# 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 m^{3 }

Known :

Pressure 1 (P_{1}) = 10 kPa = 10,000 Pa = 10,000 N/m^{2}

Pressure 2 (P_{2}) = 20 kPa = 20,000 Pa = 20,000 N/m^{2}

Volume 1 (V_{1}) = 4 liters = 4 (0.001 m^{3}) = 0.004 m^{3}

Volume 2 (V_{2}) = 10 liters = 10 (0.001 m^{3}) = 0.010 m^{3}

__Wanted :__ The work done by the gas (W)

__Solution :__

W = P (V_{2} – V_{1}) + 1/2 (P_{2} – P_{1})(V_{2} – V_{1})

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

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

Known :

Pressure 1 (P_{1}) = 1 x 10^{5} Pa = 1 x 10^{5} N/m^{2}

Pressure 2 (P_{2}) = 3 x 10^{5} Pa = 3 x 10^{5} N/m^{2}

Volume 1 (V_{1}) = 2 m^{3}

Volume 2 (V_{2}) = 4 m^{3}

__Wanted :__ Work (W)

__Solution :__

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

W = (P_{2} – P_{1})(V_{2} – V_{1})

W = (3 x 10^{5} – 1 x 10^{5})(4 – 2)

W = (2 x 10^{5})(2)

W = 4 x 10^{5} Joule

W = 400,000 Joule

W = 400 kJ

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

__Known :__

Pressure 1 (P_{1}) = 1 N/m^{2}

Pressure 2 (P_{2}) = 7 N/m^{2}

Volume 1 (V_{1}) = 1 m^{3}

Volume 2 (V_{2}) = 5 m^{3}

__Wanted :__ Work (W)

__Solution :__

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

W = ½ (P_{2} – P_{1})(V_{2} – V_{1})

W = ½ (7-1)(5-1)

W = ½ (6)(4)

W = ½ (24)

W = 12 Joule