Question

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

Answer 1

As in both baloon amount of gas taken and temperature is same.

$\therefore P _{1} V _{1}= P _{2} V _{2}$ can be applied

$48 \times 10^{-3} \times \frac{4}{3} \pi(3)^{3}= P _{2} \times \frac{4}{3} \pi(12)^{3}$

$48 \times 10^{-3} \times 3^{3}= P _{2} \times\left(3^{3} \times 4^{3}\right)$

$\frac{48}{64} \times 10^{-3} bar = P _{2}$

$\left(\frac{48}{64} \times 1000\right) \times 10^{-6} bar = P _{2}$