Problem Set #17 : Isothermal Expansion Entropy Analysis

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

2024
Views:

161

Pages:

3
Author:

Alexis H.

Problem Set #17 Assigned: February 14, 2024 Due: Feb. 16, 2024 at 2:30 pm 1. Find the change in entropy of a system that occurs when 15 g of CO2 are expanded isothermally and reversibly from 1.0 L to 3.0 L at 300 K. 2. In question #1, you were asked to find the change in entropy when the expansion occurs isothermally and reversibly. Consider the same expansion, but this time it takes place against zero external pressure (i.e., free expansion). What is the change in entropy of the system? Keep in mind that entropy is a state function, and the change in entropy of the system needs to be calculated along the reversible path. 3. One mole of a perfect gas at 27 °C is allowed to expand isothermally from a pressure of 3.00 atm down to 1.00 atm three different ways: (a) Perfectly reversibly, where maximum work is done, (b) against a constant external pressure of 1.00 atm, and (c) against zero external pressure (free sys surr tot for each of the three cases. for the three different cases? Explain what this means. (Note: I will go over the meaning of this result in Friday’s lecture). sys 1 1 40 Δ0 q w Δ5 85 Δ5 007 D8 Δ0 Cudt Cudt Pdv ΔS nRIn Y Δ5 n 19no 0.34mol 8.314 monk In Δ5 3.105 g molik n dT Δ5 0 W pdV ΔS gtfo 40 9 9 ΔU w Δg tot differ surr 0.34no ΔU nCuT o 3 a ΔU W pdv ΔH O NRTIn NRTIn 1m01 8.314 300 in W 2740 9 w 9 2740 ΔS o ΔSsys b w ΔSsys ΔSsys Δ5sys 9.13 0 9.13 Ussurr 3 71 9.13 Pex IV PexMRTdp NRT P 1m01 w 9 9 8314 1660 300 1 13 w 1660 Δ5 ΔSson 1888 ΔStot Pey 5.53 SsystSson Stot 3.60 Ssys 9 13 5 53 Sson 9.13 5.53 3.60 c Δs nrin PE 1mol 8.314 In ΔSsys 9.13 ΔSsorr 9.13 Δstot 0 4 The common factor is that for most of the S totals the answer is 0, meaning that the total change in entropy is 0 for each of them. The change in the entropy of the system equals the change of entropy in the surroundings. This shows that it doesn’t matter how the work is done for entropy, the result will still be the same.