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.


<!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> JEE (MAIN)- 2020

Answer 1

$\text{Number of mole of He gas}=\frac{P\times V}{RT}$
$\text{Number of mole of He gas}=\frac{48\times {10}^{-3}\times \frac{4}{3}\pi (3cm{)}^3}{R\times T}$

$\text{Number of mole of He gas}=\frac{48\times {10}^{-3}\times \frac{4}{3}\pi \times 27}{R\times T}$

$\text{Number of mole of He gas}=\frac{P\times \frac{4}{3}\pi \times (12cm{)}^3}{R\times T}$

$\frac{48\times {10}^{-3}\times \frac{4}{3}\pi \times 27}{R\times T}=\frac{P\times \frac{4}{3}\pi \times (12cm{)}^3}{R\times T}$

$48\times {10}^{-3}\times 27=P\times (12cm{)}^3$
$P\times 144\times 12=48\times 9\times 3\times {10}^{-3}$
$P=\left(\frac{27}{36}\right)\times {10}^{-3}$
$P=750\times {10}^{-6}$ bar