Consider a cycle tyre being filled


Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre and at each stroke of the pump ∆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?


Following is the equation before and after the stroke:

$P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$

$P(V+\Delta V)^{\gamma}=(P+\Delta P) V^{\gamma} \Rightarrow P V^{\gamma}\left(1+\frac{\Delta V}{V}\right)^{\gamma}=P\left(1+\frac{\Delta P}{P}\right) V^{\gamma}$

$P V^{\gamma}\left(1+\gamma \frac{\Delta V}{V}\right) \approx P V^{\gamma}\left(1+\frac{\Delta P}{P}\right)$

$\gamma \frac{\Delta V}{V}=\frac{\Delta P}{P}$

Therefore, work done is given as

$W=\frac{\left(P_{2}-P_{1}\right) V}{\gamma}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now