A piston cylinder device initially contains air at 150 kpa...

A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the piston is resting on a pair of stops, as shown in the figure, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air. Air Vi = 400 L P1 = 150 kPa T1 = 27°C

Transcript

00:02 Hi, here in this given problem for the piston cylinder device which is containing air, its initial pressure, that is 150 kilo pascal, initial temperature, that is 27 degrees celsius or in kelvin that will be 27 plus 27 .23.

00:38 Means this is 300 kelvin and its initial volume was 400 liters which can also be given as 400 into 10 dash to power minus 3 meter cube or this is 0 .4 meter cube.

01:00 Now pressure necessary to move the piston is 350 kilo -pascals so it clearly means that the pressure inside the cylinder must have been increased to 350 kiosk from this 150 kiosk.

01:24 And volume has been doubled.

01:27 So v2 is twice of v1 means it will be 0 .8 meter cube.

01:36 Now in the first part of the problem we have to find temperature at which we will get this situation.

01:41 So using relation among the pressure and volume and temperature of an ideal gas together, p1 v1 by t1 is equal to p2 v2 v2 by t2 for pressure this is 150 kilo -pascal into volume 0 .4 meter cubed divided by temperature that was 300 kelvin.

02:08 Then p2 that is 350 kilo -pascal.

02:11 Volume 0 .8 divided by temperature which is missing and here this temperature is calculated to be equal to 1400 kelvin or in degree celsius just by subtracting 273 from it this is 1 ,127 degrees celsius which is the answer for the first part of this problem.

02:42 Now in the second part of the problem we have to find work done by the air that will be given by the change in volume multiplied by the pressure the final pressure.

03:04 So here it will be 0 .8 minus 0 .4 for pressure this is 350 kyl or 350 into 10 dash par 3 pascal and finally this work here comes out to be equal.

03:20 To 140 into 10 to bar 3 joules or we can say this is 140 kilojoules which becomes the answer for the second part of this problem.

03:38 Now in the third part of the problem we have to find heat input to this cylinder and to find it first of all we should have an idea about the mass of the air so to find the mass of the air in the cylinder, we use an ideal gas equation pv is equal to here this m, mass of the gas into rt.

04:15 So m means mass of the gas will be given by pv...

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