Pressure exerted by the mixture of nonreactive gases
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:

$$P_2=P_1\frac{V_1T_2}{V_2T_1}=\pu{1.00 atm}\cdot\frac{\pu{10.0 L}\cdot\pu{298.13 K}}{\pu{2.0 L}\cdot\pu{273.13 K}}\approx\pu{5.46 atm}\tag{3}$$