Question

48 gms of o3 and 80gms of so3 are placed in a vessel and the total pressure is found to be 800mm.then the ratio of their partial pressure is

Answer 1

The ratio of the partial pressure of O₃ and SO₃ in the vessel is equal to 1.

Given:

Mass of O₃ present = 48 grams

Mass of SO₃ present = 80 grams

Total Pressure (P) = 800 mm

To Find:

The ratio of the partial pressures of O₃ and SO₃

Solution:

Number of moles of O₃ = 48/48 = 1 mole

Number of moles of SO₃ = 80/80 = 1 mole

According to Dalton's Law of Partial Pressures:

• In a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

• The partial pressure of each gas is directly proportional to its mole fraction in the mixture.

So, the Partial Pressure of a gas is equal to the product of the mole fraction of the gas in the mixture and the total pressure (P).

∴ Partial Pressure of O₃ = (Mole Fraction of O₃) × (Total Pressure ‘P’)

→ Partial Pressure of O₃ =  $(\frac{1}{1+1})800=(\frac{1}{2} )800=400mm$

→ Partial Pressure of SO₃ = (Mole Fraction of SO₃) × (Total Pressure ‘P’)

∴ Partial Pressure of SO₃ =$(\frac{1}{1+1})800=(\frac{1}{2} )800=400mm$

→ The partial pressures of O₃ and SO₃ are same and equal to 400 mm.

Therefore the ratio of the partial pressures of O₃ and SO₃ is equal to 1.

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