If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, $$w$$, is $$-x$$ J. The value of $$x$$ is (Given $$R = 8.314$$ J K$$^{-1}$$ mol$$^{-1}$$)

For an ideal gas undergoing a reversible isothermal expansion, the work done is given by the formula
$$w = -\,nRT \,\ln\!\left(\frac{V_f}{V_i}\right)$$

Step-1: Identify the data.
Number of moles, $$n = 5$$ mol
Temperature, $$T = 300$$ K
Initial volume, $$V_i = 10$$ L
Final volume, $$V_f = 100$$ L
Universal gas constant, $$R = 8.314$$ J K$$^{-1}$$ mol$$^{-1}$$

Step-2: Compute the volume ratio inside the logarithm.
$$\frac{V_f}{V_i} = \frac{100\ \text{L}}{10\ \text{L}} = 10$$

Step-3: Calculate the numeric factor $$nRT$$.
$$nRT = 5 \times 8.314 \times 300$$
$$= 5 \times 2494.2$$
$$= 12471.0 \text{ J}$$

Step-4: Evaluate the natural logarithm.
$$\ln 10 \;=\; 2.302585$$

Step-5: Substitute in the work formula.
$$w = -\,12471.0 \times 2.302585$$
$$w = -28720.8 \text{ J}$$

Step-6: Write the answer in the required form $$w = -x$$ J.
Hence, $$x \approx 28721$$.

Final answer: $$x = 28721$$