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(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by: $W = nRT \ln\left(\frac{V_2}{V_1}\right)$ Given: $W =

Vedclass · Accepted Answer

$3$

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(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by: $W = nRT \ln\left(\frac{V_2}{V_1}ight)$ Given: $W = 5.187 \ kJ = 5187 \ J$ $V_1 = 10 \ m^3$ $V_2 = 20 \ m^3$ $T = 300 \ K$ $R = 8.314 \ J \ mol^{-1} \ K^{-1}$ Substituting the values: $5187 = n 	imes 8.314 	imes 300 	imes \ln\left(\frac{20}{10}ight)$ $5187 = n 	imes 8.314 	imes 300 	imes 0.6931$ $5187 = n 	imes 1728.5$ $n = \frac{5187}{1728.5} \approx 3$ Therefore,the number of moles of gas used is $3$.

An ideal gas expands isothermally and reversibly from $10 \ m^3$ to $20 \ m^3$ at $300 \ K$,performing $5.187 \ kJ$ of work on the surroundings. Calculate the number of moles of gas used.

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