Calculate the work done when 5 moles of an ideal gas expand isothermally and reversibly from a volume of 2 L to 20 L at 27^ C . (Given R = 8.314 J K^ -1 mol^ -1 )

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Quantrex Academy · Accepted Answer

Correct answer: A. -28.72 kJ

For an isothermal reversible expansion of an ideal gas, the work done is given by:

W = -2.303 nRT \log \left( \dfrac{V_2}{V_1} \right)

Given:

n = 5 \text{ mol}

R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}

T = 27^{\circ}\mathrm{C} = 27 + 273 = 300 \text{ K}

V_1 = 2 \text{ L}

V_2 = 20 \text{ L}

Substituting the values into the formula:

W = -2.303 \times 5 \times 8.314 \times 300 \times \log \left( \dfrac{20}{2} \right)

W = -2.303 \times 12471 \times \log(10)

W = -28720.7 \text{ J}

W \approx -28.72 \text{ kJ}

Answer:

-28.72 \text{ kJ}

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