Temperature Change in Partitioned Tank with Ideal Gas Upon Removal of Partition

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When a partitioned tank containing an ideal gas has its partition removed, the temperature change is zero due to the process of free expansion, where no work is done and no heat is exchanged with the environment. The internal energy remains constant, leading to no change in temperature since it is solely dependent on internal energy. In contrast, when an aerosol can is discharged, the gas must work against atmospheric pressure, resulting in a drop in temperature as it loses kinetic energy. This temperature decrease occurs even in a vacuum because the energy for the expelled gas comes from the remaining gas, leading to a loss of heat energy. The phenomenon is more pronounced with liquids due to the significant heat absorption during vaporization.
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Hi all,

Sorry if this is posted in the wrong place, I'm new here.
I have a thermodynamics question. Why is the temperature change 0 in the case of a partitioned tank containing an ideal gas when the partition is removed? This seems to run counter to the idea of an aerosol spray can getting cold when it is discharged.

Thank you
 
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74baja said:
Hi all,

Sorry if this is posted in the wrong place, I'm new here.
I have a thermodynamics question. Why is the temperature change 0 in the case of a partitioned tank containing an ideal gas when the partition is removed? This seems to run counter to the idea of an aerosol spray can getting cold when it is discharged.

Thank you

Welcome to PF, 74baja! :smile:

It is called "free expansion" (look it up).

The expanding gas does not do any work, since there is no counter pressure.
Since there is also no exchange of heat with the environment, the change in internal energy is zero (dU=dQ+dW=0).
This is conservation of energy.

Since the internal energy of an ideal gas depends only on T (U=nCvT), T must have remained constant.
 
I see, thank you. So, the change in temperature when a can is discharged into the atmosphere is due to the opposing pressure of the atmosphere, and the work that the can gas must do to overcome it?

Thanks
 
Yes.
The expanding gas from the can has to push the opposing air away.
As a consequence the molecules lose kinetic energy, which means that the temperature drops.
 
74baja said:
I see, thank you. So, the change in temperature when a can is discharged into the atmosphere is due to the opposing pressure of the atmosphere, and the work that the can gas must do to overcome it?

Thanks

No. Even if you sprayed into a vacuum, you would feel a colder can.

By conservation of energy the kinetic energy acquired by the ejected propellant molecules has to come from somewhere, and that somewhere is the loss of heat energy (3kT/2 per molecule) of the remaining pressurized propellant.

If the propellant is a liquid, the effect is much more pronounced since then the latent heat of vaporization is the major absorber of environmental heat. A can of freon (like you can't get any more) is a good example. So is butane lighter fluid.
 
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