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Question: 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 c. Assume ideal gas behaviour given that R= 8.31 J . (6 Marks)

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.

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