Successfully reported this slideshow.
Upcoming SlideShare
×

# Ch8 - thermodynamics

1,432 views

Published on

This is a basic, basic power point on the Laws of Thermodynamics for Conceptual Physics students.

Published in: Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

### Ch8 - thermodynamics

1. 1. Thermodynamics Chapter 10
2. 2. 0th Law of ThermodynamicsIf system A is in thermal equilibrium with system B, andsystem B is in thermal equilibrium with system C, thensystem A must be in thermal equilibrium with system C.
3. 3. 1 Law of Thermodynamics stConservation of Energy for Heat!Involves three energies: Work, Heat, Internal Energy Internal energy = heat added to the system + the work done on the system. ∆U = Q + W Sign Conventions: Heat added + Work on system + Heat lost – Work by system –
4. 4. U, Q, and WU = Internal energy  depends on change in temperatureQ = Heat  depends on transfer of heat energyW = Work  W = F•d = P A d = P ∆V (P = F/A)
5. 5. 4 Different EventsIsobaric  Constant PressureWork is done by expanding volume Example: ∆U = Q + W Brick on top of sealed ∆U = Q + P∆V canister. Same force over same area ∴ pressure doesn’t change.
6. 6. Isochoric (isometric)  Constant volumeAll heat changes intointernal energyIf ∆V = 0, then W = 0 ∆U = Q + W ∆U = Q + 0 Examples: Inside Pressure Cooker, ∆U = Q Mist above soda. Volume stays the same. Pressure rapidly decreases. Temperature rapidly decreases causing the gas to be pushed out
7. 7. Isothermal Constant TemperatureHeat is converted tomechanical workIf ∆T = 0, then ∆U = 0 ∆U = Q + W Example: 0=Q+W Boiling Water Q = –W
8. 8. Adiabatic (GreekAdiabatos: Impassable)  No heat transferSystem is extremely wellinsulated or processhappens so fast that heatdoesn’t have time to flowin or outIf no heat transfer, Q = 0 Example: Stretching rubber band∆U = Q + W quickly – Not enough time∆U = 0 + W for heat transfer, so the∆U = W work done goes into internal energy.
9. 9. Sample 2The internal energy of the gas in a gasolineengine’s cylinder decreases by 195 J. If 52.0 J ofwork is done by the gas, how much energy istransferred as heat? Is this energy added to orremoved from the gas?∆U = Q + W-195 = Q + -52.0Q = -143 J  because it is negative it is removed
10. 10. 2nd Law of ThermodynamicsHigh Temp Engine WorkLow TempHeat Engines  any device that changes heat energy to mechanical energy
11. 11. Entropy (S)  measure of the disorder of a systemEntropy of a system tends to increase. (Become moredisordered)Another statement of the 2nd Law Natural processes tend to move toward a state ofgreater entropy.R.J. Clausius (German physicist, 1822-1888)Said 2nd Law deals with the direction a process will goClausius’ Statement of 2nd Law of Thermo.  Heat flows from hot  cold
12. 12. Most general form:Natural processes tend to have a preferred direction in which they tend to move. ie. An apple doesn’t jump up to a tree, hot water does not get cold over a fire, etc.3rd Law of ThermodynamicsIt is impossible to reach Absolute Zero.