Calculate the change in entropy accompanying isothermal expansion of 5 moles of ideal gas at 330K until is volume raised to 6 times of its initial.

The change in entropy (\(\Delta S\)) for the isothermal expansion of an ideal gas can be calculated using the formula:

\[
\Delta S = nR \ln \frac{V_f}{V_i}
\]

where:
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( V_f \) is the final volume,
- \( V_i \) is the initial volume.

Since the final volume \( V_f \) is 6 times the initial volume \( V_i \), we have \( \frac{V_f}{V_i} = 6 \).

Given dataThe change in entropy (\(\Delta S\)) for the isothermal expansion of an ideal gas can be calculated using the formula:

\[
\Delta S = nR \ln \frac{V_f}{V_i}
\]

where:
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( V_f \) is the final volume,
- \( V_i \) is the initial volume.

Since the final volume \( V_f \) is 6 times the initial volume \( V_i \), we have \( \frac{V_f}{V_i} = 6 \).

Given 
- \( n = 5 \) moles,
- \( T = 330 \, \text{K} \) (though temperature doesn’t directly affect \(\Delta S\) here as it’s only in the formula implicitly through the gas constant),
- \( R = 8.314 \, \text{J/(mol·K)} \).

Now, substituting values:

\[
\Delta S = (5 \, \text{moles}) \times (8.314 \, \text{J/(mol·K)}) \times \ln(6)
\]

Calculating each part:

1. Calculate \(\ln(6)\):
   \[
   \ln(6) \approx 1.7918
   \]

2. Substitute into the equation:
   \[
   \Delta S = 5 \times 8.314 \times 1.7918
   \]

3. Solving:
   \[
   \Delta S \approx 74.5 \, \text{J/K}
   \]

### Answer:
The change in entropy, \(\Delta S\), is approximately **74.5 J/K**.

Would you like further details on the derivation of this formula or any additional steps?

---

1. How does temperature affect the entropy change in an isothermal process?
2. What would the entropy change be if the volume increased by a different factor?
3. How does entropy change during an isothermal compression compared to expansion?
4. How is entropy calculated for non-ideal gases?
5. What are practical implications of entropy change in thermodynamic cycles?

**Tip:** Always check if the process is reversible; entropy change formulas often assume reversibility.

Calculate the change in entropy accompanying isothermal expansion of 5 moles of ideal gas at 330K until its volume is raised to 6 times of its initial.

Calculate the entropy change during the isothermal expansion of 5 moles of an ideal gas at 330K, with the final volume increased to six times the initial volume. This solution uses the formula ΔS = nR ln(Vf/Vi) and explores core concepts in thermodynamics and entropy, suitable for high school students in Grades 11-12.

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