Is it possible to lower the temperature of a gas by pressurizing it to an extreme degree?

Question

If we look at temperature as a measure of the kinetic energy present in moving gas molecules, could we lower a gas's temperature by making it impossible for the molecules to move by pressurizing it to an extreme degree? Or are lower temperatures necessary for higher pressures to be reached?

Answer 1

No. You increase the temperature of the gas as you pressurise. This assumes no heat exchange with the surroundings (adiabatic compression). The gas will become harder and harder to compress and may liquify or even solidify - you simply can't squeeze the energy out of it unless you let heat flow from the fluid into the surroundings.

Addendum

you can understand this more by looking at some reversible compression processes for an ideal gas

• The mean kinetic energy of gas molecules $\langle E_K\rangle$ is directly proportional to its temperature $T$: $\boxed{\langle E_K\rangle =\frac{3}{2}k_BT}$
• The macroscopic properties of ideal gases are given by the relationship $\boxed{\frac{pV}{T}=\mathbb{constant}}$
• It's also helpful to consider conservation of energy; the gain in intrinsic energy $U$ is equal to the net heat supplied $Q$ plus the net work input to the gas $W$: $\boxed{U=Q+W}$

Isobaric compression $U=Q+W$

We could compress a gas by cooling it at a constant pressure. This will reduce the temperature and thus the kinetic energy of the molecules. Here the pressure $p$ is constant therefore the change in temperature is given by: $$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{-1}$$ If we compress the volume by a half we must cool it to half the temperature.

Isothermal compression $0=Q+W$ since $U=0$

We could compress a gas while keeping it at a constant temperature. From the ideal gas law the initial and final states will follow the relationship $p_1V_1=p_2V_2$ and $T_1=T_2$. Here the pressure of the gas will rise in accordance to ideal gas law: $$\frac{p_2}{p_1}=\frac{V_1}{V_2}$$ If we compress the volume by a half we double the pressure. Here the internal energy of the gas is unchanged because the energy we've put into compressing the gas has been passed out into the surroundings.

Adiabatic compression $U=0+W$ since $Q=0$

So suppose we don't allow any heat exchange with the surroundings. All the work we use to compress the gas will increase the internal energy (temperature) of the gas. It can be shown that here $pV^\gamma = \mathbb{constant}$ where $\gamma$ is the adiabatic ratio. Now we have $$\frac{p_2}{p_1}=\left(\frac{V_1}{V_2}\right)^\gamma$$ and $$\frac{T_2}{T_1}=\left( \frac{V_1}{V_2}\right)^{\gamma-1}$$

If we compress the volume of air ($\gamma \approx 1.4$) then we increase both the pressure and the temperature. All the work we put in increases the molecular energy of the gas.