Question 2.3: Air is compressed inside the inner tube of a bike using a ma...

Air is compressed inside the inner tube of a bike using a manual bicycle pump. The handle of the pump is brought down from an initial position x_2 to a final position x_1 where x_1 < x_2 and the norm of the force is assumed to be given by,

F (x) = F_{max} \frac{x_2 − x}{x_2 − x_1}.

The process is assumed to be reversible and the cylinder of the pump has a cross section A. Determine in terms of the atmospheric pressure p_0,
a) the work W_h performed by the hand on the handle of the pump,
b) the pressure p (x),
c) the work W_{12} performed on the system according to relation (2.42).

W_{if}=\int_{i}^{f}{\delta W} =-\int_{V_i}^{V_f}{p(S,V)}dV     (reversible process)
Numerical Application:
F_{max} = 10 N, x_1 = 20 cm, x_2 = 40 cm, A = 20 cm^2 and p_0 = 10^5 Pa.

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a) The work performed by the hand on the handle of the pump is,

W_h = -\int_{V_2}^{V_1}{p(V)dV} =-\int_{x_2}^{x_1}{F(x)dx} =- \frac{F_{max}}{x_2-x_1} \int_{x_2}^{x_1}{(x_2-x)} dx

= – \biggl(\frac{F_{max}}{x_2-x_1}\biggr) \biggl(x_2(x_1-x_2)-\frac{1}{2}(x^2_1-x^2_2) \biggr) = \frac{1}{2} F_{max}(x_2-x_1) = 1 J.

b) The net pressure p (x) exerted on the air inside the inner tube of the bike when the handle of the pump is at position x is the sum of the atmospheric pressure p_0 and the force F (x) exerted by the piston on the air divided by the surface A,

p(x) = p_0 +\frac{F(x)}{A} =p_0 + \frac{F_{max}}{A} \frac{x_2-x}{x_2-x_1} = \bigl(10^5+5\cdot 10^3(2-5x)\bigr)N m^{-2}.

c) The work performed on the system by the atmosphere and by the person pushing the handle of the pump is,

W_{12} = -\int_{x_2}^{x_1}{p_0 A dx} -\int_{x_2}^{x_1}{p(x)A dx} = p_0 A(x_2 − x_1) + W_h = 41 J.

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