00:01
Okay, so here is a heat engine that takes one mole of gas.
00:06
The process it undergoes is shown in the pivot diagram.
00:10
Now the question asks us in the first part a, which should show that the process a is isothermal compression.
00:19
So for a, just look at the product of p .a.
00:24
V a which is is equal to if you multiply p a and v a it will be two into the power three and if you multiply pv v v v again you're gonna get two into the power three so when is up when is p v constant so pv is equal to constant heat temperature constant no that is it's ball's law so you can say that the process ab is an isothermal process this implies ab is isothermal now for the second part we have to calculate which of the processes in which of the processes is heat absorbed by the gas and when it is rejected so for this let us first to do question number part see of the question by finding the temperatures then that will be that will make it easier to solve question number c so question number b so first let us to question number c simply we have to calculate the temperatures 8 point a b and c and this we can use the idle gas equation and is pb is equal to n r t that will give us t is equal to p v by nr right so at every points like at point a so t will be equal to b a b a b a b a and r where p and b a are pressure and volume 8a and n is the number of moles and r is the universal gas constant whose value is equal to 8 .314.
02:23
Let me write this down here.
02:26
8 .314 is unity joule for more per kelvin.
02:37
So if you calculate this value for ta you will get 240 .5 kelvin, you will get 240 .5 kelvin.
02:47
So we are done with a similarly temperature at point b will be p v b b b v upon n r again we have the value for all these four parameters right so just plug these values in here and you're gonna get again 240 .5 otherwise you can just use your answer it question number a which says that av is at athermal process so temperature at a and temperature at b should be the same now tc will be equal to just use pcvc by an r which will give us the sturgess value in here and you get 481 .11 kelvin also okay we are done with question number c now let's move back to question number b now it will be easier to solve question number b now let us see every process turn by turn let us first see a to b when it goes from a to b the heat exchanged in this process qa b is equal to since it is an is a thermo process we can write an r t which t is the temperature at a and b log e v b by va now as you see v v b is less than the a so this logarithm will give us a negative value which makes that q is negative and if q is negative this implies that the heat is released in this case so heat is released right now from b to c to b c q b c is equal to this is at constant pressure so you can just write n cp remember constant pressure so ncb d t so t is the temperature difference between point c and b so tc minus tv now as you see from your answers in question on answers in c here it says that tc is greater than t v which implies that q is positive which implies that it is absorbed similarly if you go from c to a qca will be equal to n cv now remember cv okay now c2a means final temperature is a temperature is c so again in this case now compare t a and tc is greater than t a right so tc is greater than t a so tc is less than tc which means that q will be negative in this case k is negative so in this case, heat is released.
06:37
So finally we can say in the process from a to b and c to a, it is released, whereas in the process from b to c, the heat is absorbed.
06:52
So let's move on to part d, which asks us the net heat exchanged with the surroundings and network done by the engine...