Dalton's law for a gas mixture - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-21T19:23:43Z https://chemistry.stackexchange.com/feeds/question/86246 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/86246 0 Dalton's law for a gas mixture user175089 https://chemistry.stackexchange.com/users/54813 2017-11-23T09:17:25Z 2017-11-23T10:41:31Z <blockquote> <p>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 $\pu{25 ^\circ C}$?</p> </blockquote> <p>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> <p>$$p(\ce{N2}) = X(\ce{N2}) \cdot \pu{0.3289 atm}$$</p> <p>where $X$ is the mole fraction. How do I find it? </p> https://chemistry.stackexchange.com/questions/86246/-/86248#86248 2 Answer by andselisk for Dalton's law for a gas mixture andselisk https://chemistry.stackexchange.com/users/41328 2017-11-23T10:41:07Z 2017-11-23T10:41:07Z <p>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> <p>$$p = \sum_i{p_{2i}} = \sum_i{\frac{p_{1i}V_{1i}}{V_{2i}}}$$</p> <p>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>