Answer to Question #156511 in Mechanical Engineering for Rodel

Question

A piston-cylinder arrangement contains a gas at P₁ = 600 KPa, V₁ = 0.10m³ and expand to V₂ = 0.5 m³. Compute the work done by the gas if the process follows the relation (a) pV^1.4 = c, and (b) p = -300V + 630, where p is in KPa and V in m³.

Answer 1

a)

PV^1.4 =c

V =1.4

C= p₁V₁^r

=600 *(0.1)^1.4

C=23.886

work done w = c(V₂^1-r-V₁^1.r)/(1-r)

=23.886(0.5^(1-1.4)-0.10^(1-1.4))/1-1.4

answer = 71.203 kg

b)p- -300V + 630

w=$\int$pdv

=$\int$( -300V + 630)dv.................. from 0.1 to 0.5

=[(-300V²/2)/2 +630V]................. from 0.1 to 0.5

=[(-300/2)(0.5²-0.1²)/2 +630(0.5-0.1)]

answer w=216 kg