When you compress air it heats, and when you decompress it it cools, but how much?(adsbygoogle = window.adsbygoogle || []).push({});

Working from http://en.wikipedia.org/wiki/Gay-Lussac's_law I find the equation

pressure/temperature=constant

so I derived for example

if the air is 10°C (283.16°K) and the pressure is 14PSI so...

14/283.16 = 0.04944

increase the pressure by 90 PSI

104/2103.5 = 0.04944

maintian pressure, but decrease temperature to almost ambient (16.84°C)

104/290 = 0.3586

decrease pressure to ambient

14/39.04 = 0.3586

i.e. I compress a tank of air to 90psi above ambient air pressure of 14psi at 10 degrees celcius, I let it cool to almost ambient (just for easier math) and when I shoot the air out it decompresses and cools to 39 Kelvin. It doesn't sound right.

Then I realised that equation required mass and volume to remain constant.

http://en.wikipedia.org/wiki/Combined_gas_law gives me a more complete equation

(pressure x volume)/temperature = constant

but that would mean when I increassed the pressure by compacting the air I would, for example, decrease the volume by ten times thus increasing the pressure ten times and the temperature would remain constant.

Please, what am I missing???

**Physics Forums - The Fusion of Science and Community**

# Air pressure/temperature relationship

Have something to add?

- Similar discussions for: Air pressure/temperature relationship

Loading...

**Physics Forums - The Fusion of Science and Community**