The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :

Solution

Physics sir · 3 steps

Pressure depends on the number of moles, not the grams, so turn each gas into moles first.

1. Dalton's law. In a mixture each gas pushes on the walls as if it were alone, and the partial pressures add. So the total pressure is P = nₜₒₜₐₗRT/V, with nₜₒₜₐₗ the sum of all the moles.

   Gas	Mass	Molar mass	Moles n = m/M
   O₂	16 g	32 g mol⁻¹	0.5
   N₂	28 g	28 g mol⁻¹	1
   CO₂	44 g	44 g mol⁻¹	1

2. Add the moles. nₜₒₜₐₗ = 0.5 + 1 + 1 = 2.5 = 5/2. Oxygen is diatomic, so 16 g is only half a mole.

   P = nₜₒₜₐₗRT/V = 5/2 RT/V

3. So. The pressure of the mixture is 5/2RT/V, which is option (d).

The idea to keep. For any gas mixture, convert every mass to moles using the molecular (not atomic) molar mass, add the moles, and use PV = nRT once.

Answer from the NTA final answer key. Solution by PrepWiser.