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Adiabatic Expansion of A Gas

6 Answers
If a gas expands adiabatically, how do you calculate the temperature of the gas? The combined gas law states that P1V1/T1 = P2V2/T2 giving T2 = P2V2T1/P1V1 If a gas is under an initial pressure of 1 bar for instance and then expands to atmospheric pressure without the transfer of heat so that Boyle's law doesn't apply, what is the new temperature T2? In the equation above, P2 = (1/2)P1 since the pressure P1 reduces from 2 bar absolute to atmospheric pressure P2 which is 1 bar absolute. However there are two unknowns in the equation, V2 and T2, so how can T2 be calculated?
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For an ideal gas the expression PV^gamma = constant is well known. Gamma is the ratio of specific heats Cp/Cv, which are the specific heats at constant pressure and constant volume, respectively. This ratio can be shown to be equal to (f+2)/f where f is the number of degrees of freedom of the molecules of the gas. For a mono-atomic gas f=3, and gamma = 1.66, for example. The original expression can be rewritten in terms of T and P using the ideal gas equation resulting in

T^gammaP^(1-gamma) = constant.

If you halve the pressure, for gamma = 1.66, the temperature will drop by a factor of 0.76. If the gas started at room temperature (300 K), the final temperature would be 227.7 K.

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I tried to do a quick experiment with my air compressor by blowing up a plastic bag quickly to explore this.

The tank was at a pressure of 8 bar gage = 9 bar absolute, the temperature was 10 deg C or 283 K.
For air, gamma is 1.4

This should result in the temperature dropping to 150K as the air expanded to atmospheric pressure. I used a digital thermometer with a probe on the end of the cable to check the temperature. I only noticed a drop of a few degrees. So presumably the air was able to heat up through the bag and the time constant of the probe was too large and the air had warmed before the thermometer was able to respond.

I'll try this again using some expanded polystyrene to act as an insulator.

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The pressure of the gas reduces to atmospheric pressure as it expands. So yes, the volume increases. However there is no heat transfer through the walls of the vessel. So you could think of a vertical cylinder and piston arrangement where initially, the piston is weighted from above. The gas is compressed and heat energy flows out into the surroundings. The compression is isothermal so the process follows Boyle's law. Now the weight is removed allowing the gas to expand and the piston to move upwards. The pressure within the cylinder eventually reaches atmospheric. This expansion is adiabatic with no heat transfer. So there is a drop in temperature, just like when air is released from an air compressor tank or an aerosol is sprayed.

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This is not clearly worded. Adiabatic implies no heat transfer, which seems to include your situation, but you say it is non-adiabatic. For an adiabatic process (PV)^gamma will be constant and gamma can be calculated for different molecules. It is 1.66 for single atom molecules (ideal gas). So please clarify the question.

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Now you are suggesting that the volume remains constant since the gas stays in the same vessel as the pressure is reduced. Is this accomplished by reducing the amount of gas then? Or does it somehow expand into a larger vessel perhaps also doing work and the pressure goes down at the same time?

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Ok, I should have said adiabatic or non-isothermal. So the gas is thermodynamically isolated with no transfer from its containment vessel or surroundings.

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