# What happens to the temperature of a gas when it is compressed?

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Richard Feynman described it nicely. Put your finger over the hole at the end of a bike pump and push the handle in. This forces a piston (the sliding disc inside that traps the air and squashes it) to compress the air. Air particles are constantly moving and when they bounce off a stationary wall they have the same speed as before. If the wall is moving towards the particles (like when the piston is pushed in) they bounce off with a slightly greater speed than they had before. If you make particles move faster you have increased their temperature.

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According to the ideal gas law the pressure P, volume V, amount of substance n and temperature of a gas are related as P*V = n*R*T, where R is a universal constant.

As a gas is compressed, work is being done on the gas. The result of the work is first a decrease in its volume with the temperature not being affected. Once the gas has reached a volume where it cannot be compressed further into a smaller volume, we have the parameters V, n and R in the equation taking on constant values. This reduces the equation to P = k*T. Now temperature is directly proportional to the pressure.

As the gas is compressed, the work put into the system goes to increasing the temperature of the gas.

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To solve we must examine the ideal gas law, PV=NRT where P is pressure, V is volume, N is the number of moles in the gas, R is your boltzman constant, and T is your temperature.

N and R are automatically assumed constant. We are going to fix volume as a constant. Then all that is left to change is pressure and temperatue. Due to the constants, we can ignore V, N, and R hence P is in an equal relationship with T.

This yields that as pressure increases, so does temperature for an ideal gas.