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### Course: AP®︎/College Physics 2 > Unit 1

Lesson 3: Thermodynamic processes- First law of thermodynamics equation
- First law of thermodynamics problem solving
- First law of thermodynamics: word problems
- What are PV diagrams?
- PV diagrams - part 1: Work and isobaric processes
- PV diagrams - part 2: Isothermal, isometric, adiabatic processes
- Thermodynamic processes

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# First law of thermodynamics problem solving

First law of thermodynamics problem solving. Created by David SantoPietro.

## Want to join the conversation?

- in the last question we are to find the heat added but the answer is in negative so that means we have calculated the heat loss, which isn't exactly the answer. We needed to find heat added not lost?(4 votes)
- You had to find the heat....Since the work is positive (work done on the system) and you know that the change in internal energy is negative (the average kinetic energy of the gas molecules is lower than it previously was) then you are expected to get a negative q (heat lost)...(3 votes)

- At1:10, you say there is positive work done on the gas. If a gas is being compressed isn't the work required to do that negative? If you define the system as the gas itself? Because wouldn't the opposite be true that if the gas expands it does positive work?

I am confused based on a chart I am looking at in a Kaplan book that says work done by the system (expansion) = positive work and work done on the system (compression) = negative work(5 votes)- Work done by the system (compression) is positive work after all by doing so the internal energy of the gas increases (a positive DU)....work done by the gas (expansion) is negative....(1 vote)

- Determine ∆H2°298 of burning H2S process using meaning of the standard enthalpy change

H2S+3/2O2=So2+H2O(0 votes) - a steam turbine recieves 32 kg of steam per minute with an enthalpy of 3714KJ/kg and a velocity of 30m/s. it leaves the turbine at 27 m/s and 3064 KJ/kg enthalpy. the radiation loss es 88620 kJ/hr. find the KW output.(0 votes)
- The pressure of a Boiler is 9.5 kg/cm2. The barometric pressure of the atmosphere is 768 mm Hg.

Find the absolute pressure in the Boiler?(0 votes) - Two liquids of different densities (ρA = 1,500 kg/m3, ρB= 500 kg/m3) are poured together into a 100-liter tank, filling it. If the resulting density of the mixture is 800 kg/m3,

find the following;

a. the respective quantities of liquids used in kg,

b. the weight of the mixture in kgf; local g = 9.675 mps2.(0 votes)

## Video transcript

- 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.