On the surface of the earth 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 the temperature of the atmosphere remains constant and the pressure decreases 1 mm every 100-cm rise height-

Question

Toppr Admin · Accepted Answer

From the Boyle-s Law-P-8321-V-8321- - P-8322-V-8322-Given-160-P-8321- - 1 atm- V-8321- - 500 mL and-160-V-8322- - 600 mL -500 x - -frac-6-5-8756-160-P-8322- - - -dfrac-P-1V-1-V-2-dfrac-1-times 500-600-dfrac-5-6- - atm-160-or 633-33 mm of Hg-Pressure difference - -760 - 633-33-160- 126-67 mm

From the Boyle's Law,P₁V₁ = P₂V₂Given, 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

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