- #1
arpon
- 235
- 16
Homework Statement
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.
Homework Equations
$$\begin{align}c_v&=\left(\frac{dU}{dT}\right)_v\end{align}$$
$$\begin{align}TdS &=c_v dT+ T \left(\frac{\partial P}{\partial T}\right)_v dv \end{align}$$
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?