One mole of an ideal diatomic gas `(C_(v)=5cal) ` was transformed form initial `25^(@)C` and `1L` to the state when the temperature is `100^(@)C` and volume 10L. The entropy change of the process can be express as: `(R=2cal//mol//K)`

Options

A. `3 "ln"(298)/(373)+2ln10`
B. `5 "ln"(373)/(298)+2ln10`
C. `7 "ln"(373)/(298)+2"ln"(1)/(10)`
D. `5 "ln"(373)/(298)+2"ln"(1)/(10)`

Explanation

To solve the problem of calculating the entropy change for the transformation of one mole of an ideal diatomic gas, we will follow these steps:

### Step 1: Identify Given Data
- Initial temperature \( T_1 = 25^\circ C = 298 \, K \)
- Final temperature \( T_2 = 100^\circ C = 373 \, K \)
- Initial volume \( V_1 = 1 \, L \)
- Final volume \( V_2 = 10 \, L \)
- Number of moles \( n = 1 \)
- Heat capacity at constant volume \( C_v = 5 \, cal/(mol \cdot K) \)
- Universal gas constant \( R = 2 \, cal/(mol \cdot K) \)

### Step 2: Use the Entropy Change Formula
The formula for the change in entropy \( \Delta S \) for an ideal gas during a transformation is given by:

\[ \Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right) \]

### Step 3: Substitute the Values into the Formula
Substituting the known values into the formula:

\[ \Delta S = 1 \cdot 5 \cdot \ln\left(\frac{373}{298}\right) + 1 \cdot 2 \cdot \ln\left(\frac{10}{1}\right) \]

### Step 4: Calculate Each Term
1. **Calculate \( \ln\left(\frac{373}{298}\right) \)**:
- \( \frac{373}{298} \approx 1.2517 \)
- \( \ln(1.2517) \approx 0.2231 \)

2. **Calculate \( \ln\left(\frac{10}{1}\right) \)**:
- \( \ln(10) \approx 2.3026 \)

### Step 5: Substitute Back and Calculate \( \Delta S \)
Now substituting these values back into the equation:

\[ \Delta S = 5 \cdot 0.2231 + 2 \cdot 2.3026 \]
\[ \Delta S = 1.1155 + 4.6052 \]
\[ \Delta S \approx 5.7207 \, cal/K \]

### Final Answer
The change in entropy \( \Delta S \) for the process is approximately \( 5.7207 \, cal/K \).

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