Question

# calculate the molar entropy change when (a) water and (b) benzene are evaporated at their boiling...

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?

#### Homework Answers

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

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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

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