition and also with another book on statistical mechanics, I believe there are some sign errors in the 2nd edition for the virial expansions of $f, c_v$ and $u$. These are equations 13.28, 13.29, and 13.30. In each of these equations, the sign in front of $RT$ is wrong.

As usual, it looks like I should have referred to the first edition! Thanks for the tip.

In this case (all equations referred to are second edition), it appears that if one takes (13.29) as given, then one can argue that
$$du = Tds - Pdv = c_vdT + T\frac{P}{T}dv -Pdv = c_vdT \implies u = \int c_v dT.$$
I'm not sure I see how integrating (13.29) term-wise gets me to (13.30) though. Would you be able to see this? Is the integral at constant $v$?

 

[Chestermiller]

The virial expansion $\frac{P}{T} = \frac{R}{v}\left[1+\frac{B(T)}{v} + \frac{C(T)}{v^2} + ...\right]$ is one equation of state of the gas. This equation of state alone is not sufficient information to derive the virial expansions for the other quantities $f$, $c_v$, and $u$ that Callen gives in the second edition of the text. See section 3-3 in the 2nd edition for a general discussion of the information content of equations of state.

As a simple illustration, for the classical ideal gas, we have the equation of state $\frac {P}{T} = \frac {R}{v}$. This expression does not depend on the molecular structure (monatomic, diatomic, etc.) But the expression for the molar energy $u(T)$ does depend on the molecular structure. So, it is clear that you can't derive $u(T)$ from the $\frac{P}{T}$ equation alone.

In the first edition, Callen gives the virial expansions only for $\frac{P}{T}$ and $c_v(T,v)$. After giving $\frac{P}{T}$, he states,

“The specific heat $c_v$ of a real gas also depends upon the virial coefficients. It is found by statistical mechanical calculations that $c_v = …$".

So, he’s referring to results of statistical mechanics rather than implying that $c_v$ can be derived from the $\frac{P}{T}$ equation using just thermodynamics. In the 2nd edition, he is not as explicit in indicating how the results are obtained.

It is possible to determine Cv at arbitrary specific volume using classical thermodynamics by just taking the partial derivative of U with respect to T at constant v from his final equation for U(T,v) derived from the viral expansion.

 

[TSny]

It is possible to determine Cv at arbitrary specific volume using classical thermodynamics by just taking the partial derivative of U with respect to T at constant v from his final equation for U(T,v) derived from the viral expansion.

I don't see how you would derive $u(T,v)$ from the virial expansion for $\frac{P}{T}$. But, yes, once you have $u(T,v)$, it is easy to obtain $c_v(T,v)$ by differentiation.

 

[Chestermiller]

$$\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]=-RT^2\left[\frac{1}{v^2}\frac{dB}{dT}+\frac{1}{v^3}\frac{dC}{dT}+...\right]$$

 

[TSny]

I'm not sure I see how integrating (13.29) term-wise gets me to (13.30) though. Would you be able to see this? Is the integral at constant $v$?

It helps to first verify $$T\frac{d^2}{dT^2}\left(BT\right) = \frac{d}{dT}\left(T^2\frac{dB}{dT}\right)$$ Similarly for the other coefficients $C$, $D$, etc.

 

[TSny]

OK. How about this: $$U=U_{IG}(T)+\int_v^{\infty}\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$where V is a dummy variable integration for v.

I don't see how the above follows from

$$dU=C_vdT+\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$

 

[EE18]

It helps to first verify $$T\frac{d^2}{dT^2}\left(BT\right) = \frac{d}{dT}\left(T^2\frac{dB}{dT}\right)$$ Similarly for the other coefficients $C$, $D$, etc.

Just doing this out for psterity.

We have
$$T\frac{d^2}{dT^2}\left(BT\right) = T\frac{d}{dT}\left(B'T + B\right) = T^2B'' + 2TB' = \frac{d}{dT}\left(T^2\frac{dB}{dT}\right).$$

In so doing, I suppose I've verified that $u$ as written represents an antiderivative of $c_v$ (with respect to #T#). I think that's good enough. Thank you!

 

[Chestermiller]

I don't see how the above follows from

There should be a minus sign before the 2nd term. My mistake.

 

[TSny]

There should be a minus sign before the 2nd term. My mistake.

That's OK. I didn't notice the sign error.

But, I don't see how $U_{IG}(T)$ arises in the expression for $U$ in post #6.

 

[Chestermiller]

That's OK. I didn't notice the sign error.

But, I don't see how $U_{IG}(T)$ arises in the expression for $U$ in post #6.

Are you familiar with Hess’ law?

 

[Chestermiller]

That's OK. I didn't notice the sign error.

But, I don't see how $U_{IG}(T)$ arises in the expression for $U$ in post #6.

If U=U(T,v) represents the internal energy at temperature T and molar volume v, then, in the limit of ideal gas behavior $$U_{IG}(T)=U(T,v\rightarrow \infty)$$and $$C_v^{IG}(T)=\frac{dU_{IG}}{dT}$$Applying Hess' Law, we have $$U(T,v)=U_{IG}(T_{Ref})+\Delta U_1+\Delta U_2$$where $$\Delta U_1=U_{IG}(T)-U_{IG}(T_{Ref})=\int_{T_{Ref}}^T{C_v(T')dT'}$$and $$\Delta U_2=U(T,v)-U(T,v\rightarrow \infty)=U(T,v)-U_{IG}(T)$$$$=\int_v^{\infty}{\left[P-T\frac{\partial P}{\partial T}\right]_VdV}$$where V is a dummy variable of integration for v.

 

[TSny]

If U=U(T,v) represents the internal energy at temperature T and molar volume v, then, in the limit of ideal gas behavior $$U_{IG}(T)=U(T,v\rightarrow \infty)$$and $$C_v^{IG}(T)=\frac{dU_{IG}}{dT}$$Applying Hess' Law, we have $$U(T,v)=U_{IG}(T_{Ref})+\Delta U_1+\Delta U_2$$where $$\Delta U_1=U_{IG}(T)-U_{IG}(T_{Ref})=\int_{T_{Ref}}^T{C_v(T')dT'}$$and $$\Delta U_2=U(T,v)-U(T,v\rightarrow \infty)=U(T,v)-U_{IG}(T)$$$$=\int_v^{\infty}{\left[P-T\frac{\partial P}{\partial T}\right]_VdV}$$where V is a dummy variable of integration for v.

Thanks, Chestermiller. I follow and agree with all of this. It's very nice!

I believe you have shown that given only $P(T,v)$ for any real gas, you can obtain $U(T,v)$ for the gas. The argument does invoke the assumption that the behavior of a real gas must approximate an ideal gas for $v \rightarrow \infty$. Although this assumption cannot be justified by purely macroscopic thermodynamic reasoning, it is a natural assumption based on the microscopic description of a gas.

I find it very interesting that this assumption appears to be all you need in order to essentially derive all of the equations of state of a real gas from just the one equation of state for $P(T, v)$. Of course, I could be overlooking something.

 

[Chestermiller]

Thanks, Chestermiller. I follow and agree with all of this. It's very nice!

I believe you have shown that given only $P(T,v)$ for any real gas, you can obtain $U(T,v)$ for the gas. The argument does invoke the assumption that the behavior of a real gas must approximate an ideal gas for $v \rightarrow \infty$. Although this assumption cannot be justified by purely macroscopic thermodynamic reasoning, it is a natural assumption based on the microscopic description of a gas.

I find it very interesting that this assumption appears to be all you need in order to essentially derive all of the equations of state of a real gas from just the one equation of state for $P(T, v)$. Of course, I could be overlooking something.

You also need the heat capacity in the ideal gas limit.

We engineers always regard an ideal gas as the limiting behavior of a real gas at low pressure.

 

[TSny]

You also need the heat capacity in the ideal gas limit.

Yes, good point.
$C_{v_{IG}}$ is certainly not obtainable from $P(T,v)$ for the real gas. So, to get $U(T,v)$ from $P(T,v)$, you need to know $C_{v_{IG}}(T)$. And $C_{v_{IG}}$ depends on the molecular structure: monatomic, diatomic, etc.

 

[EE18]

You also need the heat capacity in the ideal gas limit.

We engineers always regard an ideal gas as the limiting behavior of a real gas at low pressure.

This was fascinating. Thank you to you and @TSny for this very interesting addendum!