00:01
In this problem, we have four grams of a gas that undergoes this cycle, and we're going to be asked a couple questions about the temperature of the gas at certain points and specific heat, but we'll take one at a time.
00:16
Before we move on, though, notice the units, atmosphere for pressure, liters, for vine.
00:23
Now, we're going to need the ideal gas law to get the first question we want is the temperature, the answer is the temperature at b.
00:29
And they give us the temperature a is 400 degrees celsius.
00:34
We're going to be using the ideal gas lot.
00:36
We cannot use celsius.
00:37
It must be in kelvin.
00:40
And the way to go from celsius to kelvin is to add 273 .15 to be 100%.
00:52
Correct.
00:54
And this gives you 673 .15 kelvin.
01:05
We don't need that many digits.
01:06
I'll just keep three in my calculations.
01:09
Now, we need tb.
01:15
How are we going to do that? p -a -v -a is equal to n -r -t -a.
01:22
That's the ideal gas law for point a.
01:26
This gives me that pa, va, and the gas is confined.
01:35
You're not gaining gas or losing gas.
01:38
So pa -v -a over ta is equal to n times r r is universal gas constant that that is a constant and in this particular case is a constant so this is a constant and that would be equal then to p b b b over t b so i can i know a relationship between p a v a t b and t b i don't need to find n i'll show you how to do that as an alternate in a second but i don't need to and i don't have to worry about converting units anything like that, if i was concerned about that.
02:24
Just solve for tb now.
02:26
Tb, p -b, b -b, p -a, p -a, f -a, t -a, just algebraically solved.
02:44
And we can put in our values now, and p -b is four -tmosphere, pa, six atmosphere, v -b is nine liters, v -a is two liters.
03:06
And multiplied by 673, calvin.
03:11
Keep, i'm always, in the end, we want they say to report your answers to two.
03:17
So i keep an extra one, then round down.
03:23
And this turns out to be 2 .02 times 10 to the 3 calvin, but meaning their criteria, 2 .0 times 10 to the 3.
03:35
Calvin is what i report as my answer.
03:41
So that's the temperature at b.
03:46
There is, you may be more used to trying to find, to calculate, you know, from this equation to get n.
04:00
It's fine.
04:01
But you just got to be careful about one thing.
04:04
The units of r are not.
04:07
You got these unusual units for you.
04:09
You got liters for volume and atmosphere for pressure.
04:12
So what you need to do is use r equals 0 .0824 liter atmosphere per mole kelvin and find n.
04:37
So you'd find n from the pa -v -a equation, stick that into the pbvb equation and solve for tb.
04:46
That's it.
04:48
If you do that, you get 0 .217 moles.
04:55
I won't do the rest.
04:56
It's straightforward enough.
04:58
It's going to give you the same exact answer.
05:01
I'll give you the exact same answer.
05:03
Okay.
05:05
Part b wants to know for this gas what its specific heat, a constant volume is.
05:13
So if you have a constant volume pressure, what would be the specific heat you would use? and this is going to be a specific heat per.
05:20
It's going to be not in terms of mole, but in terms of grams.
05:25
You couldn't find it in terms of moles, too, especially once you have this.
05:29
But seeing they gave it to you, when they give you a problem in terms of grams, usually they want things in terms of that.
05:37
So that's what i'm going to do.
05:40
Now, they tell you that there's a certain amount of heat entering process ab.
05:50
Qab is equal to 2 .20.
05:56
That's the amount of heat that enters.
05:59
Now, remember, heat and work are process dependent.
06:05
They're path dependent.
06:07
They are not state functions, not like the internal energy is.
06:11
So it matters.
06:13
It matters what process we're looking at.
06:18
A, b, the work a .b is different than the work c .d.
06:21
It's different than the work b .c.
06:23
And so, well, b .c and d .a are zero work.
06:25
But depends on the process.
06:29
It depends on the process.
06:30
They are path dependent.
06:33
Path dependent.
06:37
So let's calculate the work.
06:40
W -a -b.
06:43
This is the area under the curve.
06:51
So that will be the area of this triangle plus the area of this rectangle.
07:06
That's the area under the curve.
07:08
So, that's one half, nine liters minus two liters, six atmospheres, minus four atmospheres, which you'd just be reading these off the pv diagram.
07:24
Plus, that's for the triangle, now for the rectangle, nine liters minus two liters times four atmosphere.
07:33
And this works out to be 35 liters atmosphere.
07:42
That is a unit of work.
07:44
It is a unit of work.
07:45
I mean, it may be unusual to you, but it isn't a unit of work.
07:51
Now, seeing that they gave us the other part in calories, killer calories, but calories.
07:58
Let me convert this to calories, times 24 .2 calories, 1 liter atmosphere.
08:11
And this works out to be 847 calories, three digits...