A balloon is inflated with helium gas at room temperature of 25 °C a

1.	A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
Answer» Given : P₁ = Initial pressure = 1 bar V₁ = Initial volume = 2.27 L P₂= Final pressure = 0.3 bar To find : V₂ = Final volume Formula : P₁V₁ = P₂V₂ (at constant n and T) Calculation : According to Boyle’s law, P₁V₁ = P₂V₂ (at constant n and T) ∴ V₂ = \(\frac{P_1V_1}{P_2}\) = \(\frac{1\times 2.27}{0.3}\) = 7.566667 L ≈ 7.567 L ∴ The balloon can stay inflated below the volume of 7.567 L.

A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?

Answer»

Given :

P₁ = Initial pressure = 1 bar

V₁ = Initial volume = 2.27 L

P₂= Final pressure = 0.3 bar

To find :

V₂ = Final volume

Formula :

P₁V₁ = P₂V₂ (at constant n and T)

Calculation :

According to Boyle’s law,

P₁V₁ = P₂V₂ (at constant n and T)

∴ V₂ = \(\frac{P_1V_1}{P_2}\)

= \(\frac{1\times 2.27}{0.3}\)

= 7.566667 L ≈ 7.567 L

∴ The balloon can stay inflated below the volume of 7.567 L.