The work done during the process is 359 btu
Step-by-step explanation:
Given-
P1 = 15psia
P2 = 140 psia
V1 = 7ft³
a = 5 psia/ft³
b = C
P = aV +b
Work done, W = ?
P1 = aV1 + b
15 = 5 (7) + b
b = -20 psia
P2 = aV2 + b
140 = 5 ( V2) - 20
V2 = 32 ft³
The work done by the process is the area under the curve which is trapezoidal.
Therefore,
Work done, W = area of trapezoid
= (P2 + P1 / 2) (V2 - V1)
= ( 140 + 15 / 2 ) ( 32 - 7)
= 1937.5 psia ft³
= 1937.5/ 5.4039 = 359 btu
Therefore, the work done during the process is 359 btu