A quantity of 4.0 moles of an ideal gas at `20^(@)C` expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 10.0L. What is the entropy change of the system (in cals)?

Answer

To solve the problem of finding the entropy change of the system during the isothermal expansion of an ideal gas, we can follow these steps:

### Step 1: Identify the given values
- Number of moles (n) = 4.0 moles
- Initial volume (V1) = 1.0 L
- Final volume (V2) = 10.0 L
- Temperature (T) = 20°C = 293 K (convert Celsius to Kelvin by adding 273)
- Universal gas constant (R) in calories = 2 cal K⁻¹ mol⁻¹

### Step 2: Write the formula for entropy change
The formula for the entropy change (ΔS) for an ideal gas undergoing isothermal expansion is given by:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]

### Step 3: Convert logarithm base
Since we want the answer in calories, we will use the base 10 logarithm. The conversion from natural logarithm (ln) to common logarithm (log) is:
\[ \ln(x) = 2.303 \cdot \log(x) \]
Thus, we can rewrite the entropy change formula as:
\[ \Delta S = nR \cdot 2.303 \cdot \log\left(\frac{V_2}{V_1}\right) \]

### Step 4: Substitute the values into the formula
Now, substitute the known values into the formula:
\[ \Delta S = 4.0 \, \text{moles} \times 2 \, \text{cal K}^{-1} \text{mol}^{-1} \times 2.303 \cdot \log\left(\frac{10.0 \, \text{L}}{1.0 \, \text{L}}\right) \]

### Step 5: Calculate the logarithm
Calculate the logarithm:
\[ \log\left(\frac{10.0}{1.0}\right) = \log(10) = 1 \]

### Step 6: Complete the calculation
Now, plug this value back into the equation:
\[ \Delta S = 4.0 \times 2 \times 2.303 \times 1 \]
\[ \Delta S = 4.0 \times 2 \times 2.303 = 18.424 \, \text{cal K}^{-1} \]

### Final Answer
The entropy change of the system is:
\[ \Delta S = 18.424 \, \text{cal K}^{-1} \]

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