A spherical weather balloon has volume of 12.0 L when inflated in STP. Assuming it could stay anchored and intact as a hurricane passed overhead lowering the pressure to 720.0 mm Hg what would be the new volume of the balloon assuming the temperature stays at 0 C?

V₂ = 12.63 L

Explanation:

Given data:

Initial volume of balloon = 12.0 L

Initial temperature = standard =  273 K

Initial pressure = standard = 1 atm

Final pressure = 720.0 mmHg (720/760 = 0.95 atm)

Final volume = ?

Final temperature = 0°C ( 273 k)

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 12.0 L × 273 K / 273 K × 0.95 atm

V₂ = 3276 atm .L. K / 259.35 k.atm

V₂ = 12.63 L

According to Avagadros law 1 mole of a substance contains 6.02 × 10^23 particles.
Thus, 1 mole of C6H14 contains 6.02 × 10^ 23 molecules,
Therefore, 9.25 × 10^24 molecules will have; 
      (9.25 × 10^24)/ (6.02 ×10^23) 
   =  15.365 moles.