In a gaseous mixture of a $2.0L$ volume, at $755"mm Hg"$ , and a temperature of $346K$ , $P_"dinitrogen"=355mm*Hg$ . What are the molar quantities of helium and dinitrogen?

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Answer 1 · 2017-05-08T12:38:09

$"Dalton's Law of Partial Pressures tells us that........."$ , I get approx. $150*mg$ of helium............

$"In a gaseous mixture, the pressure exerted by a component gas"$
$"is the same as the pressure if it ALONE occupied the container."$

$"The total pressure is the sum of the individual partial pressures."$

And this $P_"mixture"=P_"He"+P_("N"_2)$

And so $P_"He"=(755-355)*mm*Hg=400*mm*Hg$ .

Given that we now $P_"He"$ we can calculate its equivalent mass by means of the Ideal Gas Equation.......

$n=(PV)/(RT)=((400*mm*Hg)/(760*mm*Hg*atm^-1)xx2.0*L)/(0.0821*(L*atm)/(K*mol)xx346*K)=0.0371*mol$ .

Which represents a mass of $0.0371*molxx4.0*g*mol^-1$ with respect to helium gas.

In a gaseous mixture of a 2.0*L2.0*L volume, at 755*"mm Hg"755*"mm Hg", and a temperature of 346*K346*K, P_"dinitrogen"=355*mm*HgP_"dinitrogen"=355*mm*Hg. What are the molar quantities of helium and dinitrogen?