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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?
Solution
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`
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