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?

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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.