Two moles of ideal gas at `27^(@)C` temperature is expanded reversibly from `2` litre to `20` liter. Find entropy change `(R=2cal//molK)`.

Options

A. `92.1`
B. `0`
C. `4`
D. `9.2`

Explanation

To find the entropy change (ΔS) for the reversible expansion of an ideal gas, we can use the formula:

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

Where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step-by-Step Solution:

**Step 1: Convert the temperature to Kelvin.**
- The given temperature is \( 27^\circ C \).
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Thus,
\[ T = 27 + 273 = 300 \, K \]

**Step 2: Identify the given values.**
- Number of moles, \( n = 2 \, \text{moles} \)
- Initial volume, \( V_i = 2 \, \text{liters} \)
- Final volume, \( V_f = 20 \, \text{liters} \)
- Gas constant, \( R = 2 \, \text{cal/mol K} \)

**Step 3: Substitute the values into the entropy change formula.**
- We need to calculate \( \frac{V_f}{V_i} \):
\[ \frac{V_f}{V_i} = \frac{20}{2} = 10 \]
- Now substitute into the entropy change formula:
\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \]
\[ \Delta S = 2 \times 2 \times \ln(10) \]

**Step 4: Calculate \( \ln(10) \).**
- The natural logarithm of 10 is approximately \( 2.303 \).
- Thus,
\[ \Delta S = 2 \times 2 \times 2.303 \]

**Step 5: Perform the multiplication.**
- Calculate:
\[ \Delta S = 4 \times 2.303 = 9.212 \, \text{cal/K} \]

### Final Answer:
The entropy change \( \Delta S \) is approximately \( 9.2 \, \text{cal/K} \).