If 2 text{ mole} of an ideal gas expand isoth | vedclass

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(C) Step $1$: Identify the given values.$n = 2 	ext{ mol}$, $T = 27^{\circ}C = 300 	ext{ K}$, $V_1 = 1 	ext{ dm}^3 = 10^{-3} 	ext{ m}^3$, $V_2 = 1

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$-34.46$

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(C) Step $1$: Identify the given values.$n = 2 	ext{ mol}$, $T = 27^{\circ}C = 300 	ext{ K}$, $V_1 = 1 	ext{ dm}^3 = 10^{-3} 	ext{ m}^3$, $V_2 = 1 	ext{ m}^3$, $R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$.Step $2$: Use the formula for isothermal reversible work: $W = -nRT \ln\left(\frac{V_2}{V_1}ight)$.Step $3$: Substitute the values: $W = -2 	imes 8.314 	imes 300 	imes \ln\left(\frac{1}{10^{-3}}ight)$.Step $4$: Calculate the natural logarithm: $\ln(1000) \approx 6.9078$.Step $5$: Calculate the final value: $W = -4988.4 	imes 6.9078 \approx -34446 	ext{ J} = -34.45 	ext{ kJ}$.The closest option is $-34.46 	ext{ kJ}$.

If $2 \text{ mole}$ of an ideal gas expand isothermally and reversibly at $27^{\circ}C$ from $1 \text{ dm}^3$ to $1 \text{ m}^3$, calculate the work done. [$R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$] (in $\text{ kJ}$)

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