What will the entropy change of the system when expansion of 1 mole of a gas takes place from 3 L to 6 L under isothermal conditions? Consider, `"R = 2 cal K"^(-1)" mol"^(-1)" and log 2 = 0.30 "`

Options

A. `"2.84 cal K"^(-1)`
B. `"1.386 cal K"^(-1)`
C. `"0.37 cal K"^(-1)`
D. `"5.26 cal K"^(-1)`

Answer

To calculate the entropy change (ΔS) of the system when 1 mole of gas expands isothermally from 3 L to 6 L, we can use the formula for entropy change during an isothermal expansion of an ideal gas:

\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]

Where:
- \( n \) = number of moles of gas = 1 mole
- \( R \) = universal gas constant = 2 cal K\(^{-1}\) mol\(^{-1}\)
- \( V_1 \) = initial volume = 3 L
- \( V_2 \) = final volume = 6 L

### Step-by-Step Solution:

1. **Identify the values**:
- \( n = 1 \) mole
- \( R = 2 \) cal K\(^{-1}\) mol\(^{-1}\)
- \( V_1 = 3 \) L
- \( V_2 = 6 \) L

2. **Substitute the values into the entropy change formula**:
\[ \Delta S = 1 \times 2 \times \ln\left(\frac{6}{3}\right) \]

3. **Calculate the ratio of volumes**:
\[ \frac{V_2}{V_1} = \frac{6}{3} = 2 \]

4. **Calculate the natural logarithm**:
\[ \ln(2) = 0.30 \quad (\text{as given in the question}) \]

5. **Substitute \( \ln(2) \) into the equation**:
\[ \Delta S = 1 \times 2 \times 0.30 \]

6. **Perform the multiplication**:
\[ \Delta S = 2 \times 0.30 = 0.60 \text{ cal K}^{-1} \]

### Final Answer:
The entropy change of the system is \( \Delta S = 0.60 \text{ cal K}^{-1} \).

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