Download - Thermodynamics
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
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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