# A 9.70 L container holds a mixture of two gases at 53C. The partial pressures of gas A and gas B, respectively, are 0.368 atm and 0.893 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

May 9, 2017

This is an illustration of $\text{Dalton's Law of Partial Pressures......}$

#### Explanation:

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

And thus ${P}_{\text{mixture}} = {P}_{A} + {P}_{B} + {P}_{c}$

We are given ${P}_{A}$, and ${P}_{B}$, but we must work out ${P}_{C}$, which we can do by assuming ideality and employing the equation:

${P}_{C} = \frac{0.210 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 326 \cdot K}{9.70 \cdot L} = 0.579 \cdot a t m .$

And so.....${P}_{\text{mixture}} = {P}_{A} + {P}_{B} + {P}_{c}$

$= \left(0.368 + 0.893 + 0.579\right) \cdot a t m$

$= 1.82 \cdot a t m .$