What is the work done when 2 text{ mole} of a | vedclass

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(D) Given: $n = 2 	ext{ mol}$,$V_{1} = 5 	ext{ m}^{3}$,$V_{2} = 10 	ext{ m}^{3}$,$T = 300 	ext{ K}$,$R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.

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$-3.458$

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(D) Given: $n = 2 	ext{ mol}$,$V_{1} = 5 	ext{ m}^{3}$,$V_{2} = 10 	ext{ m}^{3}$,$T = 300 	ext{ K}$,$R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log_{10} \frac{V_{2}}{V_{1}}$.Substituting the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} \frac{10}{5}$.$W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} 2$.Since $\log_{10} 2 \approx 0.3010$,we get: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes 0.3010$.$W \approx -3457.97 	ext{ J} = -3.458 	ext{ kJ}$.

What is the work done when $2 \text{ mole}$ of an ideal gas are expanded isothermally and reversibly from $5 \text{ m}^{3}$ to $10 \text{ m}^{3}$ at $300 \text{ K}$ (in $\text{ kJ}$)? $(R=8.314 \text{ J K}^{-1} \text{ mol}^{-1})$

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