# Work Done in an Isobaric Process

I am given the information that an air parcel undergoes isobaric heating from 0° C to 20° C, and that's all I'm given. I have to determine the work done by the parcel on its surroundings. I know that $dW = pdV$, and that $pV = RT$ ($V$ is the specific volume), but I don't know how to go about solving this in terms of the change in temperature and not the change in volume.

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That's easy. Since it's isobaric the pressure is constant and the work is just $P\Delta V$, where $\Delta V$ is the total volume change. You just need to calculate the change in volume from the change in temperature when $P$ is constant, and you can get this from the equation $PV = RT$ that you wrote in your question.

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Wait -- is it really as simple as $(RT)/P=V$ (which gives me the change since one is 0°)? It would be divided by $p$ and then again multiplied by that, so they cancel out. This actually gives what the book says is the right answer. – Vaindil Oct 9 '12 at 16:00
Remember to use absolute temperatures i.e. add 273.16 to temperatures given as Centigrade. Also note that using PV = RT gives you the work per mole of gas. The question doesn't say how much gas you've got. – John Rennie Oct 9 '12 at 16:12

its really easy

PV=NRT

P(V+dV)=NR(T+dT)

PV+PdV=NRT+NRdT

PdV=NRdT=work done (w)

you know N,R dT calculate w...

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