# 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.

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.

• For an non-ideal gas, the value of $\gamma$ stated in the previous answer would be other value since air is a mixture of different gasses with different number of atoms. – Ignacio Vergara Kausel Oct 21 '13 at 15:06
• Agreed. Fortunately air is a mixture of $N_2$ and $O_2$ mostly, which are both diatomic gases, so $\gamma$ is very close to $7/5$ (within 0.1% for dry air, see en.wikipedia.org/wiki/Heat_capacity_ratio ) – Nicolas Oct 21 '13 at 15:13
• Sure, but the question also asked on how non idealities would enter into the picture. Checking for variations in $\gamma$ would be a way to see it. – Ignacio Vergara Kausel Oct 21 '13 at 15:16
• Absolutely. But that's an independent question. – Nicolas Oct 21 '13 at 15:18
• Just to confirm, for T1=298, P1 = 102kPa, P2 = 714kPa, V1=1 then: $\gamma = 7/5$ or $\gamma = cp/cv = 1.35$ to get a bit more accurate V2 = .043 T2 = 916 K – Dov Oct 21 '13 at 16:13

Starting from the ideal gas law, you can get the following $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. Given that you increase the pressure as $P_2=7P_1$ you obtain $\frac{V_1}{T_1}=7\frac{V_2}{T_2}$. You would need to know the initial and final temperatures along the initial volume to be able to determine the volume or to know the volumes and one temperature to get the other.

Unless you assume that the process is isothermal (as in the other answer) you cannot advance more.

• You can, if you assume the process is adiabatic. I edited the other answer to clarify the hypothesis – Nicolas Oct 21 '13 at 15:16

The question of the non-idealities of air is a separate one, so I answer separately (as the first answer already has several comments)

I think that in this case the most deviation would be due to the presence of water in the air. Nitrogen and Oxygen get liquid at very low temperatures or high pressures so that gives you an indication that they're not so far from ideal gases. Indeed, for dry air, $\gamma$ is very close to 1.4. However, humidity being typically around 50% of saturation means that the small part of water vapor in air should be very non-ideal. In this case, the non-ideal character of moist air would come from inter-molecular forces that are responsible for condensation.

I couldn't find good quantitative sources with a quick web search though.