1. PV diagram below shows an ideal gas undergoes an isobaric process. Calculate the work is done by the gas in the process AB.
Known :
Pressure (P) = 5 x 105 N/m2
Initial volume (V1) = 2 m3
Final volume (V2) = 6 m3
Wanted : Work (W)
Solution :
W = P (V2 – V1)
W = (5 x 105)(6 – 2) = (5 x 105) (4)
W = 20 x 105 = 2 x 106 Joule
[irp]
2. What is difference of the work is done by the gas in process AB and process CD…
Known :
Isobaric process AB :
Pressure (P) = 6 atm = 6 x 105 N/m2
Initial volume (V1) = 1 liter = 1 dm3 = 1 x 10-3 m3
Final volume (V2) = 3 liters = 3 dm3 = 3 x 10-3 m3
Isobaric process CD :
Pressure (P) = 4 atm = 4 x 105 N/m2
Initial volume (V1) = 2 liters = 2 dm3 = 2 x 10-3 m3
Final volume (V2) = 5 liters = 5 dm3 = 5 x 10-3 m3
Wanted : Difference of the work is done by the gas in process AB and CD.
Solution :
Work is done by the gas in process AB :
W = P (V2 – V1)
W = (6 x 105)(3 x 10-3 – 1 x 10-3)
W = (6 x 105)(2 x 10-3)
W = 12 x 102 = 1200 Joule
Work is done by the gas in process CD :
W = P (V2 – V1)
W = (4 x 105)(5 x 10-3 – 2 x 10-3)
W = (4 x 105)(3 x 10-3)
W = 12 x 102 = 1200 Joule
Difference of the work is done by the gas in process AB and CD = 1200 – 1200 = 0.
[irp]
3. Work is done by the gas in process ABC is….
Known :
Pressure 1 (P1) = 6 x 105 Pa = 6 x 105 N/m2
Pressure 2 (P2) = 3 x 105 Pa = 3 x 105 N/m2
Volume 1 (V1) = 2 cm3 = 2 x 10-6 m3
Volume 2 (V2) = 6 cm3 = 6 x 10-6 m3
Wanted : Work is done in process ABC.
Solution :
In process AB, the volume is kept constant so that no work is done by the gas.
Work was done by the gas in the process BC.
W = P2 (V2 – V1)
W = (3 x 105)(6 x 10-6 – 2 x 10-6)
W = (3 x 105)(4 x 10-6)
W = 12 x 10-1
W = 1.2 Joule
Work is done in the process ABC = work is done in the process AB = 1.2 Joule.
