thermodynamics - Behavior of gasses, ideal and otherwise

I'm trying to help a child research a science project on refrigeration. Refreshing my incredibly rusty thermodynamics skills....

The ideal gas law: $PV=nRT$. Let's take air at STP:

$P = 101\,kPa$ $V = 1\,L = 0.001\,m^3$ $R = 8.3\,J/mol\cdot K$ $T = 298\,K$

$n=PV/RT = (101000) (.001) / (8.3 \cdot 298) = 0.04\, moles$?

If we compress the air to ~7atm adiabatically $P_2 = 7P$

I would think the volume goes to $\frac{1}{7V}$

$V_2 = \frac{V}{7}$

then I would expect the gas to be hotter. But I'm obviously confused because with that pressure and volume, the temperature is obviously the same. I'm assuming that I'm wrong about what the volume would be for an ideal gas if I compress to $7\,atm?$

$T_2 = \frac{P_2V_2}{nR} = ???$

The specific heat of air: $c_p = 1.006\,kJ/kg\cdot K$

Of course air is not quite ideal. I would also appreciate someone explaining what the non-ideal behavior is due to. Is it related to the mixing? Some kind of chemical interaction between the different components?

What I want to know is how to calculate the temperature of a gas given an initial temperature, pressure, heat capacity and final pressure, Adiabatically.

Answer

You've mixed isothermal (no change in temperature) and adiabatic (no heat exchange).

Either your compression is isothermal, then $P\cdot V$ is constant so indeed $V_2=V/7$.

Or your compression is adiabatic, then $PV^\gamma=cst$ where $\gamma=7/5$ for an ideal, diatomic gas. So $V_2=V_1 \left( \frac{P_1}{P_2} \right)^{1/\gamma}$ from where you can get the temperature if you want.