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6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at $27^{\circ} \mathrm{C}$. What is the maximum work done?
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The correct answer is:
$34.465 \mathrm{~kJ}$
$\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
Given $\mathrm{n}=6, \mathrm{~T}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$
$\mathrm{~V}_{1}=1 \mathrm{~L}, \mathrm{~V}_{2}=10 \mathrm{~L}$
$\therefore \mathrm{W}=-2.303 \times 6 \times 8.314 \times 300 \log \frac{10}{1}$
$\quad=34.465 \mathrm{~kJ}$
Given $\mathrm{n}=6, \mathrm{~T}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$
$\mathrm{~V}_{1}=1 \mathrm{~L}, \mathrm{~V}_{2}=10 \mathrm{~L}$
$\therefore \mathrm{W}=-2.303 \times 6 \times 8.314 \times 300 \log \frac{10}{1}$
$\quad=34.465 \mathrm{~kJ}$
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