One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $$\Delta S$$ for the process in $$JK^{-1}\,mol^{-1}$$ is [$$\ln 2 = 0.693$$ and $$R = 8.314$$ J/(mol K)]

For an ideal gas undergoing an isothermal (constant-temperature) reversible expansion, the entropy change is given by

$$\Delta S = nR \ln\!\left(\frac{V_2}{V_1}\right)$$

because temperature is constant and, for an ideal gas, $$PdV = nRT \frac{dV}{V}$$ integrates to the above form.

Here the gas expands until its pressure becomes one-half of the initial pressure while temperature remains fixed. For an ideal gas, $$PV = \text{constant}\times T$$ at constant $$T$$, so

$$P_1V_1 = P_2V_2 \quad\Longrightarrow\quad V_2 = V_1\frac{P_1}{P_2} = V_1\left(\frac{P_1}{\tfrac{1}{2}P_1}\right) = 2V_1$$

Hence $$\dfrac{V_2}{V_1} = 2$$.

Substituting in the entropy formula for one mole $$\left(n = 1\right)$$:

$$\Delta S = (1)\times 8.314 \,\text{J mol}^{-1}\text{K}^{-1}\; \ln 2$$

Using $$\ln 2 = 0.693$$,

$$\Delta S = 8.314 \times 0.693 = 5.76 \,\text{J K}^{-1}\text{ mol}^{-1}$$

Therefore, the entropy change is $$5.76 \,\text{J K}^{-1}\text{ mol}^{-1}$$.

Option B which is: 5.76