[refid]

I saw this compression tank for air toold, i notice the pipe leading from the pump to the storage tank gets hot when the pump runns filling the tanks... wondering if the PV/nR = T equation fit since the V is fix at 8 gallons and as pressure increases the temperature shoudl
increase

[lemonoman]

It's complicated, because if you're 'filling the tanks' then you're increasing n (number of moles...amount of substance) as well.  Generally though, an increase in P like you talk about will lead to a slight increase in T because of work done, etc.

Maybe it's getting 'hot' for another reason...

[Borek]

Generally though, an increase in P like you talk about will lead to a slight increase in T because of work done, etc.

Slight... why slight? Turbo compressor in cars compresses air just about twice - which leads to doubling the air temperature - but in Kelvins! So the air leaving compressor has over 200 deg C.

Tanks used for air tolls are rated about 7-15 atm or more - that means much higher compression level.

[mdlhvn]

I saw this compression tank for air toold, i notice the pipe leading from the pump to the storage tank gets hot when the pump runns filling the tanks... wondering if the PV/nR = T equation fit since the V is fix at 8 gallons and as pressure increases the temperature shoudl
increase

With this phenomenon, i think that we are discussing in the wrong way. In the pile if you fix the volume 8 gallons and know how many moles of air in the pile, then the temperature increase make the pressure of air on pile wall increase. The temperature increases because of the heat transfer from the pump, and small amount from the exchange of mechanic energy generated by the collision of gas with the pile wall, as your hand become hotter if you rub your hands together

[Borek]

Can't say I understand exactly what you mean, but whatever I understand is wrong. It is simple pV=nRT at work. You have fixed amount of gas (n moles) and you compress it - you inflate volume and increase the pressure. For the pV=nRT to hold gas temperature must go up. Gas is not heated by the heat from the pump (or rather amount of heat transferred this way is neglectable). Gas is heated because it is adiabatically (or close enough to adiabatically) compressed. Period.

[lemonoman]

Generally though, an increase in P like you talk about will lead to a slight increase in T because of work done, etc.

Slight... why slight? Turbo compressor in cars compresses air just about twice - which leads to doubling the air temperature - but in Kelvins! So the air leaving compressor has over 200 deg C.

I was under the impression that we're pressurizing this system not by adiabatic compression, but by adding more molecules (increasing n).

[mdlhvn]

Can't say I understand exactly what you mean, but whatever I understand is wrong. It is simple pV=nRT at work. You have fixed amount of gas (n moles) and you compress it - you inflate volume and increase the pressure. For the pV=nRT to hold gas temperature must go up. Gas is not heated by the heat from the pump (or rather amount of heat transferred this way is neglectable). Gas is heated because it is adiabatically (or close enough to adiabatically) compressed. Period.

What you said is definitely right!

However, In the question of refid, the pile can't be inflated, are you sure that you have fixed amount of gas in pile, the gas is constantly mobile. And the temperature here is the temperature of pile (as reflid said), not of the gas. 

[Donaldson Tan]

The temperature of air stream increases as it undergoes compression in the air 'pump'. The ideal gas equation does not directly describe the rise in temperature.

However, the new temperature of the compressed air can be given as:
P_initial^1-kT_initial^k  = P_final^1-kT_final^k
where k is the heat capacity ratio (Cp/Cv), which is approximately 1.4 for air.

The above equation assumes the air compressor is adiabatic.

The equation PV = nRT whereby V refers to the volume of the air tank is not applicable because this would mean that T refers to the temperature of the air tank and not the temperature of the air stream.

[Borek]

I was under the impression that we're pressurizing this system not by adiabatic compression, but by adding more molecules (increasing n).

Yep. And the compression is not perfectly adiabatic. Still, amount of heat exchanged with surroundings is (in the timescale of compression - it takes minutes at most) relatively small - thus adiabatic compression serves as a good first approximation.

Same answer works for geodome remark. We start with n moles of gas from the atmosphere, we compress it to known volume of the tank. Starting point is some large volume of gas, ending point is the same gas compressed to known volume of tank. Adiabatic gives upper limit to the final temperature.

[Borek]

are you sure that you have fixed amount of gas in pile

You have fixed amount of gas. Period. At first this gas is outside of tank (do you mean tank by the pile BTW?), than it gets compressed into tank. BUt you have the same amount of teh gase before and after.

And the temperature here is the temperature of pile (as reflid said), not of the gas.

How do you imagine tank having different temperature from the stored gas?

[mdlhvn]

After investigating more clearly the state equations of ideal gas, I think that:

1. We cant use the equation PV=nRT for the phenomenon that refid described (and also Turbo air compressor in cars as Borek said). All the state equations are applied for the reversible and adiabatic process. These process are definitely irreversible processes. So the temperature increase will be explained by other reason.

It's complicated, because if you're 'filling the tanks' then you're increasing n (number of moles...amount of substance) as well.  Generally though, an increase in P like you talk about will lead to a slight increase in T because of work done, etc.

Maybe it's getting 'hot' for another reason...

Maybe it is the most appropriate answer.

2. If you aim to carry out one reversible process like that, that is pumping air to the air tank with the infinitesimal small amount of gas, then the temperature must be constant.

[Borek]

All the state equations are applied for the reversible and adiabatic process. These process are definitely irreversible processes. So the temperature increase will be explained by other reason.

That's true, but you are missing fineprint and you apply 'process is reversible, pV=nRT doesn't hold' blindly.

If you compress the air adiabatically it will get hot. On the second end of theoretical possibilities if you compress air slow enough that all heat will be exchanged with the surroundings, it will not change temperature.

Reality is all things in between. Just because process is not adiabtic DOESN'T mean that air won't get hot. It will get hot when compressed, just a little bit less. pV=nRT tells you upper limit - and it tells you that in this case air gets hot just because it was compressed.

pV=nRT for irreversible process doesn't allow for exact calculation of the air temperature change, but it still allows you to tell that it will never get hotter then upper limit and that if air is compressed 8 times (from 1 atm to 8 atm, which is pressure commonly used in tanks) it may end up hot as hell.