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 dа ta:

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

Solution:

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