Trouble understanding why ideal gas law explains that can of compressed air gets cold when you spray it

Question

I was told that a canister of compressed air gets cold because of the ideal gas law, you are lowering the pressure and since the density of air is the same, the temperature of the can gets cold. I was trying to see this mathematically and I became confused. Can someone please help me with the intuition.

The ideal gas law equation is $PV=nRT$ where R is $8.314 \frac{J}{mol \cdot K}$ if you convert from moles to kg, you get:

$$ R = 8.314 \frac{J}{mol \cdot K} \times \frac{1 mol}{0.02897 kg} = \frac{287 J }{kg \cdot K} $$

So we can change $nR$ where n is number of moles and $R$ is joules per mole-kelvin to $mR$ where m is mass and $R$ is Joules per kg-kelvin because $nR = \text{ joules } = mR$.

This gives $PV=mRT$

Assuming $m$ is the mass of 1 liter of air, and V is 1, then we have $P=\rho RT$, where $\rho$ is density.

I'm told this equation would explain why the can of air gets cold, since the density of air does not change. But this equation was formed under the assumption of a constant mass of air. In the compressed air example, the mass of air in the can is decreasing, and the volume stays the same. So I don't think we can describe the air can with $P = \rho R T$ I think it must be described with $PV = mRT$, and if the mass goes down, the pressure can go down, and the can does not have to get cold.

Am I correct that the $P=\rho RT$ equation does not hold if the mass inside the can is changing?

I saw another stack answer, Why does the gas get cold when I spray it?, but it is insanely complicated and doesn't mention the ideal gas law. I'm mostly trying to figure out why the ideal gas law does or does not answer the question, I can't begin to understand what they are saying in that other answer.

Answer 1

Answer 1: There is some liquid in the can that starts evaporating when the button is pressed.

Answer 2: It's actually such compressed air can where there actually is compressed ideal gas in the can. The gas does work against the atmospheric pressure. The gas in the can helps the gas inside the nozzle to get out by pushing its back. So it loses energy.

Answer 2

First to answer one of your questions

Am I correct that the P=ρRT equation does not hold if the mass inside the can is changing?

No, this equation always holds true provided a small change : $V= \frac{M(gas) \cdot n}{\rho}$ so if you replace $V$ in the standard expression of the ideal gas law you'll get $P=\frac{ρRT}{M(gas)}$

You can't insert $M(gas)$ into $R$ because the value of $M(gas)$ depends on the gas.

Now here is the complete answer to your problem :

Emptying a canister filled with compressed gas is the same as extending a piston filled with air. You can consider that a plate is positioned horizontally separating the air in two : above is the air that will exit the canister and under is the air staying within.

Simple answer :

Since the plate is moving, when a particle hits it, it will bounce back but with a slower speed : the movement of the plate absorbs the impact. Since temperature is strongly linked to the speed of the particle (more accurately to their energy) the movement of the plate causes a drop in temperature.

More complex answer :

Now that the problem is a bit more formalized let's solve it. First, what you need to see is that the ideal gas law alone is not enough to answer the problem. Indeed once you've let some air out the only parameter which value you know is $V$. You have information neither about $P$ nor $T$. So based on that equation only there are infinitely many solutions.

But here are the necessary equations which you may not have seen yet but are not as hard as they seem :

1. $U = C_v \cdot T$ where $U$ is the internal energy of the gas (basically the sum of potential and kinetic energy for all particles) and $C_v$ a constant depending on the gas (usually $3/2$).
2. $\Delta U = W_{received}$ where $W$ is the work received by the gas. This basically says that the difference of energy of the gas between two states is equals to the energy received (here under the form of the work $W$ because no heat transfer)
3. $\delta W_{emitted} = P\cdot S \cdot dx = P\cdot dV$ where $S$ is the surface of the plate. It says that the work created by a pressure is the product of the force ($P\cdot S$) times the displacement ($dx$).

Using equation 2. between $t$ and $t+dt$ we get : $$\frac{dU}{dt}= \frac{dW_{received}}{dt}=-\frac{dW_{emitted}}{dt}$$

Using 3. we have : $$\frac{dU}{dt}= -P\cdot \frac{dV}{dt}$$

Using the law of ideal gas we get : $$\frac{dU}{dt}= -\frac{nRT}{V} \cdot \frac{dV}{dt}$$

Using 1. we get : $$C_v\frac{dT}{dt}= -\frac{nRT}{V} \cdot \frac{dV}{dt}$$

Rearranging the terms with $K = \frac{C_v}{nR} = \text{constant}$ we find : $$K\frac{1}{T}\frac{dT}{dt} = -\frac{1}{V}\frac{dV}{dt}$$

Finally if we integrate between $A$ and $B$ : $$K \ln{\frac{T_B}{T_A}} = - \ln{\frac{V_B}{V_A}}$$

Since $\ln$ is an increasing function, if there is an increase in $V$, then $T$ must decrease.

Answer 3

See https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect#Physical_mechanism

There are two factors that can change the temperature of a fluid during an adiabatic expansion: a change in internal energy or the conversion between potential and kinetic internal energy. Temperature is the measure of thermal kinetic energy (energy associated with molecular motion); so a change in temperature indicates a change in thermal kinetic energy. The internal energy is the sum of thermal kinetic energy and thermal potential energy. Thus, even if the internal energy does not change, the temperature can change due to conversion between kinetic and potential energy; this is what happens in a free expansion and typically produces a decrease in temperature as the fluid expands. If work is done on or by the fluid as it expands, then the total internal energy changes. This is what happens in a Joule–Thomson expansion and can produce larger heating or cooling than observed in a free expansion.

The atoms of an ideal gas are hard bouncy particles with no internal structure. If you allow an ideal gas to expand, the atoms have more space in which to bounce around. But their kinetic energy doesn't change. So the temperature doesn't change.

Molecules have bonds. These can be modeled, at least approximately, as balls on spring. When they bounce around, the springs get compressed. Some of the internal energy is potential energy of the bonds. Depending on the molecule, if you let this gas expand, the potential energy can change.