When 5 moles of He gas expand isothermally and reversibly at $$300$$ K from $$10$$ litre to $$20$$ litre, the magnitude of the maximum work obtained is ______ J. [nearest integer] (Given : $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$ and $$\log 2 = 0.3010$$)

We need to find the magnitude of the maximum work done when 5 moles of He gas expand isothermally and reversibly.

Here $$n = 5$$ mol, $$T = 300$$ K, $$V_1 = 10$$ L, $$V_2 = 20$$ L, $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$ and $$\log 2 = 0.3010$$.

For an ideal gas expanding isothermally and reversibly, $$W = -nRT \ln\left(\dfrac{V_2}{V_1}\right)$$, so the magnitude of maximum work is $$|W| = nRT \ln\left(\dfrac{V_2}{V_1}\right)$$.

Since $$\ln\left(\dfrac{V_2}{V_1}\right) = 2.303 \times \log\left(\dfrac{V_2}{V_1}\right)$$, we have $$\ln\left(\dfrac{20}{10}\right) = 2.303 \times \log(2) = 2.303 \times 0.3010 = 0.69320$$.

Thus, $$|W| = nRT \times 2.303 \times \log\left(\dfrac{V_2}{V_1}\right) = 5 \times 8.3 \times 300 \times 2.303 \times 0.3010.$$ First, $$nRT = 5 \times 8.3 \times 300 = 12450 \text{ J},$$ and $$2.303 \times 0.3010 = 0.69320.$$ Finally, $$|W| = 12450 \times 0.69320 = 8630.3 \text{ J}.$$

The magnitude of the maximum work obtained is 8630 J.