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I want to know the formula (and what does the symbols stand for) for how much work is needed to compress a certain volume of gas?

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Have you looked at any standard resources, e.g. textbooks, Wikipedia, other websites? –  David Z Sep 17 '12 at 20:29
    
they are all talking about expansion maybe it's the same –  Abdelrahman Esmat Sep 18 '12 at 6:50
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1 Answer 1

up vote 2 down vote accepted

Work is equal to force times distance. This applies to everything, not just the behaviour of gases.

Suppose we have some gas in a container with volume $V$ and surface area $A$. We apply some force normal to this area and we compress the gas by some tiny distance $dx$. We keep the distance $dx$ small enough that we can make the approximation that the area $A$ doesn't change.

Force

I've drawn the gas as a sphere but the argument applies whatever the shape. Because work is force times distance the work done in compressing the gas is simply:

$$ W = F \space dx $$

To get this into a more familiar form note that if we're applying a pressure $P$ over the whole surface of the sphere then the force is just the pressure times the area i.e.

$$ W = P \space A \space dx $$

Now note that the area $A$ times the distance moved $dx$ is the change in volume i.e. $Adx = dV$ so:

$$ W = P \space dV $$

and this is the equation we use to calculate the amount of work done in compressing the gas. The calculation can be a bit involved because as we compress the gas the pressure will change i.e. the pressure is a function of volume $P(V)$. Also unless we let the gas cool as we compress it, it's temperature is going to go up and that will also increase the pressure. If we allow the gas to cool we get an isothermal compression and if we don't allow it to cool we get an adiabatic compression. If we take the isothermal expansion (because the maths is easier) the pressure is related to volume by:

$$ P = \frac{nRT}{V} $$

where $R$ is the gas constant and $n$ is the number of moles of the gas. The work to compress from volume $V_1$ to $V_2$ is then simply:

$$ W = \int_{V_1}^{V_2} P(V)dV = \int_{V_1}^{V_2} \frac{nRT}{V}dV $$

and doing the integral gives:

$$ W = nRT ln \left( \frac{V_2}{V_1} \right) $$

The formula we've ended up with applies to expansion as well as compression. With a compression $V_2 < V_1$, and to calculate the work done by the gas when it expands we just put $V_2 > V_1$. The only difference is that in one case the work we calculate is the work we do on the gas, while in the other it's the work done by the gas.

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