# Polytropic Process for an Ideal Gas

1. Oct 14, 2013

### LunaFly

Hi,

I have a general question about polytropic processes when working with an ideal gas.

A polytropic process is one for which PV$\gamma$ = constant.

What I am wondering is if we assume that n is constant, does that mean that the temperature does not change for the process?

I see that PV$\gamma$ = nrT = constant, but for some reason it doesn't seem right that a polytropic process would be equivalent to an isothermal process if the number of moles of the system doesn't change.

A little light on the subject would be great.

Thanks,

-Luna

2. Oct 15, 2013

### Staff: Mentor

No. A good example of such a process is adiabatic expansion/compression, for which $T$ changes.

That equation is only valid if $\gamma = 1$. Otherwise, there is no simple relationship between $PV^\gamma$ and $T$.

3. Oct 18, 2013

### LunaFly

Ok this concept finally sunk in.

PV$\gamma$ does not equal nRT unless $\gamma$=1. So a polytropic process in no way implies that nRT is a constant (even when n is constant).

DrClaude, thanks for the input!

4. Oct 20, 2013

### Staff: Mentor

Consider a process for an ideal gas where $PV^\gamma = \textrm{const.}$. Using the ideal gas law, $PV=nRT$, we have
\begin{align} P_iV_i^\gamma &= P_fV_f^\gamma \\ \left( \frac{n R T_i}{V_i} \right) V_i^\gamma &= \left( \frac{n R T_f}{V_f} \right) V_f^\gamma \\ T_i V_i^{\gamma - 1} &= T_f V_f^{\gamma - 1} \end{align}
or $TV^{\gamma - 1} = \textrm{const.}$
Obviously, if $V$ changes during the process, then $T$ changes (except if $\gamma = 1$).