How to Calculate the Gibbs Free Energy and Entropy Due to Mixing 2.5 Moles of Argon with 3.5 Moles of Oxygen

To calculate the Gibbs free energy and entropy due to mixing, we can use the following equations: ∆G_mix = -nR * T * ln(X1^2 * X2^2) and ∆S_mix = -nR * (x1 * ln(x1) + x2 * ln(x2)) Here, n is the total number of moles of the two gases, R is the universal gas constant (8.31 J/mol*K), T is the temperature in Kelvin, x1 and x2 are the mole fractions of argon and oxygen respectively. Given the problem: n1 (argon) = 2.5 moles, n2 (oxygen) = 3.5 moles, P = 1 bar, T = 250°C First, let's convert temperature T from Celsius to Kelvin: T = 250 °C + 273.15 = 523.15 K Now, we can calculate the mole fractions: x1 (argon) = n1 / (n1 + n2) = 2.5 moles / (2.5 moles + 3.5 moles) = 2.5/6 = 0.4167 x2 (oxygen) = n2 / (n1 + n2) = 3.5 moles / (2.5 moles + 3.5 moles) = 3.5/6 = 0.5833 With these values, we calculate the Gibbs free energy change due to mixing (∆G_mix): ∆G_mix = -(2.5 moles + 3.5 moles) * 8.31 J/mol*K * 523.15 K * ln((0.4167^2) * (0.5833^2)) ∆G_mix ≈ -4155.14 J And the entropy change due to mixing (∆S_mix): ∆S_mix = -(2.5 moles + 3.5 moles) * 8.31 J/mol*K * (0.4167 * ln(0.4167) + 0.5833 * ln(0.5833)) ∆S_mix ≈ -31.20 J/K So, the Gibbs free energy and entropy due to mixing are approximately -4155.14 J and -31.20 J/K, respectively.