The work done when 2 , mol of an ideal gas ex | vedclass

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Question

(A) For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log \frac{V_2}{V_1}$ Given: $n = 2 \, mol$,$R = 8.3

Vedclass · Accepted Answer

$- 5.481$

Vedclass · Answer

(A) For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log \frac{V_2}{V_1}$ Given: $n = 2 \, mol$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$T = 300 \, K$,$V_1 = 5 \, L$,$V_2 = 15 \, L$. Substituting the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log \frac{15}{5}$ $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log(3)$ $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes 0.4771$ $W \approx -5481 \, J$ Since the question asks for the value in $kJ$: $W = -5.481 \, kJ$.

List-$I$ (Process)	List-$II$ (Conditions)
$A$. Isothermal process	$I$. No heat exchange
$B$. Isochoric process	$II$. Carried out at constant temperature
$C$. Isobaric process	$III$. Carried out at constant volume
$D$. Adiabatic process	$IV$. Carried out at constant pressure

The work done when $2 \, mol$ of an ideal gas expands from $5 \, L$ to $15 \, L$ at a temperature of $300 \, K$ is .............. $kJ$.

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