A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1 MPa to 0.7 MPa for which pv = constant. The initial density of air is 1.16 kg/m3. Find the work done by the piston to compress the air.
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Answers
Answer:
Given:
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
ρ = 1.16 Kg/m3
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Work done by the piston = ?
For PV = C;
WD = P1*V1*log V2/V1
ρ = m/V ---> V1 = m/ρ = 1.5/1.16 = 1.293m3
W1–2 = P1V1logP1/P2
= 105 × 1.293 loge (105/7 × 105)
= 105 × 1.293 × (–1.9459) = –251606.18J
we can yet improve or simplify our answer by dviding the result by 1000
= –251.6 KJ ( work done on the system " is -ve")
Explanation: W.D ( work done ) will always be negative if the displacement is opposite to the direction of the Force applied.
E.g--> Work was done the gravity on a rocket going perpendicular upwards.
Answer:
work done (WD)= -251606.18 J (workdone on the system is -ve)
Explanation:
w=1.5 kg
P1 =0.1mpa=0.1×10^6=10^5 N/M^2
P2=0.7mpa=0.7×10^6=7×10^5 N/M^2
PV=constant
ρ=1.16kg/m^2
WD=?
