On the surface of the earth at $$1$$ atm pressure, a balloon filled with $$H_2$$ gas occupies $$500$$ mL. This volume is $$5/6$$ of its maximum stretching capacity. The balloon is left in air. It starts rising. Calculate the height above which the balloon will burst if the temperature of the atmosphere remains constant and the pressure decreases $$1$$ mm for every $$100$$-cm rise height.
From the Boyle's Law,
P₁V₁ = P₂V₂
Given, P₁ = 1 atm, V₁ = 500 mL and V₂ = 600 mL (500 x $$ \frac{6}{5}$$)
∴ P₂ = $$ \dfrac{P_1V_1}{V_2}=\dfrac{1\times 500}{600}=\dfrac{5}{6} $$ atm
or 633.33 mm of Hg.
Pressure difference = (760 - 633.33) = 126.67 mm