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: $\mathrm{R}=8.3 \mathrm{J}{ \mathrm{K}}^{-1}{ \mathrm{mol}}^{-1}$ and $\log 2=0.3010$ )

The magnitude of the maximum work obtained is 8630 J.

To calculate the maximum work obtained when a gas expands isothermally and reversibly, we can use the formula for work done in an isothermal expansion of an ideal gas:

W=−nRTln(V2/V1​​)

Where:

• W is the work done (in joules),
• n is the number of moles of gas,
• R is the gas constant (8.314 J/mol·K),
• T is the temperature in Kelvin,
• V2​ is the final volume,
• V1 is the initial volume.

$\mathrm{n}=5 \mathrm{mol}$
$\mathrm{T}=300 \mathrm{K}$
${\mathrm{V}}_1=10 \mathrm{L}$
${\mathrm{V}}_2=20 \mathrm{L}$
$\mathrm{W}=-\mathrm{nRT}\ell \mathrm{n}\frac{{\mathrm{V}}_2}{{ \mathrm{V}}_1}$
$=-5\times 8.3\times 300\times \ell \mathrm{n}\frac{20}{10}$
$=-8630.38 \mathrm{J}$