How to Calculate the Gibbs free energy and entropy due to mixing 2.5 moles of argon with 3.5 moles of oxygen, both to 1 bar and 250 atmospheres

To calculate the Gibbs free energy and entropy due to mixing 2.5 moles of argon with 3.5 moles of oxygen at 1 bar and 250°C, we need to use the formulas for the partial molar Gibbs free energy of mixing and partial molar entropy of mixing for ideal gases. 1. Calculate the mole fractions of argon (χ_Ar) and oxygen (χ_O2). Total moles (n_total) = moles of argon (n_Ar) + moles of oxygen (n_O2) n_total = 2.5 + 3.5 = 6 moles χ_Ar = n_Ar / n_total = 2.5 / 6 = 0.4167 χ_O2 = n_O2 / n_total = 3.5 / 6 = 0.5833 2. Calculate the entropy of mixing (ΔS_mix) using the formula: ΔS_mix = -R * [χ_Ar * ln(χ_Ar) + χ_O2 * ln(χ_O2)] ΔS_mix = -8.31 * [0.4167 * ln(0.4167) + 0.5833 * ln(0.5833)] ΔS_mix ≈ -8.31 * [0.4167 * (-0.8751) + 0.5833 * (-0.5386)] ΔS_mix ≈ -8.31 * [-0.3644 - 0.3140] ΔS_mix ≈ 5.649 J/mol·K 3. Calculate the Gibbs free energy of mixing (ΔG_mix) using the formula: ΔG_mix = ΔH_mix - T * ΔS_mix For ideal gases, ΔH_mix = 0, and the temperature T should be converted to Kelvin: T = 250°C + 273.15 = 523.15 K ΔG_mix = 0 - (523.15 * 5.649) ΔG_mix ≈ -2959.1 J/mol The Gibbs free energy and entropy due to mixing 2.5 moles of argon with 3.5 moles of oxygen at 1 bar and 250°C are approximately -2959.1 J/mol and 5.649 J/mol·K, respectively, assuming ideal gas behavior.