# Dalton's law for a gas mixture

If $\pu{200 mL}$ of $\ce{N2}$ at $\ce{25 ^\circ C}$ and a pressure of $\pu{250 torr}$ are mixed with $\pu{350 mL}$ of $\ce{O2}$ at $\pu{25 ^\circ C}$ and a pressure of $\pu{300 torr}$, so that the resulting volume is $\pu{300 mL}$, what would be the final pressure in $\pu{torr}$ of the mixture at $\pu{25 ^\circ C}$?

I understand that to find final pressure, I need to add the partial pressure of $\ce{N2}$ and $\ce{O2}$. I'm not sure how to find the partial pressure using Dalton's law. To find partial pressure of $\ce{N2}$ first:

$$p(\ce{N2}) = X(\ce{N2}) \cdot \pu{0.3289 atm}$$

where $X$ is the mole fraction. How do I find it?

• Ideal gas equation PV=nRT can be used to find out "n" of both N2 and O2 and then use X=(n1)/(n1+n2) for both N2 and O2 and you are done. Nov 23 '17 at 9:29
• @SourabhYelluru what does the "resulting volume is 300ml" mean ? Do I need that for any of my calculation for partial fractions and hence final pressure ? Nov 23 '17 at 9:32
• You get total no of moles using ideal gas equation first.You can then use it again to find final pressure after volume is decreased to 300 mL Nov 23 '17 at 9:38

There is no need to use ideal gas law in $pV = nRT$ form or find molar fraction $X$. The process is isothermal ($T_1 = T_2 = \pu{25 ^\circ C}$, hence $p_1V_1 = p_2V_2$), so it's a matter of finding the sum of partial pressures of each gaseous component $i$ in the system:
$$p = \sum_i{p_{2i}} = \sum_i{\frac{p_{1i}V_{1i}}{V_{2i}}}$$
Also, converting the pressure back and forth between $\pu{torr}$ and $\pu{atm}$ is counterproductive as you are explicitly asked to provide an answer in $\pu{torr}$.
$$p = p_2(\ce{N2}) + p_2(\ce{O2}) = \frac{\pu{250 torr} \cdot \pu{200 mL}}{\pu{300 mL}} + \frac{\pu{350 torr} \cdot \pu{350 mL}}{\pu{300 mL}} = \pu{575 torr}$$