If a gas of volume 6000 cm3 and at pressure of 100 kPa is compressed
quasi-statically according to pV2= constant until the volume becomes 2000 cm3,
determine the final pressure and the work transfer
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The final pressure is 600 Kpa and the amount of work done is W = 480 kJ
Explanation:
Given data:
- Volume V1 = 6000 cm^3
- Pressure = 100 kPa
- Volume V2 = 2000 cm^3
Solution:
The final pressure will be:
P1 V1 = P2 V2
P2 = P1 V1 / V2
P2 = 100 x 6000 / 2000
P2 = 600 Kpa
Work done =
[ Since 100 x (6000)^n = 600 x (2000)^n ]
(6000 / 2000)^n = 600 / 100
(3000)^n = 6
n x ln (3000) = ln 6
n x 8 = 1.79
n = 1.79 / 8 = 0.2
Now work done = ( P1 V1 - P2 V2 ) ( n - 1 )
W = (100 x 6000 - 600 x 2000) ( 0.2 - 1)
W = (600000 - 1200000) (-0.8)
W = - 600000 x - 0.8
W = 480, 000 J
W = 480 KJ
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