What is the entropy change of the system when 17.5 g of liquid benzene (c6h6) evaporates at the normal boiling point? the normal boiling point of benzene is 80.1°c and δh vap is 30.7 kj/mol.

Answer : The entropy change of the system is, 19.5 J/K

Solution :

First we have to calculate the moles of benzene.

Now we have to calculate the entropy change of the system.

Formula used :

where,

= entropy change of the system = ?

= enthalpy of vaporization = 30.7 kJ/mole

n = number of moles of benzene  = 0.224 mole

= normal boiling point of benzene =

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

Therefore, the entropy change of the system is, 19.5 J/K

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: