A sample of argon gas occupies 100 L at 720. mmHg and 20°C. What will
be the temperature of the argon gas at 360. mm Hg and 2.14 L. assuming
the number of moles of gas remained constant?

T₂ =  3.135 K

Explanation:

Given data:

Initial volume = 100 L

Initial pressure = 720 mmHg

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

Final temperature = ?

Final volume = 2.14 L

Final pressure = 360 mmHg

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

P₁V₁/T₁ = P₂V₂/T₂  

T₂   = P₂V₂T₁ /P₁V₁

T₂ = 360 mmHg × 2.14 L × 293 K / 720 mmHg × 100 L  

T₂ = 225727.2 mmHg .L. K / 720 mmHg × 100 L  

T₂ =  3.135 K