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A mass of 1.5kg of air is compressed in a quasi static process from 0.1Mpa to 0.7Mpa for which PV = constant. The initial density of air is 1.16kg/m3. Find the work done by the piston to compress the air. |
Answer» Given data: m = 1.5kg P1 = 0.1MPa = 0.1 × 106 Pa = 105 N/m2 P2 = 0.7MPa = 0.7 × 106 Pa = 7 × 105 N/m2 PV = c or Temp is constant ρ = 1.16 Kg/m3 W.D. by the piston = ? For PV = C; WD = P1V1log V2/V1 or P1V1logP1/P ρ = m/V; i.e.; V1 = m/ρ = 1.5/1.16 = 1.293m3 ...(i) W1–2 = P1V1logP1/P2 = 105 × 1.293 loge (105/7 × 105) = 105 × 1.293 × (–1.9459) = –251606.18J = –251.6 KJ (–ive means WD on the system) – 251.6K |
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