# Answer to Question #156508 in Mechanical Engineering for Rodel

Question #156508

Determine the work done by a 1 Kgm fluid system as it expand slowly in a closed system from P1 = Mpa, V1 = 0.08 m3 to V2 = 0.22 m3 in accordance to the defining relations: (a) p = c, (b) pV = c, (c) pV3 = c, (d) pV (lnV) = c and (e) pV1.4 = c. Express your answer in calories.

1
2021-01-20T04:57:24-0500

(a) p = c

p1 = 1 MPa = 106 Pa

V1 = 0.08 m3

V2 = 0.22 m3

W = pdV = p dV = p(V2 - V1)

W = 106 x (0.22 - 0.08) = 0.14 x 106 J = 140 kJ = 33461 cal

(b) pV = c

W = pdV = p1V1/V dV = p1V1 dV/V = p1V1lnV2/V1

W = 106 Pa x 0.08 m3 x ln (0.22 m3/0.08 m3) = 81 kJ = 19360 cal

(c) pV3 = c

W = pdV = p1V1n dV/Vn = (p2V2 - p1V1) / (1 - n)

c = p1V13 = 106 Pa x (0.08 m3)3 = 512 Pa/m9

p2 = c/V23 = 512 / (0.22)3 = 48 084 Pa = 48 kPa

W = (48 084 x 0.22 - 106 x 0.08) / (1 - 3) = 34 711 J = 34.7 kJ = 8296 cal

(e)pV1.4 = c

W = pdV = p1V1n dV/Vn = (p2V2 - p1V1) / (1 - n)

c = p1V11.4 = 106 Pa x (0.08 m3)1.4 = 29 129 Pa/m9

p2 = c/V23 = 29 129 / (0.22)1.4 = 242 742 Pa = 243 kPa

W = (242 742 x 0.22 - 106 x 0.08) / (1 - 1.4) = 66 492 J = 66.5 kJ = 15892 cal

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Assignment Expert
09.02.21, 14:28

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Rodel
21.01.21, 01:50

Thank you for the solution! It means a lot :)