Multiple choice

A 3 : 2 molar mixture of $N_2$, and CO is present in a vessel at 500 bar pressure. Due to hole in the vessel, the gas mixture leaks out. The composition of mixture effusing out initially is :

  1. $n_{N_2} : n_{CO} : : 1 : 2$
  2. $n_{N_2} : n_{CO} : : 6 : 1$
  3. $n_{CO} : n_{N_2} : : 1 : 2$
  4. $n_{CO} : n_{N_2} : : 2 : 3$
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D Correct answer
Explanation

According to Graham's Law of Effusion, the rate of effusion is inversely proportional to the square root of the molar mass. Rate(N2)/Rate(CO) = (n_N2/n_CO_initial) * sqrt(M_CO/M_N2). Given n_N2/n_CO = 3/2 and M_N2=28, M_CO=28, the ratio is (3/2) * sqrt(28/28) = 3/2. Thus n_CO/n_N2 = 2/3.

AI explanation

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. For nitrogen (28 g/mol) and carbon monoxide (28 g/mol), the square root of their molar masses is identical, meaning their initial rates of effusion are completely equal at a 1:1 ratio. Consequently, the gases escape the vessel in the exact same proportion as they were originally present. The composition of the effusing mixture will be exactly the same as the initial mixture, making the composition n_CO : n_N_2 equal to 2:3. The provided answer n_CO : n_N_2 : : 2:3 is correct because it represents this identical initial ratio.

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