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Calculating internal energy and work example

Video transcript

in this video we're going to do an example problem where we calculate an eternal energy and also calculate pressure volume work so we know the external pressure is 1.0 1 times 10 to the fifth Pascal and our system is some balloon let's say it's a balloon of argon gas and initially our gas has a volume of 2.3 liters and then it transfers the gas transfers 485 joules of energy as heat to the surroundings once it does that the final volume of our system is 2.05 liters and we're assuming here that this the moles of gas didn't change the question we're going to answer is for this process what is Delta U so what's a change in internal energy for our system we can use the first law of thermodynamics to answer this the first law tells us that the change in internal energy Delta U is equal to the work done plus the heat transfer before we plug any numbers in here the first thing I want to do is make sure I have a good idea of what signs everything should be I think it's one of the trickiest things and these kind of problems so here since our system transferred energy to the surroundings and not the other way around Q should be negative because when your system transfers energy to the surroundings then its internal energy should go down work on the other hand since V 2 is less than V 1 the volume of our system went down which means the surroundings had to do work on the gas to get the volume to decrease we would expect if the surroundings did work on our system that would increase the internal energy so that means the work done here is positive we can also calculate work because we know the external pressure we know it's constant and work can be calculated as the external pressure times the change in volume and we know both of those things we know the external pressure and we know the initial and final volumes so if we start plugging that in we get that Delta U is equal to negative 485 joules so that's our heat we know it should have a negative sign because the heat was transferred to the surroundings so negative 485 joules - we should have a negative sign there - the external pressure 1.01 times 10 to the fifth Pascal so that's our external pressure times our change in volume so that's our final pressure to 0.05 liters minus our initial pressure which is 2.30 liters we could at this point be like okay we figured it all out we just have to stick all of these numbers in the calculator and we're done and that's probably what my first students take you would often be but there there's one more thing that we should check before we actually plug in the numbers and and have a party and that's our units we have our heat in terms of joules we probably want our change in internal energy in terms of joules - on this other side we're calculating our work here and we have Pascal's so Pascal's times so we have joules and Jules we have Pascal's times liters so then the question is okay we're doing joules - Pascal's times leaders we need to make sure that whatever we calculate here in terms of work also has units of joules otherwise we will be will be subtracting two things I don't have the same units and that's bad we'll have to do some sort of unit conversion first so let's just double check that Pascal's times leaders will give us joules so the way I did this is by converting everything to the same units so if you take joules which is already SI units we can actually simplify it more in terms of other SI units so a joule is equal to one kilogram meter squared per second squared so joules is equal to 1 joule is equal to one kilogram meter squared over second squared one Pascal Pascal's are also in terms of are also SI units and if we convert Pascal's to kilograms meters and seconds we get that one Pascal is one kilogram per meter second squared what this tells us is that we have to multiply this by units of volume and whatever we multiply it by should give us units of joules so what we need to do here is convert our leaders to meters cubed and if we do that everything is in terms of kilograms meters and seconds this meters cancels out with one of these and we end up with one kilogram meter squared per second squared on both sides so then everything is in terms of joules that's not the only way you could have made sure that the unit's made sense on you could have converted them to something else so basically you just have to make sure that all of the unit's you're using in your equation match each other if you're going to add them or subtract them all of this is to say that we need to make sure we convert liters to meters cubed so that everything works out in terms of joules so if we do that we get that minus 485 joules 1.0 1 times 10 to the fifth Pascal times negative 0.25 liters which is the change in volume which is negative the volume went down so the change in volume should be negative and then we have to add one more thing here to convert our leaders to meters cubed so one liter is equal to one 10 to the minus third meters cubed so now our leaders canceled out and Pascal's times meters cubed gives us joules so that gives us that Delta U our change in internal energy is negative 485 joules and then if we plug this all into our calculator to calculate the work we get positive twenty five point two five joules so if we add our key and our work here we get that the overall change in internal energy for this process is negative 460 joules so the key things to remember here for this kind of problem is to double check your signs for work and heat and also to make sure all of your units map
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