# A spherical balloon of radius

Question:

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

Solution:

(750)

At constant temperature and number of moles

$P_{1} V_{1}=P_{2} V_{2}$

$P_{1}=48 \times 10^{-3} \mathrm{bar} ; V_{1}=\frac{4}{3} \pi(3)^{3}$

$V_{2}=\frac{4}{3} \pi(12)^{3}$

$P_{2}=\frac{P_{1} V_{1}}{V_{2}}=\frac{48 \times 10^{-3} \times(3)^{3}}{(12)^{3}}$

$=\frac{48 \times 10^{-3}}{64}=7.5 \times 10^{-4}=750 \times 10^{-6} \mathrm{bar}$