Question

a rigid container holds 0.50 mol of ar(g) and 0.75 mol of o₂(g). how will the total pressure of the gases in the container and the partial pressure of o₂(g) be affected by the addition of 0.25 mol of he(g) at constant temperature? a both the total pressure and the partial pressure of o₂(g) will remain constant. b both the total pressure and the partial pressure of o₂(g) will increase. c the total pressure will increase, and the partial pressure of o₂(g) will decrease. d the total pressure will increase, and the partial pressure of o₂(g) will remain constant.

Explanation:

Step1: Recall Dalton's law of partial pressures

According to Dalton's law, the total pressure $P_{total}$ of a gas - mixture is the sum of the partial pressures of all the component gases, $P_{total}=P_1 + P_2+\cdots+P_n$. Also, the partial pressure of a gas $i$ in a mixture is given by $P_i = X_iP_{total}$, where $X_i=\frac{n_i}{n_{total}}$ is the mole - fraction of gas $i$, and $n_{total}$ is the total number of moles of all gases in the mixture.

Step2: Analyze the effect on total pressure

Initially, the total number of moles of gases in the container is $n_{total1}=n_{Ar}+n_{O_2}=0.50\ mol + 0.75\ mol=1.25\ mol$. When $0.25\ mol$ of $He$ is added, the new total number of moles is $n_{total2}=n_{total1}+n_{He}=1.25\ mol+ 0.25\ mol = 1.50\ mol$. At constant volume and temperature (since the container is rigid and temperature is constant), according to the ideal gas law $PV = nRT$ ($V$, $T$, and $R$ are constant), $P\propto n$. So, as the number of moles increases, the total pressure increases.

Step3: Analyze the effect on the partial pressure of $O_2$

The mole - fraction of $O_2$ initially is $X_{O_21}=\frac{n_{O_2}}{n_{total1}}=\frac{0.75\ mol}{1.25\ mol}=0.6$. After adding $He$, the mole - fraction of $O_2$ is $X_{O_22}=\frac{n_{O_2}}{n_{total2}}=\frac{0.75\ mol}{1.50\ mol}=0.5$. However, the partial pressure of $O_2$ is $P_{O_2}=X_{O_2}P_{total}$. Since $X_{O_2}$ changes and $P_{total}$ changes in such a way that $P_{O_2}=\frac{n_{O_2}}{n_{total}}P_{total}$, and $n_{O_2}$ remains constant, and using the ideal - gas law relationships, the partial pressure of $O_2$ is given by $P_{O_2}=\frac{n_{O_2}RT}{V}$ (because $P_{O_2}=X_{O_2}P_{total}$ and $P_{total}=\frac{n_{total}RT}{V}$, so $P_{O_2}=\frac{n_{O_2}}{n_{total}}\times\frac{n_{total}RT}{V}=\frac{n_{O_2}RT}{V}$). Since $n_{O_2}$, $R$, $T$, and $V$ are all constant, the partial pressure of $O_2$ remains the same.

Answer:

D. The total pressure will increase, and the partial pressure of $O_2(g)$ will remain constant.