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 :
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
[irp]
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/m2
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
[irp]
3. Determine the work done by the gas in process A-B-C-A.
Known :
Pressure 1 (P1) = 1 N/m2
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
[irp]
