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

May 8, 2017

$\text{Dalton's Law of Partial Pressures tells us that.........}$, I get approx. $150 \cdot m g$ of helium............

#### Explanation:

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

$\text{The total pressure is the sum of the individual partial pressures.}$

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

And so ${P}_{\text{He}} = \left(755 - 355\right) \cdot m m \cdot H g = 400 \cdot m m \cdot H g$.

Given that we now ${P}_{\text{He}}$ we can calculate its equivalent mass by means of the Ideal Gas Equation.......

$n = \frac{P V}{R T} = \frac{\frac{400 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 2.0 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 346 \cdot K} = 0.0371 \cdot m o l$.

Which represents a mass of $0.0371 \cdot m o l \times 4.0 \cdot g \cdot m o {l}^{-} 1$ with respect to helium gas.