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(a) Calculate the change in entropy when 1.00 mol of
Chapter 20, Problem 30E(choose chapter or problem)
(a) Calculate the change in entropy when 1.00 mol of water (molecular mass 18.0 g / mol ) at \(100^{\circ} \mathrm{C}\) evaporates to form water vapor at \(100^{\circ} \mathrm{C}\).
(b) Repeat the calculation of part (a) for 1.00 mol of liquid nitrogen, 1.00 mol of silver, and 1.00 mol of mercury when each is vaporized at its normal boiling point. (See Table 17.4 for the heats of vaporization, and Appendix D for the molar masses. Note that the nitrogen molecule is \(N_2\).)
(c) Your results in parts (a) and (b) should be in relatively close agreement. (This is called the rule of Drepez and Trouton.) Explain why this should be so, using the idea that entropy is a measure of the randomness of a system.
Questions & Answers
QUESTION:
(a) Calculate the change in entropy when 1.00 mol of water (molecular mass 18.0 g / mol ) at \(100^{\circ} \mathrm{C}\) evaporates to form water vapor at \(100^{\circ} \mathrm{C}\).
(b) Repeat the calculation of part (a) for 1.00 mol of liquid nitrogen, 1.00 mol of silver, and 1.00 mol of mercury when each is vaporized at its normal boiling point. (See Table 17.4 for the heats of vaporization, and Appendix D for the molar masses. Note that the nitrogen molecule is \(N_2\).)
(c) Your results in parts (a) and (b) should be in relatively close agreement. (This is called the rule of Drepez and Trouton.) Explain why this should be so, using the idea that entropy is a measure of the randomness of a system.
ANSWER:Solution 30E a) The way we have defined temperature is, dS = 1 --------------------(1) dQ T dS = dQ / T . So the change in entropy is the change in heat divided by the boiling point. For water, T = 100 C = 373 K . The heat required to vaporize 1 mol of liquid water is, Q = mL =V18 g × 2265000 J /kg = 0.018 × 2265000 = 40.77 KJ = 40770 jules. The change in entropy is, dS = 40770 / 373 = 109.30 b) Step 1: For liquid nitrogen, Boiling temperature is T = 77.34 K . Heat required to vaporize 1 mol, Q = mL = 28g × 201000 J/kg = 0.028 × 201000 = 5628 jules. V So, the change in entropy is, dS = 5628 / 77.34 = 73.09