In a previous problem, calculate `DeltaS_(gas) ` If process is carried out at constant volume:

Options

A. `5R ln 2`
B. `(3)/(2) R ln 2`
C. `3R ln 2`
D. `-3 R ln 2`

Explanation

To calculate the change in entropy (ΔS) of an ideal gas when the process is carried out at constant volume, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- Number of moles (n) = 2
- Initial temperature (T1) = 300 K
- Final temperature (T2) = 600 K
- Heat capacity at constant pressure (Cp) = \( \frac{5}{2} R \)

### Step 2: Identify the Relevant Heat Capacity
Since the process is at constant volume, we need to use the heat capacity at constant volume (Cv) instead of Cp. For a monatomic ideal gas, the value of Cv is:
\[ Cv = \frac{3}{2} R \]

### Step 3: Write the Formula for Change in Entropy
The change in entropy (ΔS) for an ideal gas at constant volume is given by:
\[ ΔS = n \cdot Cv \cdot \ln\left(\frac{T2}{T1}\right) \]

### Step 4: Substitute the Values into the Formula
Now, we substitute the known values into the formula:
\[ ΔS = 2 \cdot \left(\frac{3}{2} R\right) \cdot \ln\left(\frac{600}{300}\right) \]

### Step 5: Simplify the Expression
Calculating the logarithm:
\[ \ln\left(\frac{600}{300}\right) = \ln(2) \]
Now substituting this back into the equation:
\[ ΔS = 2 \cdot \left(\frac{3}{2} R\right) \cdot \ln(2) \]

### Step 6: Final Calculation
The terms \(2\) and \(\frac{3}{2}\) simplify:
\[ ΔS = 3R \cdot \ln(2) \]

### Final Answer
Thus, the change in entropy (ΔS) of the gas when the process is carried out at constant volume is:
\[ ΔS = 3R \ln(2) \]

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