On the other hand, on the reversible path, where I am slowly decreasing the
pressure from this initial level, again I'll get the same integral minus V1 to
V2: nRT over VdV. Here's the logarithmic solution.
Except that now V2, the final volume is greater than the initial volume.
So, if V2 over V1 is a positive number, the log is positive, the whole work is
negative, consistent with, the system doing work on the surroundings.
And the integral under that curve is obviously bigger.
The shaded area is the constant external pressure that we would add this extra
work when we follow the reversible path. Alright?
So, let's pause for a moment, I'd like you to add some different PV expansions
or compressions if you appreciate how they differ in terms of the pressure
involved. So, I'd really like to cement home this,
difference between a reversible path and a non-reversible path.
And I'd also like to emphasize, the reason we've been talking about
isothermal processes is because we want to take advantage of this
relationship for an ideal gas. But if the temperature is constant, then
P times V is a constant. Right?
It's, because PV for an ideal gas is equal to nRT.
So n and R are constants. If T is a constant, then PV is constant.
And so I'd just like to illustrate. Imagine again that we're expanding an
ideal gas from half a liter at 4 bar to 1 liter at 2 bars.
And one way we might do it, I'll actually draw an experiment here, imagine I've got
this mass and it's enough of a mass to press down with 4 bar worth of pressure.
And it's sitting on top of a piston and when I take it off, the piston rises and
it leaves 2 bar. So, the piston if you like weighs as much
as the mass. You've got 2 bar out of the piston, 2
more bar out of the mass. So, the constant external pressure in one
step would consist of just lift that first mass off.
And so, you immediately have an external pressure of only 2 bar.
That's the piston. And you will expand from 1 half to 1
liter. And so you'll have done 2 bar times half
a liter, you'll have done 1 liter bar. And if you look up the conversion of
liter bar to joules, that's minus 100 joules, minus because we're doing work on
the surroundings. So, you did minus 100 joules worth of
work. Well, let's imagine instead of a single
mass that was pressing down with two bars worth of pressure.
I have two masses, which is worth an addition, is worth a bar.
And so, the piston plus the 2 masses is still a net of 4.
But I'll take them off now in steps. So, I go from 4 to 3.
And so, I'll travel along until I hit the curve, which will be because PV must be a
constant. It'll be at 0.67 liters because 3 times
two thirds is 2. And I'm always along a curve where PV
equals 2 in this case. And so I can integrate under that curve.
And now I take the other one off. So, now I'm moving from 0.67 out to 1b
so, that's 2 thirds times 2, out of 4 thirds.
And this was 3 times a third. I guess that all comes up to, 7 thirds,
something like that. when you turn that into joules minus 117
joules. So, you got more work out of the system.
By doing it more slowly, not just ripping the bar off, the weights off.
But taking them off in two steps. So, the reversible path basically
imagines, I don't have a weight here, I have a pile of sand.
And I'm going to take that sand off one grain at a time.
It's going to be a slow experiment, but it's going to be a reversible experiment.
So, with each infinitesimal little grain relieving the pressure, the gas is going
to expand, expand, expand, expand, expand, doing work, raising that piston.
And I will end up with, if you integrate under that curve, if you actually compute
the logs. You'll get minus 139 jewels.
So, illustrating that work depends on the path.
It depends on how you change that external pressure.
All right, well that completes our first look at PV work.