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=(n* \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=(4.20mole* (34.6KJ/mole))/(307.6K)=0.36183kJ/K=0.36183* 1000=361.83J/K`

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