If an 85.0 mL container of helium gas at standard pressure is heated from 20 C to 91 C and the pressure is increased to 2.8 atm then what would the new volume be for the He gas?

V₂ = 37.71 mL

Explanation:

Given data:

Initial volume = 85.0 mL

Initial pressure = 1 atm

Initial temperature = 20°C (20 +273 = 293 K)

Final temperature = 91°C (91+273 = 364 K)

Final volume = ?

Final pressure = 2.8 atm

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× 85.0 mL × 364 K / 293 K × 2.8 atm

V₂ = 30940 atm.mL.K / 820.4 K.atm

V₂ = 37.71 mL

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