thermodynamics
DESCRIPTION
Thermodynamics. How Energy Is Transferred As Heat and Work. Animation Courtesy of Louis Moore. How Internal Energy can Increase. Heat can be added (Symbol Q) Mechanical Work can be done on the system(like a piston compressing gas) (Symbol W). The First Law. -. D U = Q + W - PowerPoint PPT PresentationTRANSCRIPT
How Internal Energy can Increase
Heat can be added (Symbol Q)Mechanical Work can be done on the system(like a piston compressing gas)
(Symbol W)
The First Law
U = Q + W U is internal energy +Q is heat added (absorbed) +W is net work done on system
The first law is a restatement of the law of conservation of energy
-
Sign Conventions
U = Q + W Work done on system is positive Heat leaving system is negative
Implication: When U = 0, Q = -W
Reminder: U = 3/2nRT = 3/2 NkT
Examples
(1) 1000 J of heat is added to the system and 500 j of work is done on the system. What is the change of internal energy?
(2) 300 J of heat escapes and 200 J of work is done by the system. What is the change of internal energy?
Answer: 1500 J
Answer: -500 J
Processes
1. Isothermal (no temperature change)T = U = 0 ; PV = constant2. Adiabatic (no heat flow in or out) Q = 0
3. Isobaric (no change in pressure)4. Isochoric (no change in volume)
Adiabatic Processes
No heat added or taken awayExamples: Quickly pushing down a bicycle pump Compression stroke in heat engine Expansion in power stroke
Why adiabatic – too fast for any appreciable heat to enter or leaveAlso if well insulated
PV Diagram for Adiabatic Process
Q = 0
Happens when system is well insulated or if process happens very quickly as in rapid expansion of gases in an internal combustion engine.
Expansions or Compressions?
If Q = 0 is U positive or negative? What about the temperature?
Compression, since V decreases
U and T positive
What happens in compression stroke of an engine?
Adiabatic since fastWork is done ON gasU increases since U = 0 + (W)T increasesIn diesel fuel-air
mixture ignites spontaneously
Work
If pressure constant W = -Fd = -PAd = -PVWork negative if expansionWork positive if compression
Remember new convention: work done ON is positive
Calculating Work P V
Work done by a gas equals area under PV curve
FIND THE WORK! Hint: 1 atm = 105 N/m2 ; 1l = 10-3 m3
Answer: About 250 Joules
Work Around a Closed Path
Equals area enclosed on PV diagramNegative for clockwise pathPositive for counter-clockwise path
First Law Example
An ideal gas is slowly compressed at constant pressure of 2.0 atm from 10L to 2L. With volume constant heat is added; pressure and temperature rise until temperature reaches its initial value.Find total work done on gas
Find total heat flow into gas
Find net change in U?
P
V
2atm
+1600 J
-1600 J
zero, since returns to initial value
Work Done in an Engine
0.25 moles of gas expands quickly and adiabatically against piston. T drops from 1150 k to 400K How much work is done?Hint: use U = Uf – Ui = 3/2nR(Tf – Ti)
Answer: 2300 J
Second Law of Thermodynamics
Heat flows naturally from hot to cold objectsNo device can simply transform heat to work(100% efficient engine impossible)A perfect refrigerator is impossibleThe total entropy(disorder) of a system and its environment increases in natural processes
Heat Engines
Heat input Q1 at a high temperature is partly transformed into work W. The waste heat Q2 is exhausted at a lower temperature
Efficiency
= useful work/ QH = W/ QH = (QH-QL)/QH
QH = W + QL ;QH is at operating temperature
Eff = 1 – QL/QH ; QL is exhaust heat
Eff ideal = 1 – TL/TH for Carnot EngineKelvin temperatures only
Examples
In one cycle of an engine 5000 j are released by burning and 3000 j are exhausted. What is efficiency?Eff = W/QH = 2/5
An ideal engine operates between 900 and 500 degrees. Find efficiencyEff = 1 – TL/TH = 4/9
Heat Engine Facts
Gasoline engines operate at 15-25% efficiencyDiesel engines operate at 30-50%. They operate at higher temperaturesMany other engine designs exist such as Sterling engine
Refrigerator
Heat engine operated in reverse
Q: Will your kitchen be hotter or colder with the refrigerator door open?
CP = Q2/W =
Q2/(Q1-Q2)
Coefficient of performance
W is work done by motor/compressor
Entropy
Change in entropy S of a system when heat Q is added to it is
S = Q/TAccording to Second Law of
Thermodynamics, the entropy of an isolated system never decreases.
Entropy Examples
Molecules do not migrate to one corner of roomDropped cup breaks; pieces do not spontaneously reassemble themselvesHammer hitting nail heats it. Hammer on top of nail does not spontaneously rise up as nail coolsMixtures do not separate by themselves