# A mixture of 1.00 g H_2 and 1.00 g He is placed in a 1.00-L container at 27°C. What is the partial pressure of each gas and the total pressure?

Dec 23, 2016

${P}_{\text{total"=P_"dihydrogen"+P_"helium}} \cong 18 \cdot a t m$

#### Explanation:

In other words, the total pressure is the sum of the individual partial pressures.

And $\text{Dalton's Law of partial pressures}$ assures us that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.

And we can use the Ideal Gas equation to estimate the pressure:
$P = \frac{n R T}{V}$

So all we have to do is to solve for ${P}_{\text{gas}}$ individually:

${P}_{{H}_{2}} = \frac{\frac{1.00 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 300 \cdot K}{1.00 \cdot L} \cong 12 \cdot a t m$

${P}_{H e} = \frac{\frac{1.00 \cdot g}{4.00 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 300 \cdot K}{1.00 \cdot L} \cong 6 \cdot a t m$.

They are spoon feeding you a bit here, in that they explicitly tell you that hydrogen is a bimolecular gas; in fact all the elemental gases SAVE THE NOBLE GASES are bimolecular: ${H}_{2} , {N}_{2} , {O}_{2} , {F}_{2} , C {l}_{2}$.

Alternatively, I could have summed the total moles of gas, and plugged that quantity into the Ideal Gas Equation. The partial pressure of each gas would be proportional to the mole fraction.

For a similar problem, see here.