Question

Calculate the change in entropy that occurs in the system when 4.20 mole of diethyl ether (\(\rm C_4H_6O\)) condenses from a gas to a liquid at its normal boiling point (\(34.6^{\circ} \rm C\)). \(\Delta H_{vap}\) = 26.5 \(\rm kJ/mol\)

Answer 1

Answer : The entropy change of the system is, 361.83 J/K

Solution :

Formula used :

$\Delta S=\frac{n\times \Delta H_{vap}}{T_b}$

where,

$\Delta S$ = entropy change of the system = ?

$\Delta H$ = enthalpy of vaporization = 34.6 kJ/mole

n = number of moles of diethyl ether = 4.20 mole

$T_b$ = normal boiling point = $26.5^oC=273+26.5=307.6K$

Now put all the given values in the above formula, we get the entropy change of the system.

$\Delta S=\frac{4.20mole\times (34.6KJ/mole)}{307.6K}=0.36183kJ/K=0.36183\times 1000=361.83J/K$

Therefore, the entropy change of the system is, 361.83 J/K