A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 40.0 psi. If the pump is a cylinder of length 18.0 in. with a cross-sectional area of 3.00 in2 . , how far down must the piston be pushed before air will flow into the tire?

1 Answer
Mar 12, 2018

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure, #40.0\ "psi"#.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be #14.7\ "psi"#, we use the ideal gas equation

# PV=nRT#.

As number of moles of air do not change during its compression in the pump, RHS of the gas equation is constant. Therefore we have

#P_1V_1=P_2V_2#
where #1 and 2# are initial and final states respectively.

Inserting various values we get

#14.7xx(18.0xx3.00)=(14.7+40)V_2#
#=>V_2=(14.7xx(18.0xx3.00))/(14.7+40)#
#=>V_2=14.512\ "in"^3#

Length of pump, measured from bottom, this volume corresponds to is

#14.512/3.00=4.8\ "in"#

Piston must be pushed down by more than

#18.0-4.8=13.2\ "in"#