An equimolar mixture of N2 (g) and Ar (g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial pressure of Ar in the mixture is 0.75atm. An additional amount of Ar was added to the container, enough to double the number of moles of Ar gas in the mixture. Assuming ideal behavior, what is the final pressure of the gas mixture after the addition of the Ar gas?

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Chemistry

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24 January, 15:28

An equimolar mixture of N2 (g) and Ar (g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial pressure of Ar in the mixture is 0.75atm. An additional amount of Ar was added to the container, enough to double the number of moles of Ar gas in the mixture. Assuming ideal behavior, what is the final pressure of the gas mixture after the addition of the Ar gas?

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Tyrone Richard · Answer 1

The final pressure of the gas mixture after the addition of the Ar gas is P₂ = 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n. Therefore

Inicial state) P₁V=n₁RT

Final state) P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂ = P₁ * n₂/n₁

now we have to determine P₁ and n₂ / n₁.

For P₁, we use the Dalton's law, where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ / n₁ = 3/2

therefore

P₂ = P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm

P₂ = 2.25 atm