(a) Calculate the change in entropy when 1.00 mol of
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
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
This textbook survival guide was created for the textbook: University Physics, edition: 13. The full step-by-step solution to problem: 30E from chapter: 20 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: calculate, change, entropy, mol. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The answer to “(a) Calculate the change in entropy when 1.00 mol of” is broken down into a number of easy to follow steps, and 10 words. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 30E from 20 chapter was answered, more than 268 students have viewed the full step-by-step answer.