Question

0.740 moles of a gas occupies 20.0 L at 700.0 mm Hg. What will the volume be when four moles of gas are added to the container and the new pressure is 800.0 mm Hg. Assume no change in temperature

Answer 1

n₁ = 0.740 moles

n₂ = 0.740 moles + 4.000 moles = 4.740 moles

P₁ = 700.0 mm Hg

V₁ = 20.0 L

P₂ = 800.0 mm Hg

V₂ = unknown

T₁ = T₂ = T = const

Solution:

The Ideal Gas equation can be used.

The Ideal Gas equation can be expressed as: PV = nRT

(1): P₁V₁ = n₁RT

(2): P₂V₂ = n₂RT

Combine equations (1) and (2):

RT = P₁V₁ / n₁ = P₂V₂ / n₂

P₁V₁n₂ = P₂V₂n₁

To find the final volume, solve the equation for V₂:

V₂ = P₁V₁n₂ / P₂n₁

V₂ = (700.0 mm Hg × 20.0 L × 4.740 mol) / (800.0 mm Hg × 0.740 mol) = 112.095 L = 112 L

V₂ = 112 L

Answer: the volume will be 112 liters.