Equal weights of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1. — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution: Partial Pressure of Light Gas **Step 1: Find the number of moles of each gas** Let equal weight = $w$ grams For gas 1 (M₁ = 4): $$n_1 = \frac{w}{4}$$ For gas 2 (M₂ = 40): $$n_2 = \frac{w}{40}$$ **Step 2: Calculate mole fraction of light gas** Total moles: $$n_{total} = \frac{w}{4} + \frac{w}{40} = \frac{10w + w}{40} = \frac{11w}{40}$$ Mole fraction of light gas: $$\chi_1 = \frac{n_1}{n_{total}} = \frac{w/4}{11w/40} = \frac{w}{4} \times \frac{40}{11w} = \frac{10}{11}$$ **Step 3: Apply Dalton's Law** Partial pressure = Mole fraction × Total pressure $$P_1 = \chi_1 \times P_{total} = \frac{10}{11} \times 1.1 \text{ atm}$$ $$P_1 = \frac{10}{11} \times \frac{11}{10} = 1.0 \text{ atm}$$ **Answer: The partial pressure of the light gas is 1.0 atm** The light gas (M = 4) has the larger number of moles, so it contributes more to the total pressure.