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If you have a deodorant can, it is obvious that the gas leaving the can has cooled as it expands. However, is there also a change of temperature of the gas that remains within the canister?

My assumption is that it would also cool down, but by a lesser amount. My reasoning being that as the gas is released, there is now less gas remaining in the canister, so the pressure within the canister has reduced and the gas within the canister has expanded. However, as the change in pressure is less than that of the released gas, the change in temperature will also be less.

Is my assumption correct?

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  • $\begingroup$ I highly doubt your deodorant is just compressed air. Also, check out this Related video $\endgroup$ – Aaron Stevens Apr 10 at 22:25
  • $\begingroup$ explainthatstuff.com/aerosolcans.html $\endgroup$ – Gert Apr 10 at 22:50
  • $\begingroup$ Sorry, I didn't mean to say you said it's only air. I just meant you might need to consider other things in your analysis instead of just the gas. $\endgroup$ – Aaron Stevens Apr 11 at 1:18
  • $\begingroup$ @AaronStevens No problem, I understood what you meant straight after I wrote the comment so I deleted it. $\endgroup$ – PhysicsGuy123 Apr 11 at 12:00
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Your assessment of the temperature decrease in the canister is correct. But, it is not "by a lesser amount." It is actually the dominant cooling. The gas flow through the valve takes place at constant enthalpy, and, for an ideal gas, this would involve no temperature change. This is because the expansion cooling of the gas in the valve is offset by viscous heating in the valve. This is called the Joule-Thomson effect.

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  • $\begingroup$ Thanks for your answer. Sorry, but the "dominant cooling" still confuses me a little. If for example you spray a small amount of deodorant out of the can, you notice the spray is cold, but there is generally little or no noticeable change in temperature of the canister. Are you saying that the temperature change of the gas in the canister is greater than that of the gas leaving the canister? $\endgroup$ – PhysicsGuy123 Apr 11 at 0:45
  • $\begingroup$ That's because the deodorant is a liquid which is evaporating...after it comes out. $\endgroup$ – Chet Miller Apr 11 at 2:01
  • $\begingroup$ So if it is just a gas, the gas leaving the canister would not change in temperature, but the gas inside the canister would decrease. Have I got this correct? Is there an equation to calculate the temperature change in the canister, which takes into account the initial pressure, the amount of gas released, the initial temperature etc.? $\endgroup$ – PhysicsGuy123 Apr 11 at 12:00
  • $\begingroup$ The gas within the canister is cooling and is then passing through the valve with no further temperature change. For the gas in the canister, the relationship between the pressure p and the number of moles remaining n is $pn^{-\gamma}=const$ $\endgroup$ – Chet Miller Apr 11 at 12:58
  • $\begingroup$ Thanks for explaining this. So how does $- \gamma$ relate to the temperature of the remaining gas? $\endgroup$ – PhysicsGuy123 Apr 11 at 14:17
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Yes, it will cool, but not by a lesser amount. It's actually a much greater amount. You can test this by holding the can and spraying out a lot of deodorant. It gets colder. (Unless you ignite the spray, but let's not do that. Not again - it's dangerous!)

As pointed out in the cool (geddit) video that Aaron linked to, the propellant is a combination of liquid and gas, and it's the endothermic changing of liquid to gas phase to maintain equilibrium in the can that cools the can. As you will see, the pressure does not decrease until the can is almost empty.

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  • $\begingroup$ Thanks for your answer. Having watched the video, there is a demonstration of a tyre and the temperature at the valve changing due to the expansion of the air. However, there is no demonstration of the temperature of the tyre itself (the canister in this case). Therefore, if the canister wasn't a deodorant can and did just hold a gas rather than a liquid, would the canister still cool down? $\endgroup$ – PhysicsGuy123 Apr 11 at 0:54
  • $\begingroup$ Yes, it would. The gas remaining within the tire at any time has experienced an adiabatic reversible expansion, in doing work to push the gas ahead of it out through the valve (just as if it were pushing against a piston). As with any other adiabatic reversible expansion, the gas within the canister or tire cools. On the other hand, according to the open system version of the first law of thermodynamics, the flow of the gas through a valve (or a porous plug) experiences a decrease in pressure at constant enthalpy. For an ideal gas, there is no change in temperature. $\endgroup$ – Chet Miller Apr 11 at 1:48

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