compressing air

van der Waals gas is a set of more accurate [empirical] equations for representing real gases, than the ideal gas equations.

P1V1^n=P2V2^n=c=constant   where n=Cp/Cv  it should be around n=1.4

V;Volume
P :Pressure

Work=integral (PdV)  where P=c/V^n

W=c* (V2^(-n+1)-V1(-n+1)) / (-n+1)

should be;

W=c* (V2^(-n+1)-V1^(-n+1)) / (-n+1)

I think you are going to have to have a temperature in here somewhere. A kg of
 air at a very low temperature (even at a high pressue) is not going to have much energy to do external work. A kg of air at a very high temperature will have lots of energy.

>>...is not going to have much energy to do external work.

The sign of the work will determine if the work is externally given (-) or the system itself produces the work (+)

Hello grg99

If the pressure is released quickly, so there is no heat exchange between the (compressed) gas reservoir and the environment it's called an adiabatic process, as was mentioned above.

Look at the diagram and equation at the top of this page:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

The work W is given in terms of intial and final Volume, which you can approximate as follows:
(I'll do it in S.I. units, thats easier for me)

ambient air has a pressure of 1 bar.
Say, the compressed gas has  pressure of 100 bar.
Then the amount of compressed gas inside a cylinder is 100 times that of uncompressed gas insinde that same cylinder.
Releasing it completely (p: 100bar -> 1bar) will increase the volume of the gas like
V: V(cylinder) -> 100*V(cylinder)

In general p1 * V1 = p2 * V2
 
One more comment:
If the compressed gas is initially hotter than the environment that makes no diffference as long as this excess heat is not turned into work. It is true, that hot gas has more internal energy than a cool gas, however the temperature is already taken into account when you have your gas describhed in terms of pressure and volume. 1 kg of hot gas will simply have higher pressure than 1 kg of cool gas. It makes no difference what initially caused the pressure to build.

Cheers,
Sebastian

there is actually a form on that page that will calculate the Work :-)

Hi,

W = (p_1 * V_1 - p_2 * v_2) / (k - 1)


Depending on your gas, as Macroland already pointed out this constant is the ratio of the Molar Heat Capacity at constant pressure to the Molar heat capacity at constant volume.

For monatomic ideal gasses that ratio is usually around 1.67
For diatomic ideal gasses that ratio is usually around 1.40
For polyatomic ideal gasses that ratio is around 1.30

As has been pointed out the problem conserns an adiabatic expansion. As was also pointed out the problem CANNOT be solved without a temperature someplace. All of the equations presented for work so far have involved a volume, which was not given in the problem.
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By the way
"In general p1 * V1 = p2 * V2"
only for isothermal changes

pV = const .. for T=const that is true.I figured we can neglect the cooling of the expanding/working gas (Thomson effect I belive).
Also, I think, the volume is a required quantity to determine the Work. Or the density and pressure as we need to consider the amount of expanding gas.
The volume before expansion need be given. The Volume after expansion is implicitely given by the ambient (or resultant) pressure - neglecting the temp. i.e.

Cheers,
Sebastian

"pV = const .. for T=const that is true.I figured we can neglect the cooling of the expanding/working gas (Thomson effect I belive). "
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The problem (and the physics) plainly states that in this case the temperature is NOT constant.
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Also, I think, the volume is a required quantity to determine the Work. Or the density and pressure as we need to consider the amount of expanding gas."
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In this problem. to know a volume is completely equivaltnt to knowing a temperature.

If the gas is not 'working' on anything you have the case of free adiabatic expansion and dT is approximately zero for the ideal gas (Joule expansion)

http://www.pha.jhu.edu/~broholm/l35/node6.html.
But you are right for the case of the working gas.

And of course you can replace any of the state variables with temperature if you will. But why would you if you have an expression W(p,V)