Pressure exerted by the mixture of nonreactive gases


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> $\pu{2.0 L}$ of $\ce{O2 (g)}$ and $\pu{8.0 L}$ of $\ce{N2}$, each at $\pu{0.00^\circ C}$ and $\pu{1.00 atm}$, are mixed together. The nonreactive gaseous mixture is compressed to occupy $\pu{2.0 L}$ at $\pu{298 K}$. What is the pressure exerted by this mixture?
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I tried to solve this problem by first using


$$P\_\mathrm{tot}=\frac{n\_\mathrm{tot}RT}{V}\quad(T=\text{constant}; V=\text{constant})$$


then using


$$V\_\mathrm{tot}=\frac{n\_\mathrm{tot}RT}{P}\quad(T=\text{constant}; P=\text{constant})$$


Plugging in the numbers for the second equation


$$10=\frac{n\_\mathrm{tot}\cdot 8.314\cdot 273}{101325}$$
$$n\_\mathrm{tot}=\pu{0.45 mol}$$


Then plugging in the numbers for the first equation


$$P\_\mathrm{tot}=\frac{0.45\cdot 8.314\cdot 298}{2}=\pu{6.76 atm}$$


Whereas the answer is $\pu{5.5 atm}$. What am I doing wrong here?




You did too many unnecessary steps and also didn't use unified units (e.g. liters instead of cubic meters, since you decided to use $R=\pu{8.314 J mol-1 K-1}$). I denoted both total pressures as $P\_1$ and $P\_2$ for simplicity since we don't need partial pressures here. From the ideal gas law:


$$P\_2=\frac{n\_2RT\_2}{V\_2}\tag{1}$$


Since both gases are not reacting at these conditions,


$$n\_1=n\_2=\frac{P\_1V\_1}{RT\_1}\tag{2}$$


and (1) can be rewritten as following:


$$P\_2=P\_1\frac{V\_1T\_2}{V\_2T\_1}=\pu{1.00 atm}\cdot\frac{\pu{10.0 L}\cdot\pu{298.13 K}}{\pu{2.0 L}\cdot\pu{273.13 K}}\approx\pu{5.46 atm}\tag{3}$$


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