Question

A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 65.0 lb/in². If the pump is a cylinder of length 19.2 in. with a cross-sectional area of 3.00 in.², how far down must the piston be pushed before air will flow into the tire? Assume the temperature is constant.

_____ inches

Answer 1

Pressure inside the bicycle = 65.0 lb/in^2 = 65.0 psi

Now, in order that air flows into the tire, the pressure in the pump must be more than the tire pressure, 65.0 psi.

Assume ideal gas condition and temperature of the compressed gas as constant.

Take 1 atmospheric pressure = 14.7 psi

Use ideal gas equation -

P*V = n*R*T

On the right hand side, T is constant so all the terms are constant.

Therefore,

P1*V1 = P2*V2

=> 14.7 * (19.2 x 3.0) = (14.7 + 65)*V2

=> V2 = (14.7 * 19.2 * 3.0) / 79.7 = 10.62 in^3

Therefore, length of piston measured from the bottom, this volume corresponds to is,

10.62 in^3 / 3.00 in^2 = 3.54 in.

So, the piston must be pushed down by a distance of,

19.2 - 3.54 = 15.66 inch (Answer)