# Work done by an ideal gas

Could someone please prove why the work done by a gas is the area under a $PV$ diagram?

That is, I know that $dW=P \text{ }dV$, but why is that true physically? I realize that $W=f \cdot d \text{ and } F=P \cdot A$ .Thanks.

## Answers and Replies

You have more or less done it for yourself!!
W = P x V......P = F/A and V = A x d
W = F/A x A/d ....= F x d

Well yes, I know that the equation is true, but I was wondering why it's physically true. For example, the area under a velocity-time graph is displacement. Why is the area under the graph the work done?

It is because the graph is of P against V so the area under this graph must be the PxV quantity.....which is Fxd = work done.

Oh...wow I'm silly. Thanks!

you are in good company!!

Ken G
Gold Member
Another way to say all this is that, you understand the area under the v vs. t curve is displacement x because you think of v as dx/dt. Thus, to understand why the area under a P vs. V curve is work done, you simply need to think of pressure as dW/dV, i.e., pressure is the amount of work done per unit volume change. In other words, instead of thinking of work as something that comes from pressure, think of pressure as a concept that stems directly from work. That is very much what pressure is-- the concept of the amount of work done per volume change. Indeed, there are situations where the easiest way to calculate pressure P is to first calculate the work W as a function of V and take dW/dV.

just remember that this is true under the assumption of quasistatic conditions, which is almost static but notquite.
think about a situation where the internal pressure is much higher than the external presssure. how do you calculate work then?

thafeera

Another little tip: look at the units of the quantities you are multiplying.
P x V units are Pa x m^3 = (N/m^2) x m^3 = Nm = energy units
v x t units are ms^-1 x s = m = distance units
F x e units N x m = energy units
V x Q units are volts x charge = V x C = (J/C) x C = J = energy units