A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of length 16.8 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.

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P. S. I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure, 41.0 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 14.6959 P. S. I, we can use the ideal gas equation

P*V = n*R*T

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

P₁*V₁ = P₂*V₂ ⇒ V₂ = P₁*V₁ / P₂

where

1 and 2 are initial and final states respectively,

V₁ = A*L = (3.00 in²) * (16.8 in) ⇒ V₁ = 50.4 in³

P₁ = 14.6959 P. S. I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959 P. S. I) * (50.4 in³) / (55.6959 lb/in²)

⇒ V₂ = 13.2985 in³

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

h = V₂ / A = (13.2985 in³) / (3.00 in²)

⇒ h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in