Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. What is the relationship

Question

Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. What is the relationship between their molecular masses?

Prashant Kotian · Accepted Answer

Correct answer: C. Partial pressure of B is P_B = P_total - P_A = 3 - 2 = 1 bar. Since P = nRT/V, P_A / P_B = n_A / n_B => 2 / 1 = (1 / M_A) / (2 / M_B) = M_B / (2 * M_A) => M_B = 4 * M_A.

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