Question

A hand-driven tire pump has a 2.00 cm diameter piston and a maximum stroke of 32.0 cm.

(a) How much work do you do in one stroke if the average gauge pressure is 2.40 ✕ 10⁵ N/m² (about 35 psi)? (You may consider this an isothermal process.)
J
(b) What average force do you exert on the piston, neglecting friction and gravity?
N

Answer 1

given that :

radius of the tire pump piston, r = 1 cm = 0.01 m

maximum stroke displacement, d = 32 cm = 0.32 m

(a) For an isothermal process, work done in one stroke which is given as ::

W = P V                             { eq. 1 }

where, V = A . d

A - area of the tire pump piston

V = r² d { eq. 2 }

inserting the values in eq.2,

V = (3.14) (0.01 m)² (0.32 m) = 0.00010048 m³ = 1.0048 x 10^-4 m³

using eq.1, W = P V       

where, P = average gauge pressure = 2.4 x 10⁵ N/m²

inserting the values in eq.1,

W = (2.4 x 10⁵ N/m²) (1.0048 x 10^-4 m³)

W = 24.115 J

(b) neglecting friction and gravity, the average force which exert on the piston will be given as ::

F_avg = W / d                          { eq. 3 }

inserting the values in eq.3,

F_avg = (24.115 J) / (0.32 m)

F_avg = 75.3 N