00:01
Hi students, here in this question we have two parts of questions.
00:05
In the first part of the question, we have to find out the work done.
00:15
So work done and here it is an isobaric process.
00:20
So the work done w is equal to p into delta v where p is the pressure and delta v is the change in volume, change in volume.
00:40
So here it is given that pressure p is equal to 4 .8 into 10 raised to 5 pascal and initial volume can be taken as vi that's equal to 4 meter cube and final volume is given as 8 meter cube.
01:11
So delta v will be vf minus vi that's equal to 8 minus 4 that's equal to 4 meter cube.
01:22
So this is the change in volume and pressure is the so w is equal to 4 .8 into 10 raised to 5 into 4.
01:36
So that's equal to and it's equal to 1 .92 into 10 raised to 6 joule.
01:51
Therefore work done by the ideal gas is 1 .92 into 10 raised to 6 joule.
02:06
Now for the second part of the question, we have to calculate the change in internal energy.
02:15
So change in internal energy.
02:21
So that can be calculated by using the first law of thermodynamics.
02:27
So that's equal to delta u is equal to q minus w...