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

To find the free energy of mixing for the gases, you can use the following equation for the change in Gibbs free energy (\( \Delta G \)) of mixing: \[ \Delta G = nRT \ln\left(\frac{V_f}{V_i}\right) \] Where \( n \) is the total number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. In your case, \( n = 2 + 3 = 5 \) moles, \( R = 8.314 \, \frac{\mathrm{J}}{\mathrm{mol \cdot K}} \), and \( T = 298\, K \). If we assume ideal mixing and that volume \( V_f \) is greater than \( V_i \), you can determine specific volumes or pressures to replace those in the equation. However, if we don't have specific volume or pressure data, we can say: \[ \Delta G \approx 0 \text{ for ideal gas mixtures, suggesting that mixing is spontaneous.} \] Remember, when mixing two gases, the final state tends to favor more disorder or higher entropy, leading to a negative change in Gibbs free energy, promoting spontaneity. To compute specific values, make sure you have the mixing ratios or volumes that reflect how the gases behave under the conditions you have!