Tyre of a bicycle has volume 2 × 10 - 3   m 3 . Initially the tube is filled 75 % of its volume by…

Given, 

Initial volume, $V_0=\left(\frac{75}{100}\right)\times 2\times {10}^{-3} {\mathrm{m}}^3=1.5\times {10}^{-3} {\mathrm{m}}^3$

and net pressure on tyre is, 

$P=\frac{mg}{A}+P_0=\frac{120\times 10}{24\times {10}^{-4}}+{10}^5=6\times {10}^5 \mathrm{N} {\mathrm{m}}^{-2}$

Using the ideal gas equation,

$$\begin{gathered} PV=nRT \\ \Rightarrow n=\frac{PV}{RT}=\frac{6\times {10}^5\times 2\times {10}^{-3}}{RT}...\left(1\right) \end{gathered}$$

again using the ideal gas equation,

$P_0{V}_1=nRT$, from equation $\left(1\right)$,

$$\begin{gathered} {10}^5\times V_1=6\times {10}^5\times 2\times {10}^{-3} \\ \Rightarrow V_1=12\times {10}^{-3} {\mathrm{m}}^3 \end{gathered}$$

So change in volume is,

$∆V=\left(V_1-V_0\right)=\left(12-1.5\right)\times {10}^{-3}=10.5\times {10}^{-3} {\mathrm{m}}^3$

So total air stokes,

$=\frac{10.5\times {10}^{-3}}{0.5\times {10}^{-3}}=21$