Find free energy of mixing. When two moles of gas mixed
witin 3 moles of gas at 298 kelven
given

Ask by Powell Vaughn. in India

Jan 22,2025

Question

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

UpStudy AI · Accepted Answer

To calculate the **free energy of mixing (ΔG_mix)** for two ideal gases, we'll use the following formula:

$$
\Delta G_{\text{mix}} = RT \left( n_1 \ln x_1 + n_2 \ln x_2 \right)
$$

where:
- $$ R $$ is the gas constant = $$ 8.314 \, \frac{\text{J}}{\text{mol·K}} $$
- $$ T $$ is the temperature in Kelvin = $$ 298 \, \text{K} $$
- $$ n_1 $$ and $$ n_2 $$ are the number of moles of each gas
- $$ x_1 $$ and $$ x_2 $$ are the mole fractions of each gas

**Given:**
- $$ n_1 = 2 \, \text{mol} $$
- $$ n_2 = 3 \, \text{mol} $$
- Total moles, $$ n_{\text{total}} = n_1 + n_2 = 5 \, \text{mol} $$

**Step 1: Calculate Mole Fractions**

$$
x_1 = \frac{n_1}{n_{\text{total}}} = \frac{2}{5} = 0.4
$$
$$
x_2 = \frac{n_2}{n_{\text{total}}} = \frac{3}{5} = 0.6
$$

**Step 2: Plug Values into the Formula**

$$
\Delta G_{\text{mix}} = (8.314 \, \frac{\text{J}}{\text{mol·K}})(298 \, \text{K}) \left( 2 \ln(0.4) + 3 \ln(0.6) \right)
$$

**Step 3: Calculate the Natural Logarithms**

$$
\ln(0.4) \approx -0.9163
$$
$$
\ln(0.6) \approx -0.5108
$$

**Step 4: Compute the Terms Inside the Brackets**

$$
2 \ln(0.4) + 3 \ln(0.6) = 2(-0.9163) + 3(-0.5108) = -1.8326 - 1.5324 = -3.3650
$$

**Step 5: Calculate ΔG_mix**

$$
\Delta G_{\text{mix}} = 8.314 \times 298 \times (-3.3650)
$$
$$
\Delta G_{\text{mix}} \approx 2477.772 \times (-3.3650) \approx -8340 \, \text{J/mol}
$$

**Converting to Kilojoules:**

$$
\Delta G_{\text{mix}} \approx -8.34 \, \text{kJ/mol}
$$

**Conclusion:**
The free energy of mixing for the given gases at 298 K is approximately **-8.34 kJ/mol**.
The free energy of mixing is approximately **-8.34 kJ/mol**.

Find free energy of mixing. When two moles of gas mixed
witin 3 moles of gas at 298 kelven
given

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