At STP,a closed vessel contains 1 mole each | vedclass

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(D) Initial moles of $He \ (n_1) = 1 \ mol$. Initial moles of $CH_4 \ (n_2) = 1 \ mol$. Moles of $He$ escaped $(n_1') = \frac{2 \ L}{22.4 \ L/mol} \ap

Vedclass · Accepted Answer

$0.488, 0.512$

Vedclass · Answer

(D) Initial moles of $He \ (n_1) = 1 \ mol$. Initial moles of $CH_4 \ (n_2) = 1 \ mol$. Moles of $He$ escaped $(n_1') = \frac{2 \ L}{22.4 \ L/mol} \approx 0.089 \ mol$. Moles of $CH_4$ escaped $(n_2') = \frac{1 \ L}{22.4 \ L/mol} \approx 0.044 \ mol$. Remaining moles of $He \ (n_1'') = 1 - 0.089 = 0.911 \ mol$. Remaining moles of $CH_4 \ (n_2'') = 1 - 0.044 = 0.956 \ mol$. Total remaining moles $= 0.911 + 0.956 = 1.867 \ mol$. Mole fraction of $He \ (\chi_{He}) = \frac{0.911}{1.867} \approx 0.488$. Mole fraction of $CH_4 \ (\chi_{CH_4}) = \frac{0.956}{1.867} \approx 0.512$.

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