Download Chapter 19 The first law of thermodynamics

Chapter
p 19
The first law of thermodynamics
19.1 Thermodynamic systems
19.2 Work done during
g volume changes
g
19.3 Paths between thermodynamic states
19 4 Internal energy and first law of
19.4
thermodynamics
19 5 Kinds of thermodynamic processes
19.5
19.6 Internal energy of an ideal gas
19 7 Heat capacities of an ideal gas
19.7
19.8 Adiabatic processes for an ideal gas
Thermodynamics
• The study of relationships involving heat,
mechanical work, and other aspects of energy
and energy transfer
• Name few example of thermodynamics
processes
• First law of thermodynamics deals with
conversion of energy in a thermodynamic
system
• Name an example of mechanical energy transfer
• Name an example of heat transfer
• First law will combine two forms of energy
Thermodynamic systems
• Thermodynamic systems: Any collection of objects
that is convenient to regard as a unit
unit, and that may
have potential to exchange energy with its
surroundings
g
• Thermodynamic processes: any process during
which there are changes in the state of the
thermodynamic system
• State of a thermodynamic system is defined by its
state variables such as T, P, V
• Thermodynamics does not depend on details of the
structure of matter so both microscopic and
macroscopic treatment lead to the same result.
Sign convention for exchange of energy between
a thermodynamic system and environment
Thermodynamics: work done by the system is +
Mechanics: work done by the forces acting on the body is +
Sign of work in thermodynamics and mechanics are opposite
Sign of the work done during a volume change
(whoever does the work has the positive W)
Contraction:
W(molecules)<0
Expansion:
p
W(molecules)>0
A
Work done in a volume change
F = pA
A
dW = Fdx = pAdx
Adx = dV
Ad
dW = pdV
W=
V2
∫ p(V )dV
V1
Always V1 is initial state V2 is final
If p is a variable (not a constant)
we need to know its functionality
with V to solve the integral.
Work done in constant pressure:
W = p (V2 − V1 )
Name an example
of a PV relationship
Calculating work using PV diagrams
Work is the area under the curve that represents
pressure change as a function of volume change
W = ∑ pi ΔVi = Area under the pV curve
i
• Verify sign of the work during an expansion and
contraction process with the convention set before
Example:
Isothermal expansion of an ideal gas
• An ideal g
gas undergoes
g
an isothermal
(constant-temperature) expansion at T,
during
g which its volume changes
g from V1
to V2. How much work does the gas do?
• How much is the volume expansion of an
ideal gas if the work is 2nRT?
Paths between thermodynamic
states: work
• Calculate the work done by the system to get from P1V1 to
P2V2 on the paths 132, 142,12.
• Does
D
th
the work
kd
depend
d on th
the path?
th?
Paths between thermodynamic
states: heat
Rapid removal of the wall and
free expansion
Gradual expansion and
moving the piston up
Q1
W1
Q2=0
W2=0
0
ΔT=0
• How much is the amount of the work done by the system
in a and b for one mole of an ideal gas
Stop and think
• What do you think about “the
the
amount of work contained in a
system”? Can it be defined
clearly?
• Can you name a system that the
work done by it does not depend
on the path?
Internal energy and the first law of
thermodynamics: a generalization
of the conservation of energy
•
•
•
•
Internal energ
energy, U,
U of a system
s stem is ssum
m of the kinetic energies of all of its
constituents plus sum of the potential energy of the interaction among
these particles.
When heat added to a system,
system its internal energy increases ΔU
ΔU=Q
Q
When work is done by a system its internal energy decreases by the
amount of work by ΔU=-W
Total change in internal energy of a system is ΔU
ΔU=Q-W
QW
How we determine the internal
energy off a system?
?
• We can’t determine absolute value of the
internal energy of a system (similar to the
potential energy).
• We choose an state of a system as reference
and give an arbitrary value to its internal energy.
• Then we can measure change in internal energy
using
i th
the h
heatt exchange
h
and
d work
kd
done b
by/on
/
the system ΔU=Q-W
e dQ a
and
dd
dW bot
both depe
depend
do
on tthe
e pat
path,, itt
• While
turns out that ΔU does not depend on the
path. ΔU only depends on the state of a system
which is a function of its state variables (p
(p, V
V, T)
Change in internal energy of
human body in 24 hours
Work done during a change of sate
Summary
Summary: The first law of
thermodynamics
h
d
i
Summary: Internal energy
Example
• You intake 900 calorie and run up a cliff
cliff.
How high do you have to climb to burn all
calories consumed? (your mass is 60 kg)
(ans. 6410 m)
• Evaluate your answer
answer.
A cyclic process
• State of a system
y
starts and
ends at a. The total work is 500 J.
– Why is the work negative?
– Find the change in internal
energy and the heat added to
the system in this process
process.
• For pV diagrams total work is
positive for clockwise
processes and negative for
counterclockwise. Can you
contradict it?
Comparing thermodynamic
processes
Kinds of thermodynamic processes
•
•
•
•
Adiabatic: no heat transfer
Isobaric: constant pressure
I
Isochoric:
h i constant
t t volume
l
Isothermal: constant temperature
Adiabatic processes and the first law
• No heat transfer: Q=0 can be achieved in
– Isolated systems
– Fast processes
• First Law: ΔU
ΔU=-W
W
• In adiabatic expansion system does positive work and
hence looses energy. W>0 and ΔU<0
• In
I adiabatic
di b ti compression
i system
t
d
does negative
ti
work
k
and hence gains energy. W<0 and ΔU>0
• Example:
p compression
p
stroke in an internal combustion,
expansion of the fuel during the power stoke
(temperature rise or drop happens during the process)
Isochoric processes and the first law
• No change of volume: W=0
W 0 is zero
• First Law: ΔU=Q;
• All the energy added to the system remains in it
and cusses temperature change
• Example: heating gas in a cylinder (don’t
(don t try it at
home)
• Although
t oug tthere
eea
are
e ot
other
e kinds
ds o
of work
o tthat
at do
not involve volume change but “isochoric” is
used for the systems that no work is done at all.
Isobaric processes and the first law
• No change of pressure: none of the ΔU,
ΔU Q
Q,
or W is zero
• First Law: ΔU=Q-W;
ΔU=Q W;
• Work is calculated easily W=p(V2-V1)
Isothermal processes and the first law
• No change of temperature: process has to
occur slowly and none of the ΔU, Q, or W
is zero
• First Law: ΔU=Q-W;
• In
I special
i l cases ΔT=0
ΔT 0 means ΔU=0
ΔU 0
(Example: ideal gas) but generally ΔU is
nott zero.
Thermodynamic processes on PV
diagrams
di
Internal energy of an ideal gas
• Example: adiabatic expansion of a
gas If U depended on V or p then
gas.
we would have observed a change
in T.
• Experiment shows no change in T
for expansion of low density
gasses.
• Internal
I t
l energy off an ideal
id l gas
depends only on its temperature,
not on its pressure or volume.
• For a non-ideal gas adiabatic
expansion is accompanied with
p in temperature.
p
drop
• Does internal energy of a solid
depend on its volume?
Heat capacities of an ideal gas
• Cp molar heat capacity
p
y
under constant pressure
(usually for gasses)
• Cv molar
l h
heatt capacity
it
under constant volume
(for solids or liquids)
• For a given system (ideal
gas) why these should be
different? Which one is
greater?
Cp=Cv+R molar heat capacities of an ideal gas
Usually Cv=.4Cp For water Cv<Cp between 0 and 40C
More about heat capacities
• Ratio of heat capacities γ=Cp/Cv
• For an ideal gas internal energy change only
depends on T so ΔU=nCvΔT no matter what
the volume change is.
• Example
E
l 19
19.6:
6 Find
Fi d the
th change
h
iin iinternal
t
l
energy of a dorm room containing 2400
moles
l off air
i when
h it iis cooled
l d ffrom 23
23.9
90C tto
11.60C at a constant pressure of 1.00 atm.
T t the
Treat
th air
i as an ideal
id l gas with
ith γ=1.4.
14
Adiabatic Process for an ideal gas
• Q=0 and ΔU=-W<0
• An adiabatic curve on PV
diagram is always steeper
than an isotherm.
• Adiabatic process has to be
fast for not letting heat
exchange to happen.
• If we use ideal g
gas law to
deduce any results the
process has to be slow to
allow equilibrium condition to
b maintained.
be
i t i d
• In reality it is hard to meet
both of these conditions
simultaneously So whatever
simultaneously.
we discuss here is an
approximation.
Adiabatic Process for an ideal gas
• For an adiabatic process
in an ideal gas:
• TVγ-1=constant so
T1V1γ-1= T2V2γ-1
• PVγ=constant so
P1V1γ= P2V2γ
• W=-ΔU=nCv (T1-T2)
• Use PV=nRT
PV nRT
• W=Cv (P1V1-P2V2) /R
(γ )
• W= ((P1V1-P2V2) /(γ-1)
Summary
Summary
Summary
Summary
Example: 19.7
19 7
• The compression ratio of
a diesel engine is 15 to 1
1.
If the initial pressure is
1.01x105pa and the initial
t
temperature
t
is
i 270C,
C find
fi d
the final pressure and the
temperature after
compression. Air is a
mixture of mostly diatomic
oxygen and
d nitrogen
it
th
thatt
is an ideal gas with
γγ=1.40. ((T2=886K,,
p2=44.8x105Pa=44 atm)
Work done in an adiabatic process
• In the previous example how much work is
done by the gas do during the compression
if the initial volume of the cylinder is 1.00
L=1.00x10-3m3? Assume that CV for air is
20.8 J/mol.K and γ=1.40