# Temperature dependence of Cv at very large volume

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1. Oct 15, 2016

### arpon

1. The problem statement, all variables and given/known data
In the case of a gas obeying the equation of state
\begin{align}\frac{Pv}{RT}&=1+\frac{B}{v}\end{align}
where $B$ is a function of $T$ only, show that,
\begin{align}c_v&=-\frac{RT}{v}\frac{d^2}{dT^2} (BT)+\left(c_v\right)_0\end{align}
where $\left(c_v\right)_0$ is the value at large volume.
2. Relevant equations
\begin{align}c_v&=\left(\frac{dU}{dT}\right)_v\end{align}
\begin{align}TdS &=c_v dT+ T \left(\frac{\partial P}{\partial T}\right)_v dv \end{align}

3. The attempt at a solution
\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & =\left(\frac{\partial}{\partial v}\left(\frac{\partial U}{\partial T}\right)_v\right)_T \\ & =\left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial v}\right)_T\right)_v \end{align}
Now,
\begin{align}dU &=TdS-Pdv\\ & = c_v dT+ T \left(\frac{\partial P}{\partial T}\right)_v dv - Pdv \end{align}
So, we have,
\begin{align} \left(\frac{\partial U}{\partial v}\right)_T & = T \left(\frac{\partial P}{\partial T}\right)_v - P \end{align}

Using equation (6) & (9),
\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & = \left(\frac{\partial}{\partial T}\left(T \left(\frac{\partial P}{\partial T}\right)_v - P\right)\right)_v\\ & = T \left(\frac{\partial^2 P}{\partial T^2}\right)_v + \left(\frac{\partial P}{\partial T}\right)_v - \left(\frac{\partial P}{\partial T}\right)_v \\ &= T \left(\frac{\partial^2 P}{\partial T^2}\right)_v \end{align}

Using the equation of state (1), we obtain,
\begin{align} \left(\frac{\partial^2 P}{\partial T^2}\right)_v &= \frac{R}{v^2}\frac{d^2}{dT^2} (BT) \end{align}

So, using (12) and (13), we have,
\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & = \frac{RT}{v^2}\frac{d^2}{dT^2} (BT) \end{align}
Now
\begin{align} dc_v &= \left(\frac{\partial c_v}{\partial v}\right)_T dv + \left(\frac{\partial c_v}{\partial T}\right)_v dT \end{align}

Integrating from $T=T,~V=\infty$ to $T=T,~V=V$,
\begin{align}c_v&=-\frac{RT}{v}\frac{d^2}{dT^2} (BT)+\left(c_v\right)_0\end{align}

Here, $\left(c_v\right)_0$ is the value for $v=\infty$ and $T=T$.
My question is: Is $\left(c_v\right)_0$ independent of $T$, and why?

2. Oct 16, 2016

### Staff: Mentor

At the limit for infinite volume, the B/v term disappears, therefore the gas law gets independent of T (apart from the usual factor of T).

3. Oct 17, 2016

### Staff: Mentor

For a real gas (which this is supposed to represent), in the limit of very large v, $c_v$ is observed to be a function of T.

4. Oct 17, 2016

### Staff: Mentor

Then you need a different equation of state, not the one from post 1.

Actual gases often have additional complications from frozen-in degrees of freedom, dissociation and so on, but those are all not considered here.

5. Oct 17, 2016

### Staff: Mentor

Why not?

6. Oct 17, 2016

### Staff: Mentor

The one from post 1 does not lead to a temperature-dependent c_v in the limit of infinite volumes.

7. Oct 17, 2016

### Staff: Mentor

I don't see anything in the mathematics that precludes Cv from being a function of temperature at large volume.

Last edited: Oct 17, 2016
8. Oct 17, 2016

### Staff: Mentor

Hmm... v->infinity is equivalent to B->0. Setting B=0 leads to c_v = c_v, so we cannot make any statement about c_v just with the given information.

Why exactly would a temperature-dependent c_v be problematic?

9. Oct 17, 2016

### Staff: Mentor

We could say that at large v, c_v is temperature-dependent. That's certainly consistent with the mathematics. We engineers (I'm an engineer), unlike physicists, regard a temperature-dependent heat capacity as a key aspect of the definition of an ideal gas. In all engineering thermodynamics books, the heat capacities of ideal gases are regarded as being temperature dependent. They are the same functions of temperature that the heat capacities of real gases approach in the limit of low pressures and high volumes. So, in this problem, $c_{v0}(T)$ would be the (temperature-dependent) heat capacity of the real gas in the ideal gas region.
In my judgment, it would not be problematic at all. Rather, it would be expected.

10. Oct 17, 2016

### Staff: Mentor

Wait, what? For an ideal monoatomic gas it should be 3/2 R.

11. Oct 17, 2016

### Staff: Mentor

As I said, engineers regard an ideal gas an entity which, in general, features a temperature-dependent viscosity. In the special case of an ideal monoatomic gas, the heat capacity would be virtually constant at 3/2R. For the present problem, however, no information was provided concerning the molecular nature of the gas. Therefore, it is prudent to assume that, at large specific volumes, the heat capacity can be temperature-dependent.