- #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?