# Adiabatic process of real gas

Tags:
1. Oct 15, 2016

### arpon

1. The problem statement, all variables and given/known data
Show that for a gas obeying the van der Waals equation $\left(P+\frac{a}{v^2}\right)(v-b)=RT$, with $c_v$ a function of $T$ only, an equation for an adiabatic process is $$T(v-b)^{R/c_v}=constant$$

2. Relevant equations
$TdS=c_vdT+T\left(\frac{\partial P}{\partial T}\right)_v dv$

3. The attempt at a solution
For reversible adiabatic process, $dS=0$.
So, from the third $TdS$ equation,
$$c_vdT+T\left(\frac{\partial P}{\partial T}\right)_v dv=0$$
$$c_vdT+T\left(\frac{R}{v-b}\right)dv=0~~~$$ [Using equation of state]
$$c_v\frac{dT}{T}=-\frac{RdV}{v-b}$$
If $c_v$ is a constant, integrating both sides, we have:
$$T(v-b)^{R/c_v}=constant$$
But, in this case, $c_v$ is a function of $T$.
Any help would be appreciated.

2. Oct 15, 2016

### Staff: Mentor

I guess they must be assuming that, over the range of temperatures for the process, the heat capacity is nearly constant. We certainly often do this for an ideal gas, and, since the heat capacity for a van der waals gas is a function only of temperature, it must be identical to the ideal gas heat capacity.

3. Oct 15, 2016

### arpon

Could you please explain this part? The heat capacity for ideal gas is a constant while for van der waals gas, it is a function of temperature. What did you actually mean by they are identical?

4. Oct 15, 2016

### Staff: Mentor

The heat capacity of a real gas, in the limit of very low pressures, is a function of temperature. We engineers take this into account in our definition of an ideal gas (by regarding an ideal gas as having a temperature-dependent heat capacity), but physicists have idealized it further (by regarding an ideal gas as having a constant heat capacity). Now, if the heat capacity of a van der Waals gas is a function of temperature only (and not volume and pressure), it must be the same temperature-dependent function as we engineers refer to for an ideal gas.