When 5 mol of He gas expand isothermally and | vedclass

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(D) For an isothermal reversible expansion,the maximum work $(W_{max})$ is given by the formula: $W_{max} = -nRT \ln \frac{V_2}{V_1} = -2.303 nRT \log

Vedclass · Accepted Answer

$8630$

Vedclass · Answer

(D) For an isothermal reversible expansion,the maximum work $(W_{max})$ is given by the formula: $W_{max} = -nRT \ln \frac{V_2}{V_1} = -2.303 nRT \log \frac{V_2}{V_1}$ Given: $n = 5 \ mol$,$T = 300 \ K$,$V_1 = 10 \ L$,$V_2 = 20 \ L$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\log 2 = 0.3010$ Substituting the values: $W_{max} = -2.303 	imes 5 	imes 8.3 	imes 300 	imes \log \frac{20}{10}$ $W_{max} = -2.303 	imes 5 	imes 8.3 	imes 300 	imes 0.3010$ $W_{max} = -8630.38 \ J$ The magnitude of the work is $|W_{max}| = 8630.38 \ J$. Rounding to the nearest integer,we get $8630 \ J$.

When $5 \ mol$ of $He$ gas expand isothermally and reversibly at $300 \ K$ from $10 \ L$ to $20 \ L,$ the magnitude of the maximum work obtained is $....... \ J$. [nearest integer] (Given: $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ and $\log 2 = 0.3010$)

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