Why is the work done by an ideal gas the area under a PV diagram?

[goodcow]

Could someone please prove why the work done by a gas is the area under a [itex] PV [/itex] diagram?

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

 

[truesearch]

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

 

[goodcow]

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?

 

[truesearch]

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.

 

[goodcow]

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

 

[truesearch]

you are in good company!

 

[Ken G]

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.

 

[thafeera]

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

 

[truesearch]

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