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 $\Delta V(<<V)$ of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $P_1\text{to}{P}_2$?

Answer 1

Step 1: Find relation between pressure and volume.

Formula Used: $P{V}^{\gamma }=\text{constant}$

The process is adiabatic, $\Delta Q=0$.

Let, pressure is increased by $\Delta P$ and volume is increased by $\Delta V$ at each storke.
For just before and after a storke,

By using adibatic process,

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$ $\because \gamma =\frac{C_p}{C_v}$

$\Rightarrow P(V+\Delta V{)}^{\gamma }=(P+\Delta P)V^{\gamma }$
$\because \text{volume is fixed}$

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

As $\Delta V<<V$, so by using binomial approximation we get,

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

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

Step 2: Find work done in process.

Work done in increasing the pressure from $P_1\text{to}{P}_2$ is


$W=\int P\Delta V=\frac{V}{\gamma }{\int }_{P_1}^{P_2}\Delta P=\frac{V}{\gamma }{\left|P\right|}_{P_1}^{P_2}$

$W=\frac{(P_2-P_1)V}{\gamma }$

Final Answer: $W=\frac{(P_2=P_1)V}{\gamma }$