When 1 g of an ideal gas A is introduced into an evacuated vessel at 300 K, the pressure was found to be 1 atm. Two grams of another ideal gas B is then added to A and the pressure is now found to be

Question

When 1 g of an ideal gas A is introduced into an evacuated vessel at 300 K, the pressure was found to be 1 atm. Two grams of another ideal gas B is then added to A and the pressure is now found to be 1.5 atm. What is the ratio between the average speeds of A and B at the same temperature?

Prashant Kotian · Accepted Answer

Correct answer: A. initial pressure p_a = 1 atm. final total pressure = 1.5 atm, so p_b = 1.5 - 1 = 0.5 atm. since v and t are constant, partial pressures are ∝ moles: n_a / n_b = p_a / p_b = 1 / 0.5 = 2. this gives: (1 / m_a) / (2 / m_b) = 2 → m_b / m_a = 4. since average speed is inversely ∝ √m: c_a / c_b = √(m_b / m_a) = √4 = 2. the ratio is 2:1.

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