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?

TestHub · Accepted Answer

B contributes p B = 1.5 - 1 = 0.5 atm. Same V, T, so n A : n B = p A : p B = 1 : 0.5 = 2:1. With n A = 1/M A and n B = 2/M B : (1/M A )/(2/M B ) = 2 ⇒ M B = 4M A . Then u av,A : u av,B = √(M B /M A ) = √(4) = 2:1. Confirms Ans (a) 2:1.

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