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First law of thermodynamics problem solving

Video transcript

all right you've seen the first law thermodynamics this is what it says let's see how you use it let's looking at a particular example this one says let's say you 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 gonna 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 that means heat left of the gas so heat left the gas this must have been put into a cooler environment so that he 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 negative 150 and then plus alright how much work was done 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 energy would be negative 90 joules but why do we care why do we care about the change in internal energy of the gas we'll hear 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 internal energy went down by 90 joules overall when all was 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's 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 makeup 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 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 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 questions 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 change so we know the internal energy went up because it's positive and this is the change in internal energy internal energy went up by 110 joules that means 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 a hundred and ten joules and it started 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 you 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 solve 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 a hundred and fifty joules what was the value of the heat added to the gas nope 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 a hundred and fifty 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'ma 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 eeee 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 a hundred and fifty 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 got to be careful with your positive and negative signs you 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
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