Question

Air is being pumped from a vessel of 9 litre capacity by a pump which has a barrel volume of 1 litre if the pressure of the air inside the vessel is 76 cm hg what will be the pressure of the year present air after the piston of the barrel makes two stroke upwards

Answer 1

Hello ,


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Capacity of vessel = 9 L
Capacity of barrel  = 1 L

​When the piston is lifted for first time, the air flows from vessel to the barrel and volume occupied by air will be total 10 L.

According to Boyle's law new pressure will be:

P1V1 = P2V2
76×9 = P2×10

New pressure, P2 = 68.4 cm of Hg

Now when the piston is lowered, the air from barrel is released. But pressure inside the vessel will remain same. When piston makes the 2nd upward stroke, the air from vessel travels to the barrel and volume occupied by air will be again 10L.

Now new pressure will be:
P1V1 = P2V2
68.4×9 = P2×10

New pressure after the piston of the barrel makes 2 upward strokes,

P2 = 61.56 cm of Hg

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Hope it helps u! ! !

# Nikky