00:01
So for this problem, i've submitted it up to four stages.
00:05
We'll start with stage one, and we're going to figure out what the final pressure is in this system.
00:11
So we use the equation 1 on the left -hand side, and we can say that the pressure times volume, initially, paragama, equals the pressure times the volume to the paragraph of the final.
00:27
So then we can rearrange for the final pressure, and that is equal to the initial pressure times the power of the final.
00:35
The initial volume over the end for the final volume and that's all to the power of gamma.
00:42
So that's going to be 2 .127 times 10 to the 6 pascal times 50 times 10 to the minus 6 meters cubed divided by 400 times 10 to the minus 6 meters cubed and this is all to the power of 1 .4 because it's a diatomic gas.
01:18
For this in the calculation i end up with 115728 .8 .8155 pascal and this is 1 .25.
01:43
So this is 1 .2 times 10 to the 5.
01:57
For stage 2 we can now work out the work done by the system.
02:04
And first we can work out the work done on the system and using equation 2 is adiabatic.
02:14
So no heat change...