pressure-volume-temperature

[beginner49]

HI all

I am glad to join your community.

Just thinking on this and I do not encounter a solution:

We have gas (1 litre of air, for example) in a cylinder at 1 atm pressure and 20ºC.

Then we compress this gas using a piston very quickly (there is no time enough to dissipate any heat out) down to 0.2 litre.

I assume that now we get a pressurized gas at 5 atm, and this gas temperature has also increased a lot.

My question is:

Is this new pressure higher than those 5 atm due to the fact that the higher temperature is a gas the larger volume it occupies?

Considering air as an ideal gas and there has not been any heat dissipated, what would be this pressure and temperature after compressing it down to 1/5 of its initial volumen?

Thanks.

[Redbelly98]

Welcome to PF :smile:


My question is:

Is this new pressure higher than those 5 atm due to the fact that the higher temperature is a gas the larger volume it occupies?
Pressure is higher than 5 atm due to the temperature being higher. The temperature is higher because compressing the gas quickly adds energy to it.
Considering air as an ideal gas and there has not been any heat dissipated, what would be this pressure and temperature after compressing it down to 1/5 of its initial volumen?

Thanks.
This can be worked out. Instead of PV=constant, as you have for a constant-temperature process, we have
PV γ = constant
with γ=7/5 for air. That will give you the final pressure, and then you can get the temperature from the ideal gas law knowing P and V.

[beginner49]

Clear enough, Redbelly98

Thank you very much

[jtbell]

This is called an adiabatic process, by the way:

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/adiab.html