Work done in thermodynamics process – problems and solutions

1. An ideal gas is compressed from state 1 to state 2 as shown in the graph below. Determine the work done by the gas.

Solution

1 liter = 0.001 m3

Known :Work done in thermodynamics process – problems and solutions 1

Pressure 1 (P1) = 10 kPa = 10,000 Pa = 10,000 N/m2

Pressure 2 (P2) = 20 kPa = 20,000 Pa = 20,000 N/m2

Volume 1 (V1) = 4 liters = 4 (0.001 m3) = 0.004 m3

Volume 2 (V2) = 10 liters = 10 (0.001 m3) = 0.010 m3

Wanted : The work done by the gas (W)

Solution :

W = P (V2 – V1) + 1/2 (P2 – P1)(V2 – V1)

W = 10,000 (0.010-0.004) + 1/2 (20,000-10,000)(0.010-0.004)

W = 10,000 (0.006) + 1/2 (10,000)(0.006)

W = 10 (6) + 1/2 (10)(6)

W = 60 + 1/2 (60)

W = 60 + 30

W = 90 Joule

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2. Determine the work done by the gas in process a-b-c-d-a, as shown in graph below.

Known :

Pressure 1 (P1) = 1 x 105 Pa = 1 x 105 N/m2Work done in thermodynamics process – problems and solutions 2

Pressure 2 (P2) = 3 x 105 Pa = 3 x 105 N/m2

Volume 1 (V1) = 2 m3

Volume 2 (V2) = 4 m3

Wanted : Work (W)

Solution :

The work done by the gas = area of rectangular a-b-c-d.

W = (P2 – P1)(V2 – V1)

W = (3 x 105 – 1 x 105)(4 – 2)

W = (2 x 105)(2)

W = 4 x 105 Joule

W = 400,000 Joule

W = 400 kJ

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3. Determine the work done by the gas in process A-B-C-A.

Known :

Pressure 1 (P1) = 1 N/m2Work done in thermodynamics process – problems and solutions 3

Pressure 2 (P2) = 7 N/m2

Volume 1 (V1) = 1 m3

Volume 2 (V2) = 5 m3

Wanted : Work (W)

Solution :

The work done by the gas = area of triangle A-B-C

W = ½ (P2 – P1)(V2 – V1)

W = ½ (7-1)(5-1)

W = ½ (6)(4)

W = ½ (24)

W = 12 Joule

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