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eoneil

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## Homework Statement

If the cross-section of the barrel of a pump is 5 square centimeters, the stroke of the piston is 20 centimeters, and the volume of the receiver, in which the air is drawn from, is 1 liter, calculate the ratio of the pressure of the air in the receiver to the original pressure after three strokes of the pump.

## Homework Equations

p=F/a

p=hDg

p1V1 = p2V2

## The Attempt at a Solution

Trying to visualise the problem. Compare the pressure before to the pressure after. The piston could be going downward, drawing air from below. The piston sits on top of the gas, adding its weight to the pressure of the atmosphere pushing down on the gas. That gas, 1L (0.001m^3), would then exert hydrostatic pressure upward on the piston, at 10^5Pa.

p=F/a = (9.81m/s^2)/(0.01m^2)=981

Using p1V1=p2V2, solving for either pressure.

The volume of space the piston compresses= (0.05m^2)(0.20m)= 0.01m^3 (is it correct to assume that the cross-section of the pump, 0.05m^2, is incorporated into the volume calculation?)

The weight of the piston is unknown and the temperature doesn't factor in either. So no equation including those two variables is necessary.

The ratio involves only the volume of air displaced by the piston, and the volume of air in the receiver.

Using p1V1=p2V2, we can solve for p2.

p2=p1V1/V2=(10^5Pa)(0.01m^3)/(0.001m^3)= 14841Pa, this is the pressure the 1L of air in the receiver exerts on the piston.

Then the ratio of pressure would be 10^5Pa/14841Pa= 1/10th

Is this right? How do I incorporate 3 strokes of the piston?

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