00:01
In this problem you have n moles of an ideal gas going from state i, then it undergoes an isothermal process to a, then an adiabatic process to f.
00:12
And we specified here the p and v for state i, state f, state a.
00:21
We do not know.
00:22
We're going to have to look at that later on.
00:26
Now isothermal process, t is equal to a constant.
00:32
Adiabatic process, there's no heat in or out, q is equal to zero.
00:36
Now what we're trying to get in part a is a change in entropy as we go from i to a to f, which is change in entropy.
00:46
I to a, change in entropy, a to f.
00:51
But that is zero.
00:52
Why? because here's the general formula for general process for the change in entropy.
00:57
As you're going from a to f, this is an adiabat, you have an infinite collection of adiabatic processes.
01:05
Infinitesimal adiabatic processes.
01:06
Well by definition dq is zero in each of those.
01:09
So you've got an infinite sum of zeros.
01:11
So the change in entropy will be zero for an adiabat.
01:14
So our goal is to find the change in entropy for the isothermal and we have our answer for the whole thing.
01:24
Now where do we have, now it looks like we're going to have to work on q.
01:27
But how do we get that? well we're going to have to look at internal energy and work and that's what we're going to do now.
01:32
First let me write out property of internal energy.
01:35
I use u, you might use e sub int.
01:41
I use u for internal energy.
01:43
U, internal energy, is a state function and only depends on t for an ideal gas.
02:06
And that's what we have here.
02:09
It is a state, it is a state function.
02:14
What does that mean? does not matter how you go from i to a.
02:22
The change in internal energy will be exactly the same no matter what path you take from i to a.
02:31
Path independence.
02:35
Now since i and a are on an isotherm that means they have the same t.
02:46
So delta uia is equal to zero since ti is equal to ta.
02:55
It's an isotherm.
02:58
Does not matter how you got to a.
03:03
Path independence.
03:04
That's what a state function means.
03:05
It only cares what state you're in.
03:08
What state you came from and what state you ended up in.
03:10
How you got there, irrelevant.
03:12
Remember that.
03:16
So now let us look at the first law.
03:28
Qia minus wia.
03:31
That's the first law of thermodynamics.
03:33
So this tells me that qia is equal to wia.
03:38
So now just u may be a state function, q and w are not.
03:43
They are path dependent.
03:45
So if i could calculate the work done for an isothermal path i would have the heat.
03:53
So that's what we got to work on.
03:56
So let's do that now.
03:57
Wia is equal from i to a.
04:01
Pdv, that's the general formula.
04:07
Vi to va.
04:10
Nrti over v dv.
04:16
Now nr and ti can come out.
04:18
They're constants.
04:20
So vi, va, dv over v.
04:29
And that's a natural log.
04:31
So nrti, natural log, va over vi.
04:41
That's what we have.
04:43
So now let's, we can't write our answer in va.
04:47
We don't know anything about va.
04:48
We're going to have to work on that in a minute.
04:50
Let's get the formula for the change in entropy.
04:54
I to a, dq over t.
04:59
That's the general formula.
05:00
1 over ti, i to a, dq.
05:08
Well, but that's just adding up all the individual dqs as you go along that isotherm.
05:14
And so that's just going to be qia over ti.
05:20
And we put in our formula for q, which is equal to the work.
05:25
Nrti over ti times the natural log, va over vi.
05:36
So ti's go away and we're left with nr, natural log, va, vi.
05:45
That's what we have.
05:49
Now though, we got to find va.
05:52
Now i and a are an isotherm.
05:56
So they're related in this manner.
05:58
Pi, vi is equal to nrti and that's equal to p a, va.
06:08
Let's call this equation one.
06:10
Now a and f are on an adiabat...