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: Determine mole ratio** For equal weights $w$: - Moles of light gas (MW = 4): $n_1 = \frac{w}{4}$ - Moles of heavy gas (MW = 40): $n_2 = \frac{w}{40}$ **Step 2: Calculate mole fraction** $$x_1 = \frac{n_1}{n_1 + n_2} = \frac{\frac{w}{4}}{\frac{w}{4} + \frac{w}{40}}$$ $$x_1 = \frac{\frac{w}{4}}{\frac{10w + w}{40}} = \frac{\frac{w}{4}}{\frac{11w}{40}} = \frac{w}{4} \times \frac{40}{11w} = \frac{10}{11}$$ **Step 3: Apply Dalton's Law** Partial pressure of light gas: $$P_1 = x_1 \times P_{total} = \frac{10}{11} \times 1.1 \text{ atm}$$ $$P_1 = \frac{10}{11} \times \frac{11}{10} = 1 \text{ atm}$$ **Answer: The partial pressure of the light gas is 1.0 atm** (Option C) The light gas comprises 10/11 of all moles, so it contributes 10/11 of the total pressure.