ThermodynamicsThermodynamics

Chapter 10Chapter 10

00thth Law of Thermodynamics Law of ThermodynamicsIf system A is in thermal equilibrium with system B, and If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then system B is in thermal equilibrium with system C, then system A must be in thermal equilibrium with system C.system A must be in thermal equilibrium with system C.

11stst Law of Thermodynamics Law of Thermodynamics

Conservation of Energy for Heat!Conservation of Energy for Heat!

Involves three energies:Involves three energies:Work, Heat, Internal EnergyWork, Heat, Internal Energy

Internal energy = heat added to the Internal energy = heat added to the system + the work done on the system. system + the work done on the system.

∆∆U = Q + WU = Q + W Sign Conventions:Sign Conventions:Heat added +Heat added + Work on system +Work on system +

Heat lost –Heat lost – Work by system –Work by system –

U, Q, and WU, Q, and W

U = Internal energy U = Internal energy depends on change in temperature depends on change in temperature

Q = Heat Q = Heat depends on transfer of heat energy depends on transfer of heat energy

W = Work W = Work W = F•d = P A d = P ∆V W = F•d = P A d = P ∆V

(P = F/A)(P = F/A)

4 Different Events4 Different Events

Isobaric Isobaric Constant Pressure Constant Pressure

Work is done by Work is done by expanding volumeexpanding volume

∆∆U = Q + WU = Q + W

∆∆U = Q + P∆VU = Q + P∆V

Example:Brick on top of sealed canister. Same force over same area pressure doesn’t change.

Isochoric (isometric) Isochoric (isometric) Constant volume Constant volume

All heat changes into All heat changes into internal energyinternal energy

If ∆V = 0, then W = 0If ∆V = 0, then W = 0 ∆ ∆U = Q + W U = Q + W ∆ ∆U = Q + 0U = Q + 0 ∆ ∆U = QU = Q

Examples:Inside Pressure Cooker, Mist above soda. Volume stays the same. Pressure rapidly decreases. Temperature rapidly decreases causing the gas to be pushed out

Isothermal Constant Temperature

Heat is converted to mechanical work

If ∆T = 0, then ∆U = 0∆∆U = Q + WU = Q + W 0 = Q + W Q = –W

Example:Boiling Water

Adiabatic (Greek Adiabatos: Impassable) No heat transfer

System is extremely well insulated or process happens so fast that heat doesn’t have time to flow in or outIf no heat transfer, Q = 0

∆∆U = Q + WU = Q + W ∆U = 0 + W∆U = W

Example:Stretching rubber band quickly – Not enough time for heat transfer, so the work done goes into internal energy.

Sample 2

The internal energy of the gas in a gasoline engine’s cylinder decreases by 195 J. If 52.0 J of work is done by the gas, how much energy is transferred as heat? Is this energy added to or removed from the gas?

∆∆U = Q + WU = Q + W

-195 = Q + -52.0

Q = -143 J because it is negative it is removed

22ndnd Law of Thermodynamics Law of Thermodynamics

Heat Engines Heat Engines any any device that changes device that changes heat energy to heat energy to mechanical energymechanical energy

High Temp

Low Temp

Engine Work

Entropy (S) measure of the disorder of a system

Entropy of a system tends to increase. (Become more disordered)

Another statement of the 2nd Law Natural processes tend to move toward a state of greater 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

Most general form:Most general form:

Natural processes tend to have a preferred Natural processes tend to have a preferred direction in which they tend to move. 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.

33rdrd Law of Thermodynamics Law of Thermodynamics

It is impossible to reach Absolute Zero.It is impossible to reach Absolute Zero.