Download Internal Energy and the State of a System A system (e.g., a steam

WORK IN THE FIRST LAW OF THERMODYNAMICS
Learning Objectives
After you complete the homework associated with this
lecture, you should be able to:
• Explain what is meant by “state variables”;
• Explain why work done by a gas is called “PdV” work;
• Calculate the work done by a gas as it is compressed or
expanded (e.g., in a piston engine);
• Describe what are meant by isothermal, isobaric,
isochoric, and adiabatic processes;
• Discuss how the First Law of Thermodynamics
describes the conservation of energy in all its forms;
• Use the First Law of Thermodynamics to calculate the
transformation of energy and flow of heat energy in
various processes.
Internal Energy and the State of a System
The state of a system refers to the quintessential
properties fully define a system’s characteristics.
Internal energy U is part of the description of the state
of a system – the energy it has by just sitting there, not
macroscopically moving. Its main constituents are:
C Random motions of atomic particles (temperature)
C Interatomic and intermolecular potential energies
State variables are the quantities that specify the essential
internal state of a system:
C Temperature
C Pressure
C Volume
C Phase or internal structure (gaseous, crystalline, etc.)
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WORK DONE IN A THERMODYNAMIC PROCESS
WORK DONE BY A GASEOUS SYSTEM
A system (e.g., a steam engine) can do work on its
surroundings. We have a feeling that some type of
internal energy may be extracted or converted to do this.
BUT WE MUST BE UNAMBIGUOUS !
Wdone BY system =
IFPsys C drP
= IFsys dx =
I(PA)dx
= IP(A dx)
Work done by a system on the outside equals the
negative of the work done by the outside on a system.
Wdone BY system = S Wdone ON system
Sound confusin’ ? (It is.) We will use Wdone BY system
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This is called “P dV work” – know this phrase.
Note: Pressure depends upon Temperature and Volume.
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The work done by a gas is the “area” under the P vs. V
graph along the “path” from initial state to final state.
Work(done by gas)
Vf
'
m
P(T,V ) dV .
Ff =Pf A
VB
Fi =Pi A
Vi
expansion
ΔV > 0
compression
ΔV < 0
Work A 6 B '
m
P dV
VA
' ½ ( PA% PB ) @ ( VB& VA)
Because pressure P is always positive,
Y W > 0 if gas expands (ΔV > 0) FOR ANY REASON
Y W < 0 if gas compresses/contracts (ΔV < 0)
: Plumber’s Plunger Piston
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Consider an isothermal (constant temperature T) process in
an ideal gas: P V = n R T Y PV = constant
In expansion, ΔV is positive Y W = +Area (positive)
Important
Important: We determine sign of W from sign of ΔV.
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There are many paths to changing volume, pressure, and
temperature. Some of the most common processes are:
Isothermal (same temperature): constant T
Isobaric (same pressure): constant P
Isochoric (same volume): constant V
Adiabatic (no heat exchange): Q = 0
Fast processes are often adiabatic because they are over
But if gas contracts or is compressed FOR ANY REASON,
before significant heat energy is exchanged (Q . 0) with the
then ΔV is negative Y
surrounding environment through slower heat transfer rates;
W = S Area (negative)
for example: kicking a soccer ball.
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FIRST LAW OF THERMODYNAMICS
The internal energy of a system can change as
energy flows into or out of it.
C The change in internal energy of a system ΔU is the
additional energy it retains after some process in
which it receives heat Qnet in from its surroundings
and/or does work Wdone by system on its surroundings
ΔUsystem = Qnet in S Wdone by system to outside
C
The change ΔU / Uf S Ui is completely determined
by the initial and final values of the state variables.
If we have a gas process in which the volume does not
change (dV = 0), then there is no PdV work done; i.e. W = 0.
For a constant-volume process involving n moles of gas:
W = QV = n CV ΔT
ΔUconstant volume = Qconstant volume S V
where CV is the molar heat capacity at constant volume.
Note: A fixed variable is often indicated by a subscript.
For ideal gas, CV = (½ + N) R where N = atoms/molecule.
for monatomic gas (N=1): CV = (½ + 1) R = 3/2 R
for diatomic gas (N=2):
CV = (½ + 2) R = 5/2 R
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Change of Internal Energy for Ideal Gas
Application of First Law to an Isothermal Process
Internal energy U of an IDEAL gas only depends upon
Take an ideal gas through an isothermal process (or any
process in which initial and final temps are the same):
Tf = Ti L ΔTisothermal = 0.
temperature:
ΔUideal gas = [U(Tf ) – U(Ti )]
and VERY important:
ˆ ΔUisothermal = U(Tf ) S U(Ti ) = U(Ti ) S U(Ti ) = 0
[ alternatively ΔUisothermal = n CV ΔTisothermal = n CV (0) = 0 ]
“One can show” that for any process involving ideal gases
ΔUideal gas = n CV ΔT even if V not constant!
Note:
but
Application to a Constant Volume Process
QV = n CV ΔT
in constant V process
ΔUideal gas = n CV ΔT
even if V changes
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Consequence: using ΔU = Qnet in S Wnet done L W = Q S ΔU
ΔT = 0 L Wdone = Qnet in S ΔU = Qnet in S 0 = Qnet in
Thus in an isothermal process involving an ideal gas, the heat
energy Q that flows in is completely used to do work W, with
no change in the internal energy U of the gas.
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Example of Isobaric Process of Ideal Diatomic Gas
Q and ΔU for process A Y B?
Qconstant P = n CP ΔT
CP = CV + R = ( 1/2 +2) R + R = 7/2 R
QP = n ( 7/2 R)(TB S TA)
PV = nRT Y T = PV/nR
PV ⎞
⎛ PV
Q = 72 nR ⎜ B B − A A ⎟ = 72 ( PBVB − PAVA )
nR ⎠
⎝ nR
= 72 [ PV
o o − Po (3V0 ) ] = −7 PV
o o
CYCLIC PROCESSES
If a system undergoes a cyclic process, then it
retraces the same path through the state variables. This
means it will always return to the same state. Examples
are a car engine or a steam engine.
The change of internal energy during one cycle is:
ΔU = Uf S Ui = Ui S Ui = 0 = Q S W
For complete cycle in an engine, ΔU = 0
Wdone = Qnet in
ΔU = Qnet in – Wnet out = –7 PoVo – [– Po (2Vo)] = –5 PoVo
alternatively: ΔU = n CV ΔT = n [(½ + 2) R](TB S TA) = –5 PoVo
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For gas engine cycle, * Net Work * = area enclosed in P-V graph.
But need to consider process direction to determine SIGN.
Wcycle = Qcycle
cyclic process
process: http://webphysics.davidson.edu/physlet_resources/bu_semester1/c27_process_cycle_sim.html
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This is least amount of heat energy you must supply !
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