Air is being pumped from a vessel of 9 litre capacity by a pump which has a barrel volume of 1L. If the pressure of the air inside the vessel is 76 cm of Hg what will be the pressure of the air present inside, after the piston of the barrel makes two upwards stroke. (Hint: at each upward stroke, volume doubles and pressure halves.)

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Solution

Dear Student,
Volume of vessel = 9 L
Volume 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

____________________

Regards

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