If one mole of monoatomic ideal gas expanded from 2 atm to 0.5 atm at 27°C, then the entropy change will be

Options

A. R In 2
B. 4R In 2
C. 3 Rin 2
D. 2R In 2

Answer

To calculate the entropy change (ΔS) for the expansion of one mole of a monoatomic ideal gas from 2 atm to 0.5 atm at a constant temperature of 27°C, we can follow these steps:

### Step 1: Understand the formula for entropy change
The entropy change for an ideal gas during an isothermal process can be expressed as:
\[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right) \]
or equivalently,
\[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{P_1}{P_2}\right) \]

### Step 2: Identify the parameters
Given:
- \( n = 1 \) mole (since we have one mole of gas)
- \( P_1 = 2 \) atm (initial pressure)
- \( P_2 = 0.5 \) atm (final pressure)
- \( T_1 = T_2 = 27°C = 300 \) K (temperature is constant)

### Step 3: Simplify the equation
Since the temperature remains constant, the term \(\ln\left(\frac{T_2}{T_1}\right)\) becomes:
\[ \ln\left(\frac{T_2}{T_1}\right) = \ln(1) = 0 \]
Thus, the equation simplifies to:
\[ \Delta S = nR \ln\left(\frac{P_1}{P_2}\right) \]

### Step 4: Substitute the values
Now substituting the values into the equation:
\[ \Delta S = 1 \cdot R \ln\left(\frac{2}{0.5}\right) \]
Calculating the pressure ratio:
\[ \frac{2}{0.5} = 4 \]
So we have:
\[ \Delta S = R \ln(4) \]

### Step 5: Simplify \(\ln(4)\)
We can express \(\ln(4)\) as:
\[ \ln(4) = \ln(2^2) = 2 \ln(2) \]
Thus:
\[ \Delta S = R \cdot 2 \ln(2) = 2R \ln(2) \]

### Step 6: Final result
The entropy change for the process is:
\[ \Delta S = 2R \ln(2) \]

### Conclusion
The correct answer is option 4: \( 2R \ln(2) \).

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