One mole of an ideal gas expands from an init | vedclass

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(B) For an isothermal process,the work done by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$ Given: Number of mol

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$1728$

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(B) For an isothermal process,the work done by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}ight)$ Given: Number of moles $\mu = 1 \; mol$ Gas constant $R = 8.31 \; J/mol \cdot K$ Temperature $T = 300 \; K$ Initial volume $V_1 = 10 \; L$ Final volume $V_2 = 20 \; L$ Substituting the values: $W = 1 	imes 8.31 	imes 300 	imes \ln\left(\frac{20}{10}ight)$ $W = 2493 	imes \ln(2)$ Using $\ln(2) \approx 0.693$: $W \approx 2493 	imes 0.693 \approx 1727.65 \; J$ Rounding to the nearest integer,we get $W \approx 1728 \; J$.

One mole of an ideal gas expands from an initial volume of $10 \; L$ to a final volume of $20 \; L$ at a constant temperature of $300 \; K$. Calculate the work done by the gas in $J$. $(R = 8.31 \; J/mol \cdot K)$ (in $; J$)

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