1 mole of a diatomic is heated through isochoric process from 300 k to 500 K. The entropy is :

Options

A. 10.61
B. 38.26
C. 20.05
D. 30

Answer

To solve the problem of finding the entropy change (\( \Delta S \)) for 1 mole of a diatomic gas heated through an isochoric process from 300 K to 500 K, we can follow these steps:

### Step 1: Understand the process
In an isochoric process, the volume remains constant. The formula for the change in entropy (\( \Delta S \)) for an ideal gas during an isochoric process is given by:

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

where:
- \( n \) = number of moles
- \( C_v \) = molar heat capacity at constant volume
- \( T_1 \) = initial temperature
- \( T_2 \) = final temperature

### Step 2: Identify the values
From the question:
- \( n = 1 \) mole
- \( T_1 = 300 \, K \)
- \( T_2 = 500 \, K \)

For a diatomic gas, the molar heat capacity at constant volume (\( C_v \)) is:

\[ C_v = \frac{5}{2} R \]

where \( R \) (the universal gas constant) is approximately \( 8.314 \, J/(mol \cdot K) \).

### Step 3: Calculate \( C_v \)
Substituting the value of \( R \):

\[ C_v = \frac{5}{2} \times 8.314 = 20.785 \, J/(mol \cdot K) \]

### Step 4: Substitute values into the entropy formula
Now we can substitute the values into the entropy change formula:

\[ \Delta S = 1 \times 20.785 \ln \left(\frac{500}{300}\right) \]

### Step 5: Calculate the logarithm
Calculating the ratio:

\[ \frac{500}{300} = \frac{5}{3} \]

Now, calculate the natural logarithm:

\[ \ln \left(\frac{5}{3}\right) = \ln 5 - \ln 3 \]

Using approximate values:

\[ \ln 5 \approx 1.6094 \quad \text{and} \quad \ln 3 \approx 1.0986 \]

Thus,

\[ \ln \left(\frac{5}{3}\right) \approx 1.6094 - 1.0986 = 0.5108 \]

### Step 6: Final calculation of \( \Delta S \)
Now substitute back into the equation for \( \Delta S \):

\[ \Delta S \approx 20.785 \times 0.5108 \approx 10.61 \, J/K \]

### Conclusion
The change in entropy (\( \Delta S \)) for the process is approximately:

\[ \Delta S \approx 10.61 \, J/K \]