A second question regarding a quasistaic process

[hasan_researc]

Hi,

I am posting a new question regarding a quasistatic process. My lecturer writes in his lecture notes the following:

(Imagine a gas in a cylinder being compressed by a piston. Then,)
"Quasistatic compression: F = (P + deltaP)A, where deltaP is average pressure change as piston moves an infinitesimally small distance deltax
dW = F*deltax = P*A*x + A*P*deltax.

Take the limit deltax tending to dx, then ignore 2nd term (2nd order of smallness), and using : Adx = dV
dW = P dV."

These are my questions:

1) My lecturer writes that deltaP is the average pressure change. Does he mean that after measuring the changes in pressure from many infinitesimal changes in the distance, he used all the data to arrive at the average pressure change?

2) The distance deltax is infinitesimally small, yet he takes the limit dW as deltax tends to x. This is surprising because deltax itself is an infinitesimal quantity. Or does he mean infinitesimal in this context just to mean a very small amount and not the proper definition as used in calculus. Is dx an infinitesimal distance anyway?

3) We are ignoring the second term because of its smallness. BUT, in a strictly mathematical sense, is the product of two infinitesimally small quantities equal to zero?

I have begun my undergrad course last year and still after one year of mathematical physics, I don't understand what the answers to these questions should be.

Please help!

 

[hikaru1221]

1/ I think practically we can understand deltaP that way. But mathematically, not really. And I think what the lecturer said is more about theoretical reasoning than measuring in practice.
Theoretically, there is possibility that in dt, only 1 particle collides with the piston, but there is possibility that there are 1000 particles or so in the collision. But pressure is a quantity of macro scale. Therefore, when considering the collision, we use neither the number 1 nor 1000, we use the AVERAGE number instead. This average number of particles coming to collide in dt can be found theoretically.

dW = F*deltax = P*A*x + A*P*deltax.

I think it's wrong. It should be:
dW = F.deltax = (P + deltaP)A.deltax = PA.deltax + A.deltaP.deltax.
So there is no "x" here And both deltax and dx are infinitesimal amount.

3/ Yes. For example:
P=ab
P + \Delta P = (a+\Delta a)*(b+\Delta b) = P + a\Delta b + b\Delta a + \Delta a\Delta b
Thus:
\Delta P = a\Delta b + b\Delta a + \Delta a\Delta b (1).
\Delta P/\Delta a = a.(\Delta b/\Delta a) + b + \Delta b
For \Delta a \rightarrow 0 and \Delta b \rightarrow 0:
dP/da = lim(\Delta P/\Delta a) = lim (a.(\Delta b/\Delta a) + b + \Delta b) = a.(db/da)+b
And so: dP = adb+bda

Again, this is theoretical reasoning, so \Delta x is unmeasurable. It and dx are simply the same. This is why we can omit the term dP.dx.

Just my 2 cents