Thermodynamics Part 5

First, we determine the total number of moles $(n_{\text{total}})$in the mixture using the Ideal Gas Law: PV = nRT.Given the values:Pressure (P): $100\,\mathrm{kPa}=10^{5}\,\mathrm{Pa}$Volume (V): $8310\,\mathrm{cm}^3=8.31\times10^{-3}\,\mathrm{m}^{3}$Temperature (T): 300 KGas Constant (R): $8.31\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K}$$\;\frac{PV}{RT}=\;\frac{10^{5}\times8.31\times10^{-3}}{8.31\times300}=\;\frac{100}{300}\approx0.333\,\mathrm{mole}$1. Mole equation: $n_{1}+n_{2}=0.333$2. Mass equation: $44n_1+32n_2=13.2$ (where 44 and 32 are the molar masses of $\mathrm{CO}_2$ and $\mathrm{O}_2$ respectively)3. Solve for individual moles Using the first equation to express $n_2$ as $0.333-n_1$, we substitute this into the mass equation:$44n_{1}+32(0.333-n_{1})=13.2$$44n_{1}+10.667-32n_{1}=13.2$$12n_{1}=2.533$$n_{1}\approx0.211\,\mathrm{mol}$Then, calculate $n_{2}:$$n_{2}=0.333-0.211=0.122\,\mathrm{mol}$