00:01
So in this question, we have an ideal gas enclosed in the cylinder with boom -weval piston.
00:05
The walls of the cylinder are insulated so heat cannot enter or exits.
00:10
So q is equal to zero and gas initially occupies volume v1, having pressure p1 and temperature t1.
00:17
And now the volume, the piston is moved so that the volume v2 has become equals to 3v1.
00:24
V2 equals to 3v1.
00:25
The process happens so rapidly that the enclosed gas do not do.
00:30
Any work.
00:30
So we have w equals to 0 also.
00:33
Okay.
00:33
So we have to find p2, t2 and change in entropy of the case.
00:38
So since q equals to 0 and w equals to 0, so we can say that from the first law of thermodynamics, q that is equals to change in internal energy plus work done.
00:50
So q and w both are 0.
00:52
So from here we can say that change in internal energy, this will be equals to 0.
00:56
And if change in internal energy is 0, so we can say that temperature, temperature is constant, so we can write that t2, this will become equals to t1.
01:06
So this is the answer for the temperature t2.
01:09
Now since the temperature is constant, so from the ideal gas equation, pv equals to nrt, we can write that p and v both are inversely proportional...