Why Does Spray Deodorant Feel Cold When Applied?

[Sweeney]

Recently I was wondering why deodorant is cold when its comes out of the spray can. Some explanations say that it is because of a change in pressure. But looking at the ideal gas law T = PV/nR and both n and R are constant. And for every decrease in pressure there should be an increase in volume (Boyle's law:PV=constant). Therefore T should be constant.

The error in this logic could be that this only applies to ideal gases. If so, what property of the deodorant makes it act like this?

 

[Nugatory]

Recently I was wondering why deodorant is cold when its comes out of the spray can. Some explanations say that it is because of a change in pressure. But looking at the ideal gas law T = PV/nR and both n and R are constant. And for every decrease in pressure there should be an increase in volume (Boyle's law:PV=constant). Therefore T should be constant.

The error in this logic could be that this only applies to ideal gases. If so, what property of the deodorant makes it act like this?

It will depend on the specific formulation, but chances are that the stuff is a liquid while it's the in the can (shake a half-empty can - does it slosh?) under pressure, evaporates when it's released. Evaporation absorbs heat from (that is, cools) the surroundings fairly quickly.

And of course the gas laws, whether ideal or not, don't apply to liquids or the liquid->vapor transition.

 

[Dale]

Recently I was wondering why deodorant is cold when its comes out of the spray can. Some explanations say that it is because of a change in pressure. But looking at the ideal gas law T = PV/nR and both n and R are constant. And for every decrease in pressure there should be an increase in volume (Boyle's law:PV=constant). Therefore T should be constant.

The error in this logic could be that this only applies to ideal gases. If so, what property of the deodorant makes it act like this?

Boyle's law doesn't apply here. Boyle's law ASSUMES that T is constant, so it cannot be used to show that T is constant. In fact, in order to keep T constant you generally have to have the gas exchange heat with some reservoir.

The gas goes through the nozzle of the can so fast that it does not have time to exchange heat with anything. So the proper way to think of this is as an adiabatic expansion (http://en.wikipedia.org/wiki/Adiabatic_process). In an adiabatic expansion there is no heat exchange.

Since the volume is increasing and since the pressure is non-zero that means that ΔPV is positive, which in turn means that the gas is doing work on the surroundings. The energy to do that work comes from the thermal energy of the gas, so the thermal energy of the gas decreases and therefore its temperature decreases.

 

[fluidistic]

Since the volume is increasing and since the pressure is non-zero that means that ΔPV is positive, which in turn means that the gas is doing work on the surroundings. The energy to do that work comes from the thermal energy of the gas, so the thermal energy of the gas decreases and therefore its temperature decreases.

I'm curious if I'm getting the following right:
The gas escapes by the nozzle due to a difference of pressure (P being higher inside the bottle than outside). When the gas expands into the air, it is not a free expansion (because in a free expansion there's no air and the expansion costs no work and hence the temperature of the gas remains the same). The work that the gas does on the surroundings is basically done by pushing against the air molecules. The energy of the gas lost by doing so translates as a decrease in temperature. Is that correct?
So it's exactly the same as the case if I'm blowing air from my mouth?

 

[webboffin]

Perhaps the same way a refrigerant works in a fridge

 

[Dale]

When the gas expands into the air, it is not a free expansion (because in a free expansion there's no air and the expansion costs no work and hence the temperature of the gas remains the same).

That is a good question. I don't know about expansion into a vacuum.

 

[256bits]

I would go with the evaporation ( Nugatory ) from the skin of the carrier fluid.

 

[Chestermiller]

I would go with the evaporation ( Nugatory ) from the skin of the carrier fluid.

Ditto. Back in the day, they used to use CFC's as the propellant fluid, but now-a-days, I don't know. During the time that the CFC/0zone issue was prominent in the media, there was a book out called the Spray Can Wars. People who used spray can deodorants were considered environmentally irresponsible.

Chet

 

[rcgldr]

One aspect not mentioned yet is that at the initial moment of spray, the temperature of the can has not yet decreased, and that the sprayed output isn't cooler until the temperature within the can has decreased due to expansion of gas and/or conversion of liquid to gas over a period of time.

The other part of this is that the spray itself could be getting cooled due to evaporation and/or reduction in pressure of the tiny droplets of the spray as it travels out the nozzle and through the air.

 

[fluidistic]

According to wikipedia:

Adiabatic expansion causes the gas (with a low boiling temperature) to rapidly cool on exit from the aerosol applier. According to controlled laboratory experiments, the gas from a typical deodorant spray can drop up to sixty degrees Celsius.[2]

Here's a small study of coldburns: http://pediatrics.aappublications.org/content/126/3/e716.full.pdf+html, in which one can read

According to Amonton’s law,the pressure (P) of a gas is proportional to the absolute temperature (T) for a fixed quantity of gas in a fixed volume:
P1/T1=P2/T2. Therefore, a decline in the pressure of a gas (eg, by spraying the content of an aerosol can) results in a decline of temperature.
Cooling effects may also arise from the low boiling points of propellants (eg, 42.2°C for propane, 11.7°C for isobutane, and 0.6°C for butane).

 

[SteamKing]

Stop worrying yourself to death. Switch to a roll-on deodorant.

 

[cragar]

http://en.wikipedia.org/wiki/Joule–Thomson_effect
Throttling process