Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- First Law of Thermodynamics by Sean Clark Luinor... 2828 views
- Thermodynamics by Melchor Potestas 16270 views
- The first law of thermodynamics by Dr. Tanuja Nautiyal 581 views
- Laws Of Thermodynamics by k v 33421 views
- Thermodynamic Chapter 3 First Law O... by Muhammad Surahman 57551 views
- Thermodynamics by thirunavukk arasu 26533 views

1,796 views

Published on

Lecture 15 first law of thermodynamics

No Downloads

Total views

1,796

On SlideShare

0

From Embeds

0

Number of Embeds

2

Shares

0

Downloads

123

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Lecture 15 First law of thermodynamics.
- 2. Internal energy State variables: p, T, V They only depend on the “state” of the system They do NOT depend on the “history” Total energy of the molecules is also a state variable: Internal energy: U = KE all + PE all KE : kinetic energy of atoms, random motion PE : interactions, e.g. attraction between molecuses For ideal gases, no interaction: U = KE all
- 3. Internal energy for some systems Nk = nNAk = nR Monoatomic ideal gas: 3 U = KE all = N kT ÷ 2 3 U = nRT 2 Diatomic ideal gas: 5 U = KE all = N kT ÷ 2 U = Monoatomic solid crystal: U = KE all + PE all 3 = N kT ÷× 2 2 5 nRT 2 U = 3nRT In general, internal energy can also depend on p, V. But it never depends on the history of the system.
- 4. Changes in internal energy Temperature change is associated to heat transfer. Change in internal energy transfer of energy We already know that: When heat is absorbed/released by a system, its internal energy increases/decreases. Other ways of transferring energy? Work!! (we learned it in 221)
- 5. Work done by gas volume change A gas with pressure p expands by pushing a piston by a distance dx p Force by gas on piston F = pA (A = area of the piston) dx Work by gas: dW = pAdx = pdV As volume goes from Vi to Vf Vf W = ∫ pdV Vi
- 6. Work is the area under pV curve But be careful with the sign! Expansion (Work done by gas) > 0 Compression (Work done by gas) < 0
- 7. ACT: Three processes - Work A gas can go from state 1 to state 2 through three different processes. In which process does gas do the least amount of work ? A C B D: It’s the same for all three
- 8. Even if the initial and final states are the same, work depends on the path taken by the process A B W is NOT a state variable. It is only defined when the state changes. C
- 9. ACT: Three processes – Internal energy In which case does the internal energy of the gas change the most ? A D: It’s the same for all three C B U is a state function and does not depend on the process. ΔU = U1 − U2 for all processes.
- 10. In-class example: Work in closed cycle Which of these processes represents the most work done by the system per cycle? P P P V V A C V E P P Work = area inside cycle. Note that WE < 0 V V B D CW ⇔ W > 0 CCW ⇔ W < 0
- 11. First law of thermodynamics If heat is absorbed by a system, Q>0 ΔU > 0 If heat is released by a system, Q<0 ΔU < 0 W>0 ΔU < 0 If work is done on the gas (compression), W < 0 ΔU > 0 If work is done by the gas (expansion), ∆U = Q −W Remember the sign convention! W > 0 work done by system Q > 0 heat absorbed by system
- 12. Example: Cyclic process A gas (not necessarily ideal) goes through the cycle shown in the pV diagram below. A Process 2 3 Process 1 B Pr oc es s P Data: C VA = 2.0 m3 VC = 4.0 m3 PA = 1.0 × 105 Pa PC = 2.0 × 105 Pa a) Determine the work done by the gas in each of the three parts of the cycle. V 1 (A to B): W1 = 0 (constant volume) 5 2 (B to C): W2 = p ∆V = pB (VC −VB ) = 2 × 10 J 5 3 (C to A): W2 = − ( yellow area ) = −3 × 10 J
- 13. P A C 3 Process 2 Pr oc es s Process 1 B Data: V VA = 2.0 m3 VC = 4.0 m3 PA = 1.0 × 105 Pa PC = 2.0 × 105 Pa b) For the entire cycle, what are the work done by the gas, the change in internal energy of the gas and the heat exchanged with the surroundings? Is this heat absorbed or released by the gas? Wcycle = W1 +W2 +W3 = −1 × 105 J ∆Ucycle = Uf − Ui = UA − UA = 0 ΔUclosed cycle = 0 Qcycle = ∆Ucycle +Wcycle = −1 × 105 J Q < 0: heat is released by the gas
- 14. Overall, in this cycle: Wcycle < 0 Qcycle < 0 work Work done on the system. System releases heat. system P B C heat A We do work on the system, we obtain heat. (This could be used to warm up a room…) Reverse cycle (A→C →B): Wcycle > 0 Work done by the system. Qcycle > 0 System absorbs heat. work system heat System absorbs heat and produces work. (Like some kind of steam motor…) V P B C A V

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment