The entropy change involved in the isothermal reversible expansion of 2 moles of an idealgas from a volume of 10dm3 to a volume of 100 dm3 at 27°C ??

To determine the entropy change during the isothermal reversible expansion of an ideal gas, we can use a specific formula that relates entropy change to the volume change and temperature. Let's break this down step by step.

Understanding the Variables

In this scenario, we have:

• Number of moles (n): 2 moles
• Initial volume (V1): 10 dm³
• Final volume (V2): 100 dm³
• Temperature (T): 27°C, which is equivalent to 300 K (since we convert Celsius to Kelvin by adding 273).

Entropy Change Formula

The formula for the change in entropy (ΔS) during an isothermal process for an ideal gas is given by:

ΔS = nR ln(V2/V1)

Where:

• ΔS: Change in entropy
• n: Number of moles of gas
• R: Universal gas constant (approximately 8.314 J/(mol·K))
• V2: Final volume
• V1: Initial volume

Calculating the Entropy Change

Now, let's plug in the values into the formula:

First, calculate the ratio of the volumes:

V2/V1 = 100 dm³ / 10 dm³ = 10

Now, substituting the values into the entropy change formula:

ΔS = 2 moles × 8.314 J/(mol·K) × ln(10)

Next, we need to calculate ln(10):

ln(10) ≈ 2.302

Now, substituting this value back into the equation:

ΔS = 2 × 8.314 J/(mol·K) × 2.302

ΔS ≈ 2 × 8.314 × 2.302 ≈ 38.34 J/K

Final Result

The change in entropy (ΔS) for the isothermal reversible expansion of 2 moles of an ideal gas from 10 dm³ to 100 dm³ at 27°C is approximately 38.34 J/K.

Conceptual Insight

This increase in entropy reflects the greater disorder in the gas as it expands into a larger volume. In essence, the gas molecules have more space to occupy, leading to a higher number of possible microstates, which is a fundamental concept in thermodynamics. The isothermal nature of the process means that the temperature remains constant, allowing us to focus solely on the volume change's impact on entropy.