# My textbook says that that for a mixture of 15.2*g helium gas, and 30.6*g dioxygen gas, the partial pressure is dominated by the helium. Can you find P_"He", and P_"dioxygen" and P_"Total"?

Feb 24, 2017

Well, I think the text book is right mate. Because $\text{helium gas}$ is present in a large molar quantity, its partial pressure should dominate. Let's see what we get.

#### Explanation:

$\text{Dalton's Law of Partial Pressures}$ states UNEQUIVOCALLY that in a gaseous mixture, the partial pressure exerted by a gaseous component is the SAME as it would exert if it ALONE occupied the container.

And thus P_"He"=(n_"He"RT)/(V)=((15.2*g)/(4.0*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx295*K)/(5.00*L)

${P}_{\text{He}} = 18.4 \cdot a t m$

${P}_{{O}_{2}} = \frac{{n}_{{O}_{2}} R T}{V} = \frac{\frac{30.6 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 295 \cdot K}{5.00 \cdot L}$

${P}_{\text{dioxygen}} = 4.63 \cdot a t m$.

$\text{Dalton's Law of Partial Pressures}$ further states that the total pressure, ${P}_{\text{Total}}$ is the sum of the individual partial pressures.

So here ${P}_{\text{Total"=P_"He}} + {P}_{{O}_{2}} = \left(18.4 + 4.63\right) \cdot a t m = 23.0 \cdot a t m$, as required. Have another look at your calcuations. It is all too easy to make an error, and everybody has done this.