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- So far you've seen the
First Law of Thermodynamics.

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This is what it says.

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Let's see how you use it.

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Let's look at a particular example.

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This one says, let's say
you've got this problem,

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and it said 60 joules of
work is done on a gas,

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and the gas loses 150 joules
of heat to its surroundings.

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What is the change in internal energy?

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Well, we're going to use the First Law.

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That's what the First
Law lets us determine.

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The change in internal
energy is going to equal

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the amount of heat
that's added to the gas.

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So let's see, heat added to the gas.

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Well it says that the gas
loses 150 joules of heat

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to its surroundings.

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So that means heat left of the gas

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so heat left the gas.

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This must have been put
into a cooler environment

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so that heat could leave.

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And so it lost 150 joules.

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A lot of people just stick 150 here.

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It's got to be negative 150 because

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this Q represents the
heat added to the gas,

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if you lost 150 joules,
it's a negative 150.

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And then plus, all right
how much work was done.

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It says 60 joules of work is done on a gas

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so that's work done on the gas.

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That means it's a positive contribution

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to the internal energy.

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That's energy you're adding to the gas.

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So 60 joules has to be positive,

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and so this is plus the work
done is positive 60 joules.

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Now we can figure it out the
change in internal would be

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negative 90 joules.

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But why do we care?

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Why do we care about the change in

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internal energy of the gas?

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Well here's something important.

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Whether it's a monatomic or
diatomic or triatomic molecule

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the internal energy of the
gas is always proportional

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to the temperature.

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This means if the temperature goes up,

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the internal energy goes up.

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And it also means if the
internal energy goes up,

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the temperature goes up.

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So one thing we can say,
just going over here,

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looking that the change in
internal energy was negative.

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This means the energy
went down by 90 joules.

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Overall when all is said
and done, this gas lost

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90 joules of internal energy.

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That means the temperature went down.

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That means this gas is going to be cooler

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when you end this process
compared to when it started.

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Even though you added
60 joules of work energy

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it lost 150 joules of heat energy.

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That's a net loss.

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The temperature is going to go down.

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So this is an important key fact.

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Whatever the internal energy does,

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that's what the temperature does.

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And it makes sense since
we know that an increase in

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internal energy means an
increase in translational

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kinetic energy, rotational kinetic energy,

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vibrational energy.

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That temperature is also a
measure of that internal energy.

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Note that we cannot say exactly how low

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the temperature went.

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This is a loss of 90 joules

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but this doesn't mean
a loss of 90 degrees.

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These are proportional.

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They're not equal.

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If I go down 90 joules that doesn't mean

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I go down 90 degrees.

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I would have to know more
about the make up of this gas

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in order to do that.

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But the internal energy

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and the temperature are proportional.

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Let's try another one.

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Let's say a gas started with
200 joules of internal energy

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and while you add 180
joules of heat to the gas,

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the gas does 70 joules of work.

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What is the final internal
energy of the gas?

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All right, so the change
in internal energy

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equals Q.

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Let's see, gas starts with
200 joules of internal energy,

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that's not heat.

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While you add 180 joules of heat,

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here we go, 180 joules should
it be positive or negative?

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It's going to be positive.

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You're adding heat to that system.

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So positive 180 joules of heat are added

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plus the amount of work done on the gas,

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it says the gas does 70 joules of work.

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So most people would just
do, all right, 70 joules.

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So there we go.

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But this is wrong.

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This is wrong because this is
how much work the gas does.

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This W up here with the
plus sign represents

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how much work was done on the gas.

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If the gas does 70 joules of work,

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negative 70 joules of
work were done on the gas.

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You have to be really careful about that.

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So we can find the change
in internal energy.

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In this case it's going to
equal positive 110 joules.

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But that's not our answer.

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The question's asking us for the

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final internal energy of the gas.

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This is not the final
internal energy of the gas.

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This is the amount by which
the internal energy changed.

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So we know the internal energy went up,

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because this is positive

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and this is the change in internal energy.

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Internal energy went up by 110 joules.

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That means the temperature
is also going to go up.

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So what's the final
internal energy of the gas?

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Well, if the internal
energy goes up by 110 joules

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and the gas started with 200 joules

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we know the final
internal energy, U final,

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is just going to be 200 plus 110 is 310,

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or if you want to be
more careful about it,

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you can write this out.

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Delta U, we can call U
final, minus U initial.

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That's what delta U stands for.

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U final is what we want to find

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minus U initial is 200.

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So positive 200 joules was
what the gas started with.

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Equals, that's the change
and that's what we found,

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110 joules.

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Now you've solved this for U final.

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You would add 200 to both
sides and again you would get

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310 joules as the final
internal energy of the gas.

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Let's look at one more.

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Let's say you got this one
on a test and it said that

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40 joules of work are done on a gas,

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and the internal energy
goes down by 150 joules.

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What was the value of the
heat added to the gas?

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Note we're not solving for
the internal energy this time

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or the change in internal energy.

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We're trying to solve for the heat.

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What's heat? Heat is Q.

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So this time we're going to
plug in for the other two

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and solve for Q.

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What do we know?

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40 joules of work are done on a gas,

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so this work has got to be a positive 40

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because the work is done on
the gas and not by the gas.

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And we know the internal
energy goes down by 150 joules.

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It means the change in internal energy

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has to be negative 150.

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So if I plug in here my delta U,

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since my internal energy went down by 150,

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delta U is going to be negative 150.

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Q we don't know

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so I'm just going to
put a variable in there.

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Q I don't know.

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I'm just going to put a Q in there.

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I'm going to name my ignorance

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and I'm going to solve for it.

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Plus the work done.

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We know the work done was 40
joules and it's positive 40.

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Positive 40 because work
was done on the gas.

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Now we can solve for Q.

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The amount of heat.

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The value of the heat added
to the gas is going to be,

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if I move my 40 over here I
subtract it from both sides,

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I'm going to get negative 190 joules.

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This means a lot of heat left.

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190 joules of heat left the system

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in order for it to make
the internal energy

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go down by 150 joules.

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And that makes sense.

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40 joules of work were added.

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But we said the internal energy went down.

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That means the heat has to take away

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not only the 40 that you
added but also another 150

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to make the energy go down
overall so the heat taken away

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has to be negative 190 joules.

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All right, so those were a few examples

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of using the First Law.

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Basically, you've got to be
careful with your positive

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and negative signs.

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You've got to remember
what these things are

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that Q is the heat, W is the work, delta U

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is the change in the internal energy

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which don't forget that
also gives you an idea

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of what happens to the temperature.