6 moles of an ideal gas expand isothermally a | vedclass

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Question

(D) The work done in an isothermal reversible expansion is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$Using the base-$10$ logarithm: $W = -2

Vedclass · Accepted Answer

$34.465$

Vedclass · Answer

(D) The work done in an isothermal reversible expansion is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$Using the base-$10$ logarithm: $W = -2.303 	imes nRT \log(\frac{V_2}{V_1})$Given values: $n = 6 \, mol$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$T = 27 + 273 = 300 \, K$,$V_1 = 1 \, L$,$V_2 = 10 \, L$.Substituting the values: $W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes \log(\frac{10}{1})$$W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes 1$$W = -34464.8 \, J = -34.465 \, kJ$The magnitude of the maximum work done is $34.465 \, kJ$.

$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1$ litre to a volume of $10$ litres at $27\,^oC$. What is the maximum work done (in $, kJ$)?

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