A gas at STP occupies 1.50 L. If the pressure drops to 0.75 atm and the temperature is raised to 300 K, what is the new volume?

 V₂ = 1.82 L

Explanation:

Given data:

Initial volume of gas = 1.50 L

Initial temperature = 273 K

Initial pressure = 1 atm

Final pressure = 0.75 atm

Final temperature = 300 K

Final volume = ?

Solution:

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 × 1.50 L × 273 K / 300 K ×0.75 atm

V₂ =  409.5 atm .L. K / 225 K.atm

 V₂ = 1.82 L

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