2.
A
3). WW2
gas undergoes a cyclic process ABCDA
as shown in the figure. The part ABC of
process is semicircular. The work done by
the gas is
ro
P
uo
2 TO 14
)
60 Pa
- TT (20
2
A
200
20 Pa
C
loot
D
20m3
60m
1) 200 J 2) 100+ J 3) 150 7 J 4) 3007 J
VATOIO ROTIVE TURVATEDIA
Answers
Answered by
0
Answer:
question is incomplete.
Answered by
0
Answer:
Let P be the pressure of the gas in the cylinder, then the force exerted by the gas on the piston of the cylinder,
F=PA
In a small displacement of piston through dc, work done by the gas,
dW=F.dx=PAdx=PdV
∴ Total amount of work done ΔW=∫dW=∫
V
i
V
f
PdV=P(V
f
−V
i
)
In P−V diagram or indicator diagram, the area under P−V curve represents work done.
W= Area under P−V diagram
According to the P−V diagram given in the question,
Work done in the process ABCD= Area of rectangle ABCDA
=AB×BC=(3V
0
−V
0
)×(2P
0
−P
0
)
=2V
0
×P
0
=2P
0
V
0
Since, the cyclic process is anti clockwise, work done by the gas is negative.
That is, −2P
0
V
0
. Hence there is a net compression in the gas.
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