Problem Set #21: Entropy Analysis in Chemical Systems - Edubirdie

University:
The Pennsylvania State University
Course:
CHEM 450 | Thermodynamics
Academic year:

2024
Views:

30

Pages:

1
Author:

Alexis H.

Problem Set #21 Assigned: February 21, 2024 Due: Feb. 23, 2024 at 2:30 pm 1. A perfect crystal of benzene at 0 K is comprised of 1.00 mole of benzene molecules with 1 deuterium atom substituted for a hydrogen on each ring. What is the residual entropy for this system if we assume that all benzene molecules adopt the same configuration within the crystalline lattice? 2. The normal melting point of toluene (C7H8) is -95.0 °C and its boiling point is 110.6 °C. By contrast, n-heptane (C7H16) has a melting point of -90.5 °C and a boiling point of 98.4 °C. (A) If you have a 1 mole sample of each prepared at 100 °C and 1 bar, which sample would you expect to have the higher third law molar entropy? Explain your answer. (B) Next, both systems are heated to 800 °C and 1 bar under conditions where they remain intact. Which sample would you expect to have the higher third law molar entropy? Explain your answer. 3. A 2.00 mole sample of a perfect gas is originally placed into 4.00 L container at 291 K. The sample is allowed to expand to 10.0 L, while it is simultaneously heated to 405 K. What is the change in entropy for this system if Cv,m = 31.6 J/mol K? (Hint: Entropy is a state function.) 1 5 K In 2. (a) Toulene would have the higher entropy because it would be in the gaseous state while the n-heptane would still be in the liquid state. w 14 138 1023 w 6orientations (b) n-heptane would have the higher molar entropy because it has more moles/orientations than toulene does. 5 1.38 1023 In 6 5 2 47 10 23J k 3 ΔS Δ5 2 316 In 45 df UIPV It PLV Mcvd Nevin II pΔV 1 6

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