# Isobaric thermodynamics processes – problems and solutions

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

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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.

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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.