At 273 K the maximum work done when pressure | vedclass

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Question

(C) For an isothermal,reversible expansion of an ideal gas,the work done is given by the formula: $W = -nRT \ln(P_1/P_2)$.Given:Mass of $H_2$ gas = $1

Vedclass · Accepted Answer

$-26.09 \ kJ$

Vedclass · Answer

(C) For an isothermal,reversible expansion of an ideal gas,the work done is given by the formula: $W = -nRT \ln(P_1/P_2)$.Given:Mass of $H_2$ gas = $10 \ g$.Molar mass of $H_2$ = $2 \ g \ mol^{-1}$.Number of moles $(n) = 10 \ g / 2 \ g \ mol^{-1} = 5 \ mol$.Temperature $(T) = 273 \ K$.Initial pressure $(P_1) = 10 \ atm$.Final pressure $(P_2) = 1 \ atm$.Gas constant $(R) = 8.3 \ J \ K^{-1} \ mol^{-1}$.Substituting the values:$W = -5 \ mol 	imes 8.3 \ J \ K^{-1} \ mol^{-1} 	imes 273 \ K 	imes \ln(10/1)$.$W = -5 	imes 8.3 	imes 273 	imes 2.303 	imes \log(10)$.$W = -11331.45 	imes 2.303 	imes 1 \ J$.$W \approx -26096 \ J = -26.096 \ kJ$.Since the work is done by the system,it is negative. Thus,the correct option is $C$.

Reaction	$\Delta H \ (kJ \ mol^{-1}) / \Delta S \ (J \ K^{-1} \ mol^{-1})$
$I$	$+98.0, +14.8$
$II$	$-55.5, -84.6$
$III$	$+28.3, -17.0$
$IV$	$-40.5, +24.6$
$V$	$+34.7, 0.0$

At $273 \ K$ the maximum work done when pressure on $10 \ g$ of hydrogen is reduced from $10 \ atm$ to $1 \ atm$ under isothermal,reversible conditions is (Assume the gas behaves ideally) $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

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