Understanding Dalton's Law
Dalton's Law of Partial Pressure states that for a mixture of non-reacting gases in a container, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
Mathematically, it is expressed as: $$P_{total} = P_1 + P_2 + P_3 + ... + P_n$$
Where $P_i$ is the pressure that each gas would exert if it occupied the container alone at the same temperature.
Solving the Problem
Given:
- $V = 1.00 \text{ L}$
- $T = 25^{\circ}\text{C} = 298.15 \text{ K}$
- $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$
- Masses: $N_2 = 0.8904 \text{ g}$, $O_2 = 0.2741 \text{ g}$, $Ar = 0.0152 \text{ g}$, $CO_2 = 0.00107 \text{ g}$
Step 1: Calculate moles ($n = \text{mass} / \text{molar mass}$)
- $n_{N_2} = 0.8904 / 28.01 = 0.03179 \text{ mol}$
- $n_{O_2} = 0.2741 / 32.00 = 0.00857 \text{ mol}$
- $n_{Ar} = 0.0152 / 39.95 = 0.00038 \text{ mol}$
- $n_{CO_2} = 0.00107 / 44.01 = 0.00002 \text{ mol}$
Step 2: Calculate Partial Pressures ($P = nRT/V$)
Since $RT/V = (0.0821 \times 298.15) / 1.00 = 24.478$, we multiply each mole value by this constant:
- $P_{N_2} = 0.03179 \times 24.478 = 0.778 \text{ atm}$
- $P_{O_2} = 0.00857 \times 24.478 = 0.210 \text{ atm}$
- $P_{Ar} = 0.00038 \times 24.478 = 0.009 \text{ atm}$
- $P_{CO_2} = 0.00002 \times 24.478 = 0.0005 \text{ atm}$
Step 3: Total Pressure
$P_{total} = 0.778 + 0.210 + 0.009 + 0.0005 = 0.9975 \text{ atm}$