`1` mole of an ideal gas at `25^(@)` C is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in `J K^(-1) mol^(-1)`)

Options

A. `19.15`
B. `-19.15`
C. `4.7`
D. zero

Explanation

To calculate the change in entropy during the expansion of 1 mole of an ideal gas that expands reversibly and adiabatically to ten times its initial volume, we can follow these steps:

### Step 1: Understand the Process
The gas is expanding adiabatically and reversibly. In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). However, we can still calculate the change in entropy using the relationship for reversible processes.

### Step 2: Use the Entropy Change Formula
For an ideal gas undergoing a reversible process, the change in entropy (ΔS) can be calculated using the formula:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]
where:
- \( n \) = number of moles of the gas (1 mole in this case)
- \( R \) = universal gas constant (8.314 J/(K·mol))
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step 3: Identify the Volumes
Given that the final volume \( V_f \) is ten times the initial volume \( V_i \):
\[ V_f = 10 V_i \]

### Step 4: Substitute the Volumes into the Entropy Change Formula
Substituting \( V_f \) into the entropy change formula:
\[ \Delta S = nR \ln\left(\frac{10 V_i}{V_i}\right) \]
This simplifies to:
\[ \Delta S = nR \ln(10) \]

### Step 5: Plug in the Values
Now, substituting the values of \( n \) and \( R \):
\[ \Delta S = 1 \times 8.314 \times \ln(10) \]

### Step 6: Calculate \( \ln(10) \)
Using the approximate value of \( \ln(10) \):
\[ \ln(10) \approx 2.303 \]
Thus,
\[ \Delta S = 8.314 \times 2.303 \]

### Step 7: Final Calculation
Calculating the final value:
\[ \Delta S \approx 8.314 \times 2.303 \approx 19.15 \, \text{J/K·mol} \]

### Final Answer
The change in entropy during the expansion is approximately:
\[ \Delta S \approx 19.15 \, \text{J/K·mol} \]

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