00:01
Want to find the pressure at a, the pressure at b, and pressure at c, whether pressure at a is given, but pressure at b and pressure at c.
00:06
So we know that we have a monotomic gas, which means that our gamma value is going to be equal to five over three.
00:23
And we can say that p sub b, v, b divided by t sub b is going to equal p sub a, v sub a, ideal gas constant and so p sub b would be equal to p sub a v sub a t sub b divided by v sub a t and we can solve so the a is one atmosphere the volume at a is 22 .4 liters volume at b is 56 .0 liters and then the temperature at b would be 273 kelvin.
01:11
Temperature at a would be 273 kelvin as well.
01:14
The process from a to b is isothermal.
01:18
So we can say that the pressure at b would be equal to 0 .40 atmospheres.
01:26
Now for the pressure at c, we can say that for the adiabatic compression in this case, we can say that p sub a v sub a to the gamma power would be equal to p sub c v sub c to the gamma power so p sub c would be equal to p sub a times v sub a divided by v sub c to the gamma power this is equal in one atmosphere times 22 .4 liters divided by 56 .0 liters to the 5 thirds power this is giving us a pressure at c of 0 .217 atmospheres.
02:15
So this would be our answer for the second answer for part a and our first answer for part a.
02:25
So for part b, we want to find the temperature at c.
02:32
So we can say p sub b, v sub b divided by t sub b is equaling p sub c, v sub c divided by t c which means that t sub c would be equal to p sub c v sub c t sub b divided by p sub b v v sub b these are going to cancel out and this equals 273 kelvin the temperature at b and then times 0 .2172 atmospheres divided by 0 .40 atmospheres.
03:13
And we find that t sub c is going to equal 148 kelvin.
03:20
That would be our answer for part b.
03:23
And finally, for part c, we need to find the work done and the heat transfer, rather just find the we're in the process of the process we're going to be finding the change in internal energy.
03:39
The heat transfer associated with each process and the work done in each process.
03:45
So we can say for process ab, we can say that the change in internal energy for ab would be equal to the number of moles times the molar heat capacity, a constant volume times delta t.
04:05
And we know that this is zero because this is an is an isothermal expansion.
04:09
So we can say that the heat transfer associated with this process would be equal to the work done and this would be equal to nr t times the natural log of the product rather of the quotient of the volumes and so we can say that q a sub b which again equals work a b would be equal to one mole times 8 .314 jules per mole per kelvin times 273 kelvin multiplied by the natural log of 2 .5.
04:52
And this is equaling 2 ,080 joules.
04:57
For the change in entropy at ab, we find that this is going to be q sub ab divided by t sub ab.
05:05
So this would be equal to 2 ,079 .7 joules and then divided by 273 kelvin.
05:15
This is equaling 7 .62 joules per kelvin.
05:21
At this point, we can say for process bc, we know that the work done in bc would be equal to zero joules because this is an is an isocoric process, so constant volume, no work is done.
05:36
This is c.
05:41
And then we can say that the change in internal energy of bc would be equal to q sub bc, the heat transfer associated with that process, and this would be equal to the number of moles times the molar heat capacity at a constant volume, times the change in temperature...