# The molar quantities of THREE GASES sum to 1.02*mol. What pressure would result if the gases were heated to 301*K, and confined to a 30*L volume?

Apr 2, 2017

We can use Dalton's law of partial pressures............ and get ${P}_{\text{Total}} \cong 0.8 \cdot a t m$

#### Explanation:

Dalton's law of partial pressures states 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 because of these properties:

${P}_{\text{Total}} = {P}_{A} + {P}_{B} + {P}_{C}$

And if we assume ideality, then ${P}_{A} = \frac{{n}_{A} R T}{V}$, ${P}_{B} = \frac{{n}_{B} R T}{V}$, and ${P}_{C} = \frac{{n}_{C} R T}{V}$

But $\frac{R T}{V}$ is a common factor, so...........

${P}_{\text{Total}} = \left\{\frac{R T}{V}\right\} \left({n}_{A} + {n}_{B} + {n}_{C}\right)$

$= \left(1.02 \cdot m o l\right) \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 301 \cdot K \times \frac{1}{30 \cdot L}$

${P}_{\text{Total}} \cong 0.8 \cdot a t m$

I don't know what you mean by the $\text{modified Ideal Gas law}$.