Answer:
A.) P = 2bar, W = - 12kJ
B.) P = 0.8 bar, W = - 7.3 kJ
C.) P = 0.608 bar, W = - 6.4kJ
Explanation: Given that the relation between pressure and volume is
PV^n = constant.
That is, P1V1^n = P2V2^n
P1 = P2 × ( V2/V1 )^n
If the initial volume V1 = 0.1 m3,
the final volume V2 = 0.04 m3, and
the final pressure P2 = 2 bar.
A.) When n = 0
Substitute all the parameters into the formula
(V2/V1)^0 = 1
Therefore, P2 = P1 = 2 bar
Work = ∫ PdV = constant × dV
Work = 2 × 10^5 × [ 0.04 - 0.1 ]
Work = 200000 × - 0.06
Work = - 12000J
Work = - 12 kJ
B.) When n = 1
P1 = 2 × (0.04/0.1)^1
P1 = 2 × 0.4 = 0.8 bar
Work = ∫ PdV = constant × ∫dV/V
Work = P1V1 × ln ( V2/V1 )
Work = 0.8 ×10^5 × 0.1 × ln 0.4
Work = - 7330.3J
Work = -7.33 kJ
C.) When n = 1.3
P1 = 2 × (0.04/0.1)^1.3
P1 = 0.6077 bar
Work = ∫ PdV
Work = (P2V2 - P1V1)/ ( 1 - 1.3 )
Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )
Work = (8000 - 6080)/ -0.3
Work = -1920/0.3
Work = -6400 J
Work = -6.4 kJ
