Question

Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump ∆V (<< V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P₁ to P₂?

Answer 1

Let initial volume of air in tyre be V and after pumping one stroke it becomes (V + dP) and pressure increased from P to (P + dP).

Then,

P₁V₁^γ = P₂V₂^γ

P(V + dV)^γ= (P + dP)V^γ

PV^γ. \((1+\frac{dV}{V})^γ\) = \(P[1+\frac{dP}{P}]V^γ\)

Volume of tyre remains constant

 PV^γ. \([1+γ\frac{dV}{V}]\) = \(PV^γ[1+\frac{dP}{P}]\)

or, \( γ\frac{dV}{V}=\frac{dP}{P}\) 

or, \( {dV}=\frac{VdP}{γP}\)

or, \( {PdV}=\frac{VdP}{γ}\)

Integrating both sides,

\(\int{PdV}\)= \(\int^{p_2}_{p_1}\frac{VdP}{γ}\)

or, \(\int{dW}\) = \(\int^{p_2}_{p_1}\frac{VdP}{γ}\)  [V = constant]