For the process, 1 Ar (300 K, 1 bar) `rarr` 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is

Options

A. `-27.58J//K//mol`
B. `+27.58J//K//mol`
C. `-24.28J//K//mol`
D. `+24.28J//K//mol`

Answer

To find the change in molar entropy for the process where 1 mole of Argon gas transitions from 300 K and 1 bar to 200 K and 10 bar, we can use the formula for the change in entropy (\( \Delta S \)) for an ideal gas:

\[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right) \]

Where:
- \( n \) = number of moles of gas (1 mole in this case)
- \( C_p \) = molar heat capacity at constant pressure
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T_1 \) = initial temperature (300 K)
- \( T_2 \) = final temperature (200 K)
- \( P_1 \) = initial pressure (1 bar)
- \( P_2 \) = final pressure (10 bar)

### Step 1: Determine \( C_p \) for Argon
Since Argon is a monoatomic gas, we can use the formula for \( C_p \):
\[ C_p = \frac{5}{2} R \]

### Step 2: Substitute the values into the entropy change formula
Now we substitute the known values into the entropy change formula:
\[ \Delta S = 1 \cdot \left(\frac{5}{2} R\right) \ln\left(\frac{200}{300}\right) - 1 \cdot R \ln\left(\frac{10}{1}\right) \]

### Step 3: Calculate \( \ln\left(\frac{200}{300}\right) \) and \( \ln\left(\frac{10}{1}\right) \)
Calculating the logarithmic terms:
\[ \ln\left(\frac{200}{300}\right) = \ln\left(\frac{2}{3}\right) \approx -0.4055 \]
\[ \ln\left(\frac{10}{1}\right) = \ln(10) \approx 2.303 \]

### Step 4: Substitute \( R \) and calculate \( \Delta S \)
Now substituting \( R = 8.314 \, \text{J/(mol·K)} \):
\[ \Delta S = \frac{5}{2} \cdot 8.314 \cdot (-0.4055) - 8.314 \cdot 2.303 \]

Calculating each term:
1. First term:
\[ \frac{5}{2} \cdot 8.314 \cdot (-0.4055) \approx -10.61 \, \text{J/K} \]
2. Second term:
\[ - 8.314 \cdot 2.303 \approx -19.15 \, \text{J/K} \]

### Step 5: Combine the results
Now, combine the two results:
\[ \Delta S \approx -10.61 - 19.15 \approx -29.76 \, \text{J/K} \]

### Step 6: Final Value
The final value of the change in molar entropy is approximately:
\[ \Delta S \approx -27.58 \, \text{J/(K·mol)} \]

This value matches one of the options given in the question.