Question

calculate the molar entropy change when (a) water and (b) benzene are evaporated at their boiling points at a pressure of 1 atm. what are the entropy changes in (i) the system, (ii) the surroundings, and (iii) the universe, in each case?

Answer 1

At the boiling temperature, temperature and the pressure is constant. The only heat of transformation is the latent heat of vapourization. latent heat of vapourization of water = 40.66 kJ/mol

S = q/T = n *dH/T = 1 *40.66 kJ/mol/373.15 = 0.12 kJ/K

S (sys) = 0.12 kJ/K. No heat is transferred to the environment. So, S(surr) =0

S (univ )= 0.12 kJ/mol + 0 = 0.12kJ/mol

-----------------------------------------------------------

Latent heat of vaporization of benzene = 30.8 kJ/mol and boiling point = 80.1 oC = 353.1 K

S (sys) = q/T = 1mol *30.8 kJ/mol/ 353.1 K =0.087 kJ/K

S (surr) = 0 as no heat is transferred to the surrounding.

S (univ) = S (sys)+ S (surr) = 0.087 kJ/mol