Suppose that 1.28 g of neon gas and 2.49 g of argon gas are confined in a 9.87-L container at 27 C. What would be the pressure in the container?

Aug 29, 2016

Approx. $0.3 \cdot a t m$

Explanation:

We use $\text{Dalton's Law of Partial Pressures}$, which states that in a gaseous mixture, the partial pressure of a gaseous component 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 ${P}_{\text{Ne}}$ $=$ $\frac{{n}_{\text{Ne}} R T}{V}$ $=$ $\frac{\frac{1.28 \cdot g}{20.18 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 300 \cdot K}{9.87 \cdot L}$ $=$ $0.158 \cdot a t m$

And ${P}_{\text{Ar}}$ $=$ $\frac{{n}_{\text{Ar}} R T}{V}$ $=$ $\frac{\frac{2.49 \cdot g}{39.95 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 300 \cdot K}{9.87 \cdot L}$ $=$ $0.156 \cdot a t m$

${P}_{\text{Total}}$ $=$ ${P}_{\text{Ne"+P_"Ar}}$ $=$ ??*atm

Of course, I could have simply worked out the molar quantity of the combined gases, and solved the problem directly. This treatment illustrates $\text{Dalton's Law}$.