Physics 27 First Law of Thermodynamics (22 of 22) Work Done By A Gas

to help us understand a little bit more about the differences in the thermodynamic processes let's take a look at a simple example of work done by gas and of course in that case we can only compare three of the four processes because in the naso volumetric process work is equal to zero but let's say that we have a process where the volume is doubled by the gas work perform is 300 joules how do we find q and how do we find delta u meaning heat added to the gas and the internal energy change of the gas in the situation like that for these three processes so let's just put it side by side to get a feel for it all right so we know that work done for a isobaric process is simply equal to p times the change in the volume but how do we find q and how to find delta u because we know that heat is added to the gas and the process is isobaric so pressure stays the same that means q is equal to n times c sub v times c sub no not c sub v because volume doesn't say constant it's c sub p times delta t we also know that the change in internal energy delta u is equal to n c sub v delta t and by using the first law of thermodynamics we can also say that delta u is equal to q minus w all right so what that means is that w if i solve this for w w is equal to q minus delta u w is the difference between those two and we know what that is we know that that was equal to 300 joules so if we subtract those two from one another we can say that w which is equal to q minus delta u is equal to n c sub p delta t minus n c sub v delta t and then if i factor out an n and a delta t that becomes w is equal to n times delta t times c sub p minus c sub v and c sub v minus c sub v is equal to r that's the gas constant so w is equal to n r delta t by replacing c sub p minus c sub v by r now also remember that c sub v for depending upon what gas we're dealing with and of course i should probably make that a condition or a problem let's say that we're dealing with a diatomic ass which means that in this case c sub p is let's see c sub p would be 3 over 2 r well it doesn't matter hang on a second let's continue uh work is equal to that so this is equal to 300 joules now how do we then from that get to q and delta u now notice if i replace c sub p by since we're diatomic and let's split over here in the diatomic gas c sub p is equal to not 5 over 2 but 7 over 2 r and c sub v is equal to 5 over 2 r all right then we can plug that in here and we can then say that q is equal to n times so i'm coming over here c sub p is 7 over 2 r 7 over 2 r times delta t and delta u is equal to n times 5 over 2 r times delta t and notice that nr delta t is equal to 300 that means that q would be 7 halves times 300 and delta u would be 5 halves times 300 so that would be 7 over 2 times n r delta t and this would be equal to 5 over 2 times n r delta t and as we discovered nr delta t is equal to the work done which is equal to 300 joules so this is equal to 7 over 2 times 300 joules and this would be equal to 5 over 2 times 300 joules and so 7 times 3 is 21 divided by 2 is 1050 joules and this would be 1500 divided by 2 or 750 joules all right so now we can summarize this for the isobaric process we know that the work done is 300 joules we know that the heat added to the gas is 150 joules so since we then subtract 300 joules from 1050 joules that means there's 750 joules left over that we didn't use to do work which means that the internal energy of the gasoline will increase so then going back to our what we call our first law of thermodynamics we could then say that in this case delta u is 1050 joules is equal to the heat added to the gas oop i've got this wrong delta u is 750 joules that's equal to 750 joules is equal to 1050 joules minus 300 joules so you can see that an isobaric process is relatively complicated in the fact that there's a very complicated relation between q delta u and w but as long as you keep stuck keep not stop but keep concentrating on the first law of thermodynamics and then working out like this it becomes very straightforward process is not complicated at all because in that case in isothermic process delta u is equal to zero and so that means let me put a line here that if we use the first line of thermodynamics delta u equals q minus w since in isothermic process delta u is zero we get zero is equal to q minus w and so therefore w is equal to q or q is equal to w and so if the work done is 300 joules that means heat added to the gas is 300 joules as well so this becomes 300 joules easy to find delta u is zero and w is 300 joules for an adiabatic process it's also fairly straightforward so what happens here is that q is equal to 0 which means that delta u is equal to minus w because the q drops out and so that means that delta u is therefore equal to minus the work which is minus 300 joules and so simply stated there is that the gas is 300 joules of work but it doesn't receive any energy from the outside so it means that all the energy has to come from within and so the energy lost from the gas is the same as the amount of work that it did at 300 joules and so therefore you can compare the three processes to one another and how to find the work done the heat added or removed and the change in internal energy for the three different kinds of processes that can do work

Channel: Michel van Biezen

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Keywords: ilectureonline, ilectureonline.com, Mike, Mike van Biezen, van Biezen, ilecture, ilecture online, Physics, Thermodynamics, Specific Heat, Temperature, Pressure, Volume, Heat, Work, Internal Energy, Work Done By the Gas, Heat Added by The Gas, First Law of Thermodynamics, The First Law Of Thermodynamics, Change Of State, Adiabatic Process, Adiabatic, Isobaric, Isovolumetric, Isothermic

Id: zuRCN9ty1Zw

Length: 7min 8sec (428 seconds)

Published: Thu Jul 25 2013