A vertical cylinder of total lengh 100 cm is closed at the lower end and is fitt

Question

A vertical cylinder of total lengh 100 cm is closed at the lower end and is fitted with a movable frictionless gas-tight disc at the other end.an ideal gas is trapped under the disc.initially the height of the gas column was 90cm when the disc was in equilibrium between the gas and the atmosphere.mercury is now slowly poured on the top of the discand it just started overflowing wen the disc descended through 32cm. find the atmospheric presure .Assume the temperature of the gas constant and neglet the thickness and weight of disc?

Explanation / Answer

First of all note that the temperature of gas remains constant, so we have P1V1 = P2V2 Now initially the the pressure in atmospheric pressure only, so let it be P This is also the pressure of the gas. So, P1V1 = PA.(90) where A is the area of the cylinder When we start pouring the mercury the disc goes down by 32 cm leaving a space of 32 + 10 = 42 cm in the cylinder because the piston was at 90 cm initially. So now the total pressure from outside is P + 42 which is equal to pressure of the gas. So, P2V2 = ( P + 42 ).A.58 So using P1V1 = P2V2 we get PA.(90) = ( P + 42 ).A.58 and thus solving for P we get P = 76.125 cm of mercury.

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