Question

When air is being pumped from a vessel of 9 litres capacity by a pump which has a barrel volume of 1 litre. If the pressure of the gas inside the vessel is 76 cm of Hg. What will be the pressure of air inside after the piston of the barrel makes two upward strokes (at each upward stroke the volume doubles and the pressure becomes half)?​

Answer 1

pressure might be one-fourth of the total pressure.

Answer 2

Heya mate

                   

Given:-

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 satisfy you..........................................