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, volume is increased by ∆V and pressure is increased by ∆P by a stroke.

Just before and after a stroke, we can write

`P_1V_1^γ = P_2V_2^γ`

⇒ `P(V + ∆V)^γ = (P + ∆P)V^γ`  .....(∵ Volume is fixed)

⇒ `PV^γ (1 + (∆V)/V)^γ = P(1 + (∆P)/P)V^γ`

⇒ `PV^γ (1 + γ (∆V)/V)^γ = PV^γ (1 + (∆P)/P)`  .....(∵ ∆V << V)

⇒ `γ (∆V)/V = (∆P)/P`

⇒ `∆V = 1/γ V/P ∆P`

⇒ `dV = 1/γ V/P dP`

Hence, work done is increasing the pressure from P₁ to P₂

`W = int_(P_1)^(P_2) PdV`

= `int_(P_1)^(P_2) P xx 1/γ V/P dP`

= `V/γ int_(P_1)^(P_2)`

= `V/γ (P_2 - P_1)`

⇒ `W = ((P_2 - P_1))/γ V`