Temperature Dependence on Pressure Change in Gases

[MMD]

I have a simple question.
If we increase the pressure of a gas by decreasing its volume without heat exchange, does the temperature always change? Or does it depend how fast we change the pressure?

I tried to find the answer for this question, unfortunately I couldn’t.
Thanks for any help!

 

[KingNothing]

The ideal gas law is [tex]PV = nRT[/tex].

As you can see from it, the pressure will increase with decreasing volume, however the temperature doesn't need to change at all.

 

[MMD]

KingNothing,
The ideal gas law, [tex]PV = nRT[/tex] shows that P, V and T are dependent to each others, but it doesn't say that the temperature doesn’t need to change. For example we know that when we compress the air the temperature increases. How you can explain that?

 

[haiha]

As I remember, the heat gain is because the entropy of the system is reduced. That is the term TdS (delta S). Entropy decreases because the volume is reduced. And TdS does not depend on how fast you compress the gas. When you expand the gas, V increases so S increase then temperature decreases.
When you compress the gas very slowly, you can not see it is getting hotter just because the precess is not adiabatic any more.

 

[MMD]

KingNothing says the temperature doesn’t need to change! hayha says it changes! I am confused. I know that the internal energy of the gas is changed because of the work and I know that if we have the condition/ e.g. pressure, volume and temperature of a gas we must be able to find the final condition of the gas after the process. Is there any equation for such a process? Again the process is adiabatic.
(Let’s think of a gas that we know its volume, pressure and temperature. If just by compressing it to the half of its initial volume, what would be its temperature after the process? It can't be possible that the gas has a choice to have any temperature.)

 

[haiha]

There's nothing to be confused!. Surely in an adiabatic process, the gas temperature increases if the it is compressed and vice versa. I can give you some examples describing the phenomenon. If you ever work with compressed gas cylinder, when you use the gas with high flow rate, the regulator can get very cold, even ice is condensed on it. Or if you pump you motorbike's tyre using a hand-pump, the pump trunk and hose will be very hot after some time.
The thing I am confused here is how to explain this process more scientifically.

 

[cevdet.erturk]

Here is the equation:

"A gas with pressure P, temperature T, and chemical potential:

dU=PdV+TdS+dN;

where dV is the change in volume, dS is the change in entropy, and dN is the change in the number of particles. Note that this assumes that, prior to the changes indicated by the dierentials (i.e., prior to changing the volume, entropy, or number of particles), the gas was in equilibrium."

I hope it will help.

 

[MMD]

Alright, suppose that dN = 0 , so the equation simplifies to:

dU=PdV+TdS;

Now, suppose that the initial pressure is 2 atm, the initial volume is 2 L, the temperature is 300 K, if we by compressing an idea gas, we change the volume of air to 1 L, what would be the final temperature? (we don't know dU or dS)

 

[cevdet.erturk]

Your question is incomplete. Without both dU and dS, nothing can be found!

But, there are some more useful equations here:

http://www.grc.nasa.gov/WWW/K-12/airplane/compexp.html

Check it! It will make it easier for you!

 

[MMD]

I checked the web site. But why Cp - Cv = R?

 

[cevdet.erturk]

Check that link (it has the derivations of known equtions):

http://66.102.9.104/search?q=cache:...ntro.ppt+cp+-+cv+=+R&hl=tr&ct=clnk&cd=5&gl=tr