A closed system consisting of 10 lb of air undergoes a polytropic process from p1 = 75 lbf/in2, v1 = 4 ft3/lb to a final state where p2 = 20 lbf/in2, v2 = 12 ft3/lb. Determine the polytropic exponent, n, and the amount of energy transfer by work, in Btu, for the process.

The polytropic exponent (n) is 1.2

The amount of energy transfer by work is 552.45 Btu

Explanation:

P1V1^n = P2V2^n

P1 = 75 lbf/in^2 = 75 lbf/in^2 × 143.3 in^2/ft^2 = 10,747.5 lbf/ft^2

V1 = 4 ft^3/lb × 10 lb = 40 ft^3

P2 = 20 lbf/in^2 = 20 × 143.3 = 2866 lbf/ft^2

V2 = 12 ft^3/lb × 10 lb = 120 ft^3

(V2/V1)^n = P1/P2

(120/40)^n = 10747.5/2866

3^n = 3.75

ln 3^n = ln 3.75

n ln 3 = ln 3.75

n = ln 3.75/ln 3 = 1.2

W = (P2V2 - P1V1)/1 - n = (2866×120 - 10747.5×40)/1-1.2 = (343,920 - 429,900)/-0.2 = -85,980/-0.2 = 429,900 lbf.ft = 429,900/778.17 Btu = 552.45 Btu