=ind free energy of mixing. When twio moles of gas mixed witin 3 moles of gas at 298 kelven given \( R=8.314 \frac{\mathrm{~J}}{\text { mol. }} \)

To calculate the Gibbs free energy of mixing (\( \Delta G_{\text{mix}} \)) for two gases, we can use the following formula: \[ \Delta G_{\text{mix}} = RT \left[ n_1 \ln \left(\frac{n_1}{n_1 + n_2}\right) + n_2 \ln \left(\frac{n_2}{n_1 + n_2}\right) \right] \] **Given:** - Number of moles of Gas 1 (\( n_1 \)) = 2 mol - Number of moles of Gas 2 (\( n_2 \)) = 3 mol - Temperature (\( T \)) = 298 K - Gas constant (\( R \)) = 8.314 J/(mol·K) **Step-by-Step Calculation:** 1. **Calculate the mole fractions:** \[ \text{Total moles} = n_1 + n_2 = 2 + 3 = 5 \, \text{mol} \] \[ x_1 = \frac{n_1}{n_1 + n_2} = \frac{2}{5} = 0.4 \] \[ x_2 = \frac{n_2}{n_1 + n_2} = \frac{3}{5} = 0.6 \] 2. **Compute the natural logarithms:** \[ \ln(x_1) = \ln(0.4) \approx -0.9163 \] \[ \ln(x_2) = \ln(0.6) \approx -0.5108 \] 3. **Plug values into the Gibbs free energy equation:** \[ \Delta G_{\text{mix}} = 8.314 \times 298 \times \left[ 2 \times (-0.9163) + 3 \times (-0.5108) \right] \] \[ \Delta G_{\text{mix}} = 8.314 \times 298 \times (-3.365) \] \[ \Delta G_{\text{mix}} = 2477.772 \times (-3.365) \approx -8337.8 \, \text{J} \] 4. **Convert joules to kilojoules:** \[ \Delta G_{\text{mix}} \approx -8.34 \, \text{kJ} \] **Conclusion:** The Gibbs free energy of mixing for 2 moles of one gas with 3 moles of another gas at 298 K is approximately **\(-8.34\, \text{kJ}\)**.