Equal weights of two non-reacting ideal gases with molecular weights and are mixed in a rigid contai — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Let the weight of each gas be $w$. Moles of lighter gas $n_1 = w/4$. Moles of heavier gas $n_2 = w/40$. Mole fraction of the lighter gas $\chi_1 = n_1 / (n_1 + n_2) = (w/4) / (w/4 + w/40) = (1/4) / (11/40) = 10/11$. By Dalton's law, $p_1 = \chi_1 P_{total} = (10/11) \times 1.1\text{ atm} = 1.0\text{ atm}$.