# After 0.600 L of Ar at 1.40 atm and 210 "^oC is mixed with 0.200L of O_2 at 398 torr and 120 "^oC in a 400 -mL flask at 29 "^oC, what is the pressure in the flask?

Jul 2, 2018

Well, FIRST off we calculate the molar quantities of each gas...by means of the Ideal Gas equation...

$n = \frac{P V}{R T}$...

#### Explanation:

${n}_{\text{Ar}} = \frac{1.40 \cdot a t m \times 0.600 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 483.15 \cdot K} = 0.0212 \cdot m o l$

${n}_{{O}_{2}} = \frac{\frac{398 \cdot m m}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 0.200 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 393.15 \cdot K} = 0.00325 \cdot m o l$

And now we ADD these molar quantities, and calculate the pressure expressed under the NEW given conditions...

${P}_{\text{mixture}} = \frac{0.245 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 302.15 \cdot K}{400 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

I make this approx. $15 \cdot a t m$ but do not trust my arithmetic...and there is a lot of rithmetick here...