One mole of an ideal monoatomic gas at `27^(@)C` is subjected to a reversible isoentropic compression until the temperature reached to `327^(@)C`. If the initial pressure was `1.0 atm`, then find the value of ln `P_(2)`
`(` Given `: ln 2=0.7)`

Options

A. `1.75` atm
B. `0.176` atm
C. `1.0395` atm
D. 2.0 atm

Explanation

To solve the problem of finding the value of \( \ln P_2 \) for a monoatomic ideal gas undergoing isoentropic compression, we can follow these steps:

### Step 1: Understand the Process
The process is isoentropic, meaning that the entropy change \( \Delta S = 0 \). For an ideal gas, the relationship between temperature and pressure during an isoentropic process can be expressed using the formula:
\[ \Delta S = nC_p \ln \frac{T_2}{T_1} + nR \ln \frac{P_1}{P_2} = 0 \]
Where:
- \( n \) is the number of moles of gas (1 mole in this case),
- \( C_p \) is the molar heat capacity at constant pressure,
- \( R \) is the universal gas constant,
- \( T_1 \) and \( T_2 \) are the initial and final temperatures,
- \( P_1 \) and \( P_2 \) are the initial and final pressures.

### Step 2: Identify Given Values
- Initial temperature \( T_1 = 27^\circ C = 300 \, K \) (after converting Celsius to Kelvin),
- Final temperature \( T_2 = 327^\circ C = 600 \, K \),
- Initial pressure \( P_1 = 1.0 \, atm \),
- For a monoatomic ideal gas, \( C_p = \frac{5}{2} R \).

### Step 3: Substitute Values into the Equation
Substituting the known values into the isoentropic equation:
\[ 0 = nC_p \ln \frac{T_2}{T_1} + nR \ln \frac{P_1}{P_2} \]
This simplifies to:
\[ 0 = 1 \cdot \frac{5}{2} R \ln \frac{600}{300} + 1 \cdot R \ln \frac{1}{P_2} \]
This can be rewritten as:
\[ 0 = \frac{5}{2} R \ln 2 + R \ln \frac{1}{P_2} \]

### Step 4: Cancel \( R \) and Rearrange
Since \( R \) is a common factor, we can cancel it from both sides:
\[ 0 = \frac{5}{2} \ln 2 + \ln \frac{1}{P_2} \]
Rearranging gives:
\[ \ln \frac{1}{P_2} = -\frac{5}{2} \ln 2 \]

### Step 5: Solve for \( \ln P_2 \)
Taking the negative of both sides:
\[ \ln P_2 = \frac{5}{2} \ln 2 \]
Now substituting \( \ln 2 = 0.7 \):
\[ \ln P_2 = \frac{5}{2} \times 0.7 = \frac{3.5}{2} = 1.75 \]

### Final Answer
Thus, the value of \( \ln P_2 \) is:
\[ \ln P_2 = 1.75 \]