Question

Three glass bulbs, joined by closed stopcocks, have the following volumes and initial pressures of the specified gases. Bulb A: 150. mL of CO(g) at 190. torr Bulb B: 300. mL of Ar(g) at 0.500 atm Bulb C: 750. mL of Kr(g) at 75.994 kPa 1. After both stopcocks are opened and the gases allowed to diffuse throughout, what will be the ultimate total pressure?2. What is the partial pressure of CO(g)?
3. What is the mole fraction of CO₂(g)?

Answer 1

Answer:

1. Total pressure is 475 torr.

2. The partial pressure of CO is 23.8 torr.

3. The mole fraction of CO is 0.0501.

Explanation:

We have 3 gases in different bulbs. Once the stopcocks are opened, they share the same final volume which is the sum of all individual volumes.

V = Va + Vb + Vc = 150 mL + 300 mL + 750 mL = 1200 mL

Since we know initial pressures and volumes for each gas, we can find the final pressures using Boyle's Law. The mathematical expression is

P₁ . V₁ = P₂ . V₂

We assume that temperature remains constant and that gases behave as ideal gases.

CO

P₁ = 190 torr; V₁ = 150 mL; P₂ = ?; V₂ = 1200 mL

P₁ . V₁ = P₂ . V₂

190 torr . 150 mL = P₂ . 1200 mL

P₂ = 23.8 torr

Ar

P₁ = 0.500 atm; V₁ = 300 mL; P₂ = ?; V₂ = 1200 mL

P₁ . V₁ = P₂ . V₂

0.500 atm . 300 mL = P₂ . 1200 mL

P₂ = 0.125 atm

$P_{2}=0.125atm.\frac{760torr}{1atm} =95.0torr$

Kr

P₁ = 75.994 kPa ; V₁ = 750 mL; P₂ = ?; V₂ = 1200 mL

P₁ . V₁ = P₂ . V₂

75.994 kPa . 750 mL = P₂ . 1200 mL

P₂ = 47.5 kPa

$P_{2}=47.5kPa.\frac{7.50torr}{1kPa} =356torr$

The total pressure is the sum of partial pressures.

P = P(CO) + P(Ar) + P(Kr) = 23.8 torr + 95.0 torr + 356 torr = 475 torr

We can find the mole fraction of of CO using the following expression, based on Dalton's Law:

P(CO) = P . X(CO)

where,

X(CO) is the mole fraction of CO

Then,

X(CO) = P(CO)/P = 23.8 torr / 475 torr = 0.0501.