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

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

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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 **$$-8.34\, \text{kJ}$$**.
The Gibbs free energy of mixing is approximately **-8.34 kJ**.

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

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