First law of thermodynamics problem solving | Chemical Processes | MCAT | Khan Academy

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

This is what it says.

Let's see how you use it.

Let's look at a particular example.

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

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

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

What is the change in internal energy?

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

That's what the First
Law lets us determine.

The change in internal
energy is going to equal

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

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

Well it says that the gas
loses 150 joules of heat

to its surroundings.

So that means heat left of the gas

so heat left the gas.

This must have been put
into a cooler environment

so that heat could leave.

And so it lost 150 joules.

A lot of people just stick 150 here.

It's got to be negative 150 because

this Q represents the
heat added to the gas,

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

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

It says 60 joules of work is done on a gas

so that's work done on the gas.

That means it's a positive contribution

to the internal energy.

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

So 60 joules has to be positive,

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

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

negative 90 joules.

But why do we care?

Why do we care about the change in

internal energy of the gas?

Well here's something important.

Whether it's a monatomic or
diatomic or triatomic molecule

the internal energy of the
gas is always proportional

to the temperature.

This means if the temperature goes up,

the internal energy goes up.

And it also means if the
internal energy goes up,

the temperature goes up.

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

looking that the change in
internal energy was negative.

This means the energy
went down by 90 joules.

Overall when all is said
and done, this gas lost

90 joules of internal energy.

That means the temperature went down.

That means this gas is going to be cooler

when you end this process
compared to when it started.

Even though you added
60 joules of work energy

it lost 150 joules of heat energy.

That's a net loss.

The temperature is going to go down.

So this is an important key fact.

Whatever the internal energy does,

that's what the temperature does.

And it makes sense since
we know that an increase in

internal energy means an
increase in translational

kinetic energy, rotational kinetic energy,

vibrational energy.

That temperature is also a
measure of that internal energy.

Note that we cannot say exactly how low

the temperature went.

This is a loss of 90 joules

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

These are proportional.

They're not equal.

If I go down 90 joules that doesn't mean

I go down 90 degrees.

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

in order to do that.

But the internal energy

and the temperature are proportional.

Let's try another one.

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

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

the gas does 70 joules of work.

What is the final internal
energy of the gas?

All right, so the change
in internal energy

equals Q.

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

that's not heat.

While you add 180 joules of heat,

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

It's going to be positive.

You're adding heat to that system.

So positive 180 joules of heat are added

plus the amount of work done on the gas,

it says the gas does 70 joules of work.

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

So there we go.

But this is wrong.

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

This W up here with the
plus sign represents

how much work was done on the gas.

If the gas does 70 joules of work,

negative 70 joules of
work were done on the gas.

You have to be really careful about that.

So we can find the change
in internal energy.

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

But that's not our answer.

The question's asking us for the

final internal energy of the gas.

This is not the final
internal energy of the gas.

This is the amount by which
the internal energy changed.

So we know the internal energy went up,

because this is positive

and this is the change in internal energy.

Internal energy went up by 110 joules.

That means the temperature
is also going to go up.

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

Well, if the internal
energy goes up by 110 joules

and the gas started with 200 joules

we know the final
internal energy, U final,

is just going to be 200 plus 110 is 310,

or if you want to be
more careful about it,

you can write this out.

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

That's what delta U stands for.

U final is what we want to find

minus U initial is 200.

So positive 200 joules was
what the gas started with.

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

110 joules.

Now you've solved this for U final.

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

310 joules as the final
internal energy of the gas.

Let's look at one more.

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

40 joules of work are done on a gas,

and the internal energy
goes down by 150 joules.

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

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

or the change in internal energy.

We're trying to solve for the heat.

What's heat? Heat is Q.

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

and solve for Q.

What do we know?

40 joules of work are done on a gas,

so this work has got to be a positive 40

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

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

It means the change in internal energy

has to be negative 150.

So if I plug in here my delta U,

since my internal energy went down by 150,

delta U is going to be negative 150.

Q we don't know

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

Q I don't know.

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

I'm going to name my ignorance

and I'm going to solve for it.

Plus the work done.

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

Positive 40 because work
was done on the gas.

Now we can solve for Q.

The amount of heat.

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

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

I'm going to get negative 190 joules.

This means a lot of heat left.

190 joules of heat left the system

in order for it to make
the internal energy

go down by 150 joules.

And that makes sense.

40 joules of work were added.

But we said the internal energy went down.

That means the heat has to take away

not only the 40 that you
added but also another 150

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

has to be negative 190 joules.

All right, so those were a few examples

of using the First Law.

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

and negative signs.

You've got to remember
what these things are

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

is the change in the internal energy

which don't forget that
also gives you an idea

of what happens to the temperature.