# A 10.0 L flask contains 1.031 g O2 and 0.572 g CO2 at 18C. What are the partial pressures of oxygen and carbon dioxide?

## What are the partial pressures of oxygen and carbon dioxide? What is the mole fraction of oxygen in the mixture?

Dec 18, 2016

${P}_{{O}_{2}} = \frac{\left(\frac{1.031 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}\right) \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 291 \cdot K}{10 \cdot L}$

#### Explanation:

We know from $\text{Dalton's Law of Partial Pressures}$ 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. The total pressure is the sum of the individual partial pressures. And thus we can solve for ${P}_{{O}_{2}}$ and ${P}_{C {O}_{2}}$ individually, and then later add them together to get ${P}_{\text{Total}}$.

So all we have to do is use the $\text{Ideal Gas Law:}$

P_"component"=(n_"component"xxRxxT)/V, and use a suitable $\text{Gas constant, } R$, with appropriate units.

${P}_{{O}_{2}} = \frac{\left(\frac{1.031 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}\right) \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 291 \cdot K}{10 \cdot L}$

$= 0.0770 \cdot a t m$

${P}_{C {O}_{2}} = \frac{\left(\frac{0.572 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1}\right) \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 291 \cdot K}{10 \cdot L}$

$= 0.0311 \cdot a t m$

${P}_{\text{Total}} = {P}_{{O}_{2}} + {P}_{C {O}_{2}}$

And ${\text{mole fraction}}_{{O}_{2}} = {P}_{{O}_{2}} / \left({P}_{{O}_{2}} + {P}_{C {O}_{2}}\right)$, i.e. the partial pressure of dioxygen divided by the total pressure.