Question

A spherical balloon of radius $3$ $\mathrm{cm}$ containing helium gas has a pressure of $48\times {10}^{-3}\mathrm{bar}$. At the same temperature, the pressure, of a spherical balloon of radius $12\mathrm{cm}$containing the same amount of gas will be __________$\times {10}^{-6}\mathrm{bar}.$

Answer 1

The pressure, of a spherical balloon of radius $12\mathrm{cm}$containing the same amount of gas will be $\mathbf{750}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{bar}\mathbf{.}$

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{in}\mathbf{}\mathbf{terms}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Pressure} \\ \mathrm{moles}=\frac{48\times {10}^{-3}\times {\displaystyle \frac{4}{3\mathrm{\pi }}}{\left(3\mathrm{cm}\right)}^3}{\mathrm{R}\times \mathrm{T}} \\ \mathrm{moles}=\frac{\mathrm{P}\times \frac{4}{3\mathrm{\pi }}{\left(12\mathrm{cm}\right)}^3}{\mathrm{R}\times \mathrm{T}} \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Pressure} \\ \mathrm{P}\times 144\times 12=48\times 9\times 3\times {10}^{-3} \\ \mathrm{P}=(27/36)\times {10}^{-3} \\ \mathbf{P}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{750}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{bar} \\ \mathrm{Hence},\mathrm{the}\mathrm{pressure}\mathrm{of}\mathrm{a}\mathrm{spherical}\mathrm{balloon}\mathrm{is}\mathbf{}\mathbf{750}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{bar}\mathbf{.} \end{gathered}$$