Question

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 0.75 atm. 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?
(a) 0.75 atm, because increasing the partial pressure of ar decreases the partial pressure of n₂
(b) 1.13 atm, because 33% of the moles of gas are n₂
(c) 1.50 atm, because the number of moles of n₂ did not change.
(d) 2.25 atm, because doubling the number of moles of ar doubles its partial pressure.

Explanation:

Step1: Determine initial partial pressure of N₂

Since it's an equimolar mixture initially and partial - pressure is proportional to the number of moles for ideal gases in a mixture at constant T and V, the initial partial pressure of N₂ is equal to the initial partial pressure of Ar. So, $P_{N_2,i}=0.75$ atm.

Step2: Calculate new partial pressure of Ar

The number of moles of Ar is doubled. According to the ideal - gas law for partial pressures ($P = \frac{nRT}{V}$ at constant T and V, so $P\propto n$), the new partial pressure of Ar, $P_{Ar,f}=2\times0.75 = 1.5$ atm.

Step3: Calculate final total pressure

The partial pressure of N₂ remains the same ($P_{N_2,f}=P_{N_2,i}=0.75$ atm) as the number of moles of N₂ does not change. The total pressure of a gas mixture is the sum of the partial pressures of its components, $P_{total}=P_{N_2,f}+P_{Ar,f}$. So, $P_{total}=0.75 + 1.5=2.25$ atm.

Answer:

D. 2.25 atm, because doubling the number of moles of Ar doubles its partial pressure.