The entropy change of `x_(2)` gas, during the following irreversible process, approximately is how much? `x_(2)` gas (`27^(@)C`, 1 atm, 5 moles ) `rarr x_(2)` gas (`117^(@)C`, 5 atm, 5 moles) (`C_(P)` of `x_(2)` = 6.95 cal/mol/`""^(@)C`), log 1.3= 0.1139

Options

A. `-6.86 "cal"//""^(@)C`
B. `+3.26 "cal"//""^(@)C`
C. `+1.86 "cal"//""^(@)C`
D. `-9.82 "cal"//""^(@)C`

Answer

To find the entropy change of \( x_2 \) gas during the irreversible process, we will calculate the entropy change for both the volume change and the temperature change, and then combine these results.

### Step 1: Calculate the entropy change due to volume change (\( \Delta S_1 \))

The formula for the entropy change due to a change in volume at constant temperature is given by:

\[ \Delta S_1 = nR \ln\left(\frac{V_2}{V_1}\right) \]

However, we can also express this in terms of pressure since \( PV = nRT \):

\[ \Delta S_1 = nR \ln\left(\frac{P_1}{P_2}\right) \]

Given:
- \( n = 5 \) moles
- \( R = 1.987 \, \text{cal/mol/K} \)
- \( P_1 = 1 \, \text{atm} \)
- \( P_2 = 5 \, \text{atm} \)

Calculating \( \Delta S_1 \):

\[ \Delta S_1 = 5 \times 1.987 \times \ln\left(\frac{1}{5}\right) \]

Using \( \ln\left(\frac{1}{5}\right) = -\ln(5) \) and knowing that \( \ln(5) \approx 1.6094 \):

\[ \Delta S_1 = 5 \times 1.987 \times (-1.6094) \approx -15.98 \, \text{cal/K} \]

### Step 2: Calculate the entropy change due to temperature change (\( \Delta S_2 \))

The formula for the entropy change due to a temperature change at constant pressure is given by:

\[ \Delta S_2 = nC_P \ln\left(\frac{T_2}{T_1}\right) \]

Where:
- \( C_P = 6.95 \, \text{cal/mol/K} \)
- Convert temperatures from Celsius to Kelvin:
- \( T_1 = 27^\circ C = 300 \, K \)
- \( T_2 = 117^\circ C = 390 \, K \)

Calculating \( \Delta S_2 \):

\[ \Delta S_2 = 5 \times 6.95 \times \ln\left(\frac{390}{300}\right) \]

Calculating \( \frac{390}{300} = 1.3 \) and using \( \ln(1.3) \approx 0.2624 \):

\[ \Delta S_2 = 5 \times 6.95 \times 0.2624 \approx 9.09 \, \text{cal/K} \]

### Step 3: Combine the entropy changes

The total entropy change \( \Delta S \) is the sum of the two changes:

\[ \Delta S = \Delta S_1 + \Delta S_2 \]

Substituting the values:

\[ \Delta S = -15.98 + 9.09 \approx -6.89 \, \text{cal/K} \]

### Final Answer

The approximate entropy change of \( x_2 \) gas during the irreversible process is:

\[ \Delta S \approx -6.89 \, \text{cal/K} \]