Difference between revisions of "PChem312 f20 w13"
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:::V<sub>i</sub>T<sub>i</sub> --> V<sub>f</sub>T<sub>i</sub> | :::V<sub>i</sub>T<sub>i</sub> --> V<sub>f</sub>T<sub>i</sub> | ||
| − | :[[File:Screen Shot 2020-11-11 at 7.31.58 AM.png| | + | :[[File:Screen Shot 2020-11-11 at 7.31.58 AM.png|140px]] |
2) Constant P, V temperature change | 2) Constant P, V temperature change | ||
:::V<sub>i</sub>T<sub>i</sub> --> V<sub>i</sub>T<sub>f</sub> | :::V<sub>i</sub>T<sub>i</sub> --> V<sub>i</sub>T<sub>f</sub> | ||
:::P<sub>i</sub>T<sub>i</sub> --> P<sub>i</sub>T<sub>f</sub> | :::P<sub>i</sub>T<sub>i</sub> --> P<sub>i</sub>T<sub>f</sub> | ||
| + | |||
| + | :[[File:Screen Shot 2020-11-11 at 7.37.17 AM.png|300px]] | ||
3) Changes in both V and T, constant P | 3) Changes in both V and T, constant P | ||
Revision as of 13:37, 11 November 2020
Monday, Nov 9, 2020
We will now highlight how to calculate entropy (ΔS) for four processes involving ideal gases:
1) Reversible Isothermal Expansion or Compression
- ViTi --> VfTi
2) Constant P, V temperature change
- ViTi --> ViTf
- PiTi --> PiTf
3) Changes in both V and T, constant P
- ViTi -->V fTf
4) Changes in both P and T, constant V
- ViTi -->V fTf
Example Problem 1
- (Please write out your complete solution and email to bes by next class period)
Consider a gas undergoing the following change under constant pressure (1.00 bar) conditions; calculate the change in entropy (ΔS): Vi = 10.0 L; Ti = 298 K; Tf = 398 K
Example Problem 2
- (Please write out your complete solution and email to bes by next class period)
Consider a gas undergoing the following change under constant volume (10.0 L) conditions; calculate the change in entropy (ΔS): Pi = 1.00 bar; Ti = 298 K; Tf = 398 K