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eduard
4 years ago
12

During an expansion process, the pressure of a gas changes from 15 to 140 psia according to the relation P = aV + b, where a = 5

psia/ft3 and b is a constant. If the initial volume of the gas is 7 ft3, calculate the work done during the process.
Engineering
1 answer:
Dominik [7]4 years ago
5 0

The work done during the process is 359 btu

<u></u>

<u>Explanation:</u>

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

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Answer:

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\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

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