One mole of an ideal gas expands isothermally and reversibly from 10 $$dm^3$$ to 20 $$dm^3$$ at 300 K. $$\Delta U$$, q and work done in the process respectively are: (Given: $$R = 8.3 J K^{-1} mol^{-1}$$, $$\ln 10 = 2.3$$, $$\log 2 = 0.30$$, $$\log 3 = 0.48$$)

We are given: one mole of an ideal gas expands isothermally and reversibly from $$10 \text{ dm}^3$$ to $$20 \text{ dm}^3$$ at $$300 \text{ K}$$.

Given: $$R = 8.3 \text{ J K}^{-1} \text{mol}^{-1}$$, $$\ln 10 = 2.3$$, $$\log 2 = 0.30$$, $$\log 3 = 0.48$$.

Step 1: Find $$\Delta U$$

For an ideal gas undergoing an isothermal process, the internal energy depends only on temperature. Since $$T$$ is constant:

Step 2: Find work done (W)

For isothermal reversible expansion, the work done by the gas is:

We need $$\ln\left(\frac{V_2}{V_1}\right) = \ln\left(\frac{20}{10}\right) = \ln 2$$.

Converting: $$\ln 2 = 2.303 \times \log 2 = 2.303 \times 0.30 = 0.6909$$

Substituting the values:

The negative sign indicates work is done by the system on the surroundings (expansion).

Step 3: Find heat (q)

From the first law of thermodynamics:

Since $$\Delta U = 0$$:

The positive sign indicates heat is absorbed by the gas from the surroundings.

Summary:

$$\Delta U = 0$$, $$q = 1.718 \text{ kJ}$$, $$W = -1.718 \text{ kJ}$$

Hence, the correct answer is Option D.