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 L"`

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 find the entropy change (ΔS) of the system when 1 mole of a gas expands isothermally from 3 L to 6 L, we can use the following formula:

\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]

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

### Step 1: Calculate the ratio of final to initial volume

\[ \frac{V_f}{V_i} = \frac{6 \, \text{L}}{3 \, \text{L}} = 2 \]

### Step 2: Substitute the values into the entropy change formula

Using the logarithmic identity, we can write:

\[ \Delta S = nR \ln\left(2\right) \]

### Step 3: Convert the natural logarithm to base 10 logarithm

Using the relationship between natural logarithm and base 10 logarithm:

\[ \ln(x) = 2.303 \log_{10}(x) \]

Thus,

\[ \ln(2) = 2.303 \log_{10}(2) \]

Given that \(\log_{10}(2) = 0.30\):

\[ \ln(2) = 2.303 \times 0.30 \]

### Step 4: Calculate ΔS

Now substituting back into the entropy change formula:

\[ \Delta S = 1 \, \text{mol} \times 2 \, \text{cal K}^{-1} \, \text{mol}^{-1} \times (2.303 \times 0.30) \]

Calculating the value:

\[ \Delta S = 2 \times 2.303 \times 0.30 \]

\[ \Delta S = 2 \times 0.6909 \approx 1.38 \, \text{cal K}^{-1} \]

### Final Answer

The entropy change (ΔS) of the system is approximately:

\[ \Delta S \approx 1.38 \, \text{cal K}^{-1} \]