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

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.