One mole of an ideal gas at 350 K is in a 2.0 L vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from... One mole of an ideal gas at 350 K is in a 2.0 L vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to 3.0 L. The magnitude of change in entropy of the surroundings (ΔS) is _____ J K⁻¹ (Nearest integer). Given: R = 8.314 J K⁻¹ mol⁻¹. Read more

Steps to Solve

1. Identify the entropy change formula

For isothermal reversible processes, the change in entropy ($\Delta S$) of the surroundings can be calculated using the formula: $$ \Delta S_{\text{surroundings}} = -\frac{Q_{\text{rev}}}{T} $$

2. Calculate the work done

For isothermal expansion of an ideal gas, the heat absorbed by the gas ($Q_{\text{rev}}$) is equal to the work done by the gas during expansion. This can be calculated using: $$ Q_{\text{rev}} = nRT \ln\left(\frac{V_f}{V_i}\right) $$ Where:

• $n = 1 \text{ mol}$
• $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
• $T = 350 \text{ K}$
• $V_i = 2.0 \text{ L}$
• $V_f = 3.0 \text{ L}$

3. Insert values into the entropy formula

Calculate $Q_{\text{rev}}$: $$ Q_{\text{rev}} = 1 \cdot 8.314 \cdot 350 \ln\left(\frac{3.0}{2.0}\right) $$

4. Evaluate the natural logarithm

Calculate $\ln\left(\frac{3.0}{2.0}\right)$: $$ \ln\left(\frac{3.0}{2.0}\right) \approx 0.4055 $$

5. Compute $Q_{\text{rev}}$

Now plug the value back into the formula for $Q_{\text{rev}}$: $$ Q_{\text{rev}} = 1 \cdot 8.314 \cdot 350 \cdot 0.4055 $$

6. Calculate the entropy change

Finally, substitute $Q_{\text{rev}}$ into the entropy change formula: $$ \Delta S_{\text{surroundings}} = -\frac{Q_{\text{rev}}}{350} $$

7. Final calculations

Complete the calculations to find the exact value of $\Delta S_{\text{surroundings}}$.