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

2024
Views:

232

Pages:

3
Author:

Alexis H.

Problem Set #18 Assigned: February 16, 2024 Due: Feb. 19, 2024 at 2:30 pm 1. Calculate the increase in molar entropy for O2(g) when the temperature is increased from 298 K to 348 K at constant pressure. Note that Cp,m = 29.355 J/mol K for O2(g) and that this value remains constant over the entire temperature range. 2. In problem #1 you were asked to find S by assuming that Cp,m remained constant over the entire range. In reality the vibrational mode participate to a greater extent as the molecule is heated. Should the actual change in entropy be smaller or larger than the one you calculated when heating over the identical temperature range? Please explain your answer. 3. Two copper blocks, each of mass 1.00 kg, are placed in thermal contact with each other inside an isolated container. The first block is initially at 50 °C and the second one is at 0 °C. Calculate the final temperature of the two blocks, the change in entropy of each block, and the total entropic change. Note that the heat capacity of copper is 0.385 J/K g. 4. In problem #3 you found Stot. Explain why the value that you determined is consistent with the Clausius inequality, dS dq/T. 5. Two copper blocks are placed in thermal contact with each other inside an isolated container. The first block has a mass of 2.75 kg and is initially at 50 °C. The second block has a mass of 0.81 kg and is initially at 0 °C. Calculate the final temperature of the two blocks, the change in entropy 6. Consider an ideal heat engine. 10.00 kJ of heat is withdrawn from the hot source at 272 K and 3.00 kJ of work is done. What is the temperature of the cold sink? 7. What is the efficiency of an ideal heat engine when the hot source operates at 100 °C and the cold sink is at 10 °C? 1 Δ5 pin Rin Δ5 Cpin 29355 in Δ5 4 554m01 k 3 Q MCΔT Q Q 38 Q block1 T 323K 0 MC Tf Tic MC Tica block2 T 273K 2 It would be larger because the more the vibrational modes participate the higher the Cp,m value would be leading to a larger change in entropy MC Tf Tic Mc 1 0.385 Tf 323 110.385 Tf 273 0.385 Tf 273 124.355 0.385T 0.77Tt D 0 105.105 0 229.46 0 0.77T 2 75kg m 1 0.385 298 273 Q 9.625 92 5 ΔS block2 T 323K Q Q Q 2.75 0.385 Tf 323 0.0324k Δ52 9 625 298 ΔS 0.03201k Δ5 o 0.032 0.032 04k ΔS o m 0.81kg T 273 K D Q 2.75 0.385 Tf 323 0.81 0.385 Tf 273 0 1 05875 Tt 323 0.31185 Tf 273 0 10587511 341.98 0.31185T 85.14 0 427.12 D 137067 427.12 Tf 312 MCDT Q1 Q 0.81 0.385 Tf 273 1 3706T Q MCLT 229.46 It is consistent because the delta S value is 0 which is greater than that of S1 because that value is negative and is equal to the S2 value which is what the equality tells us it is supposed to be, greater than or equal to. 5 block1 9.625J Δs Tf 298k 4 0 85 298 323 Q Tiggy 0.385 Tf 323 0.385T I Tf 2.75 0.385 312 323 11.65 Q MCLT 02 Δs ΔS Δsto 0.81 0.385 312 273 12.16 5,625 0.037 0.039 0.037 0.039 0.002 6 If 2 1 N To 10 C 283K Th 1 1 I3 1 0.3 3 I 7 2 1 0.7 To 191.1K 100 C 373K 2 1 N 0.24 or 241