A 100.-L flask contains a mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and argon gases at \(25^{\circ} \mathrm{C}\). The mass of argon present is \(228 \mathrm{~g}\) and the mole fraction of methane in the mixture is \(0.650 .\) Calculate the total kinetic energy of the gaseous mixture.

Using the Ideal Gas Law \(PV = nRT\), we can find the moles of argon. We're given the mass of argon, the volume of the flask, and the temperature, but we don't have the pressure. However, since the mole fraction of methane is given, we can find the total moles and use it to back-calculate the moles of argon. We know: Mass of Argon = 228 g Molecular weight of Argon = 39.95 g/mol Volume = 100 L Temperature = 25°C = 298.15 K Mole fraction of methane = 0.650 R (Gas Constant) = 0.0821 L atm / (K mol) First, let's find the moles of argon: Moles of argon = Mass of argon / Molecular weight of argon Moles of argon = \( \frac{228}{39.95} \) Moles of argon = 5.707 mol

Using the mole fraction equation and the moles of argon, we can find the moles of methane: Mole fraction of methane = 0.650 = \(\frac{n_{CH_4}}{n_{CH_4}+n_{Ar}}\) 0.650 = \( \frac{n_{CH_4}}{n_{CH_4}+5.707} \) Now, solve for the moles of methane: \(n_{CH_4} = \frac{0.650 \times (n_{CH_4}+5.707)}{1-0.650}\) \(n_{CH_4} = 10.595~mol\)

To find the total kinetic energy of the gaseous mixture, we sum up the individual kinetic energies of methane and argon, using the formula: \(KE = \frac{3}{2}nRT\) Calculate the kinetic energy of methane: \(KE_{CH_4} = \frac{3}{2} \times 10.595 \times 0.0821 \times 298.15\) \(KE_{CH_4} = 7789.7~J\) Calculate the kinetic energy of argon: \(KE_{Ar} = \frac{3}{2} \times 5.707 \times 0.0821 \times 298.15\) \(KE_{Ar} = 4148.1~J\) Now add these two kinetic energies for the total kinetic energy of the gaseous mixture: Total kinetic energy = \(KE_{CH_4} + KE_{Ar}\) Total kinetic energy = \(7789.7+4148.1\) Total kinetic energy = \(11937.8~J\) The total kinetic energy of the gaseous mixture is approximately 11,937.8 Joules.