# Air machine and compressed air

• Karol

## Homework Statement

Air at pressure 20[atm] enters an air machine and leaves at 1[atm]. what must be the temperature of the air at the entrance so that when it leaves it is 00C.

PV=nRT

## The Attempt at a Solution

The air expands and only the mass per second is unchanged. i take 1[mol] at the exit:
$$PV=nRT\rightarrow 1[atm]\cdot V_2=1\cdot 0.08208\cdot 273\rightarrow V_2=22.4[liter]$$
$$P_1V_1^\gamma=P_2V_2^\gamma \rightarrow 20\cdot V_1^{1.4}=22.4^{1.4}\rightarrow V_1=2.63[liter]$$
$$P_1V_1=nRT_1\rightarrow 20\cdot 2.63=1\cdot 0.08208\cdot T_1\rightarrow T_1=642^0K=369^0C$$
Is it correct?

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an air machine
Sound like it might be doing work? Is there any mention of a heat source? Could this be an expansion process?

Sound like it might be doing work? Is there any mention of a heat source? Could this be an expansion process?
The answer looks right to me. During an Adiabatic expansion (q = 0) a gas will cool down. Since Delta E is negative (Delta E = Cv Delta T), w <0 -- work is done by the gas.

If you have ever deflated a bicycle tire quickly before, you may have noticed this. The gas coming out of the valve stem is cold, and will actually cool the valve stem.

Thanks

If you have ever deflated a bicycle tire quickly before, you may have noticed this. The gas coming out of the valve stem is cold, and will actually cool the valve stem.
Is it because the expansion is against pressure of 1[atm]? since in adiabatic free expansion to vacuum there should be no change in temperature, right?

Why can you just simply assume that you are working with 1 mol of air? This doesn't seem quite right to me...

Edit: Well, I completed it without assuming 1 mol, and had the same answer as you did, so apparently you can. Is this since the only important thing here are ratios and not actual values?

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Is it because the expansion is against pressure of 1[atm]? since in adiabatic free expansion to vacuum there should be no change in temperature, right?

Yes.

For ideal gases, expansion into a vacuum (the familiar two bulb 'experiment', one with gas, the other with vacuum, separated by a stopcock) the change in temperature is zero when the gas expands into the empty container. For real gases, there can be a change in temperature (due to non-zero intermolecular interactions).

In some physical chemistry experiments, researchers use a supersonic expansion of a gas from high pressure to essentially vacuum through a small orifice to create samples with very low temperatures. This is more complicated, but you can read about it here: http://www.chem.utah.edu/_documents/faculty/morse/84.pdf

Why can you just simply assume that you are working with 1 mol of air? This doesn't seem quite right to me...
You can assume 1 mol, because the heat capacity is proportional to the mass. Increasing the amount of mass in the sample increases the heat capacity by the same proportion. The final temperature will be the same, regardless of how much material you deal with.