One mole of an ideal monoatomic gas expands isothermally against constant external pressure of `1 atm` from intial volume of `1L` to a state where its final pressure becomes equal to external pressure. If initial temperature of gas in `300 K` then total entropy change of system in the above process is:
`[R=0.0082L atm mol^(-1)K^(-1)=8.3J mol^(-1)K^(-1)]`

Options

A. `0`
B. `R l n(24.6)`
C. `R l n(2490)`
D. `3//2R l n(24.6)`

Explanation

To solve the problem, we need to calculate the total entropy change (ΔS) of the system during the isothermal expansion of one mole of an ideal monoatomic gas.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial volume (V1) = 1 L
- External pressure (P_ext) = 1 atm
- Initial temperature (T) = 300 K
- R = 8.3 J mol⁻¹ K⁻¹

2. **Understand the Process:**
- The gas expands isothermally against a constant external pressure until its final pressure equals the external pressure. Since the process is isothermal, the temperature remains constant throughout the process.

3. **Use the Formula for Entropy Change:**
- The formula for the change in entropy (ΔS) for an isothermal process is given by:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]
- Where:
- \( n \) = number of moles (1 mole in this case)
- \( V_f \) = final volume
- \( V_i \) = initial volume

4. **Determine Final Volume (V_f):**
- Since the gas expands isothermally against a constant external pressure of 1 atm, and the final pressure becomes equal to the external pressure, we can use the ideal gas law:
\[ P_1 V_1 = P_2 V_2 \]
- Here, \( P_1 = 1 \) atm, \( V_1 = 1 \) L, \( P_2 = 1 \) atm. Therefore:
\[ 1 \, \text{atm} \times 1 \, \text{L} = 1 \, \text{atm} \times V_f \]
- This implies that \( V_f = V_1 = 1 \) L.

5. **Calculate Entropy Change (ΔS):**
- Now substituting \( V_f \) and \( V_i \) into the entropy change formula:
\[ \Delta S = 1 \times 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \ln\left(\frac{1 \, \text{L}}{1 \, \text{L}}\right) \]
- Since \( \ln(1) = 0 \):
\[ \Delta S = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \times 0 = 0 \]

6. **Final Result:**
- The total entropy change of the system during this isothermal expansion is:
\[ \Delta S = 0 \, \text{J K}^{-1} \]