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?

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 23 '17 at 9:29
  • $\begingroup$ @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 ? $\endgroup$
    – user175089
    Nov 23 '17 at 9:32
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    $\begingroup$ 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 $\endgroup$ 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 $\pu{torr}$.

$$p = p_2(\ce{N2}) + p_2(\ce{O2}) = \frac{\pu{250 torr} \cdot \pu{200 mL}}{\pu{300 mL}} + \frac{\pu{350 torr} \cdot \pu{350 mL}}{\pu{300 mL}} = \pu{575 torr}$$

  • 1
    $\begingroup$ This method is quicker and more efficient than using ideal gas equation. $\endgroup$ Nov 23 '17 at 10:43

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