Thermodynamics :
Temperature, Heat Transfer, and
First Law of Thermodynamics,
Yohanes Edi Gunanto
Dept. of Math. Educ.
UPH
Temperature
Apakah Suhu = Panas ?
• Apa yang dimaksud dengan Suhu ?
• Apa yang dimaksud dengan Panas ?
Apa yang terjadi ketika benda padat,
cair dan gas menerima panas ?
Ketika menerima panas, molekulmolekul bergerak makin lama makin
cepat !
• Inilah yang disebut dengan pemuaian
• Jadi apakah pemuaian itu ?
Mengapa termoemeter dapat
digunakan untuk mengetahui panas
suatu benda ?
Bandingkan skala-skala termometer di
bawah ini
Bagaimana cara panas berpindah?
Panas berpindah dari benda bersuhu panas ke
benda bersuhu dingin
Panas berpindah dengan cara konveksi,
konduksi dan radiasi
Konduksi
Perpindahan panas tanpa memindahkan
penghantarnya
Konveksi
• Perpindahan panas dengan memindahkan
perantaranya
Radiasi
• Perpindahan panas tanpa memerlukan
perantara
THERMODYNAMICS BASICS
Zeroth Law
A
B
C
If A and B and B and C are in
thermal equil, then A and C are
in thermal equil. [ie. At same T]
First Law of Thermodynamics
Conservation of Energy for Thermal
Systems
Joule Equivalent of Heat
• James Joule showed that mechanical energy
could be converted to heat and arrived at the
conclusion that heat was another form of
energy.
• He showed that 1 calorie of heat was
equivalent to 4.184 J of work.
1 cal = 4.184 J
Energy
• Mechanical Energy: KE, PE, E
• Work is done by energy transfer.
• Heat is another form of energy.
Need to expand the conservation of energy
principle to accommodate thermal systems.
1st Law of Thermodynamics
• Consider an
example system of
a piston and
cylinder with an
enclosed dilute
gas characterized
by P,V,T & n.
1st Law of Thermodynamics
• What happens to
the gas if the
piston is moved
inwards ?
1st Law of Thermodynamics
• If the container is
insulated the
temperature will
rise, the atoms
move faster and the
pressure rises.
• Is there more
internal energy in
the gas?
1st Law of Thermodynamics
• External agent
did work in
pushing the
piston inward.
• W = Fd
•
=(PA)Dx
• W =PDV
Dx
1st Law of Thermodynamics
• Work done on
the gas equals
the change in the
gases internal
energy,
W = DU
Dx
1st Law of TD
• Let’s change the
situation:
• Keep the piston fixed
at its original location.
• Place the cylinder on a
hot plate.
• What happens to gas?
Heat flows into the gas.
Atoms move faster,
internal energy
increases.
Q = heat in Joules
DU = change in internal
energy in Joules.
Q = DU
1st Law of TD
• What if we added
heat and pushed
the piston in at
the same time?
F
1st Law of TD
• Work is done on the
gas, heat is added to
the gas and the
internal energy of the
gas increases!
Q = W + DU
F
1st Law of TD
Some conventions:
For the gases perspective:
• heat added is positive, heat removed is
negative.
• Work done on the gas is positive, work done
by the gas is negative.
• Temperature increase means internal energy
change is positive.
First Law of Thermodynamics
“Energy cannot be created or destroyed. It
can only be changed from one form into
another.”
Rudolf Clausius 1850
First Law of Thermodynamics
• Conservation of Energy
• Says Nothing About Direction of Energy
Transfer
1st Law of TD
• Example: 25 L of gas is enclosed in a
cylinder/piston apparatus at 2 atm of pressure
and 300 K. If 100 kg of mass is placed on the
piston causing the gas to compress to 20 L at
constant pressure. This is done by allowing
heat to flow out of the gas. What is the work
done on the gas? What is the change in
internal energy of the gas? How much heat
flowed out of the gas?
• Po = 202,600 Pa, Vo = 0.025 m3, To = 300 K, Pf =
202,600 Pa, Vf=0.020 m3, Tf=
n = PV/RT.
W = -PDV
DU = 3/2 nRDT
Q = W + DU
W =-PDV = -202,600 Pa (0.020 – 0.025)m3
=1013 J energy added to the gas.
DU =3/2 nRDT=1.5(2.03)(8.31)(-60)=-1518 J
Q = W + DU = 1013 – 1518 = -505 J heat out
Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
P
W 1 2  0
2
PV= NkBT2
PV= NkBT1
1
V1,2
V
Q 1 2 
3
2
Nk B T 2  T1   0
(see the last slide)
dU  Q1 2
isobaric
P
 C V D T 
( P = const )
2
W 1 2    P (V , T ) dV   P V 2  V1   0
2
1
V1
1
PV= NkBT2
PV= NkBT1
V2
V
Q 1 2 
5
2
Nk B T 2  T1   0
dU  W1 2  Q1 2
 C P D T 
Isothermal Process in an Ideal Gas
P
isothermal ( T = const ) :
PV= NkBT
W
V2
dU  0
V2
V1
V
W 1 2    P (V , T ) dV   Nk B T
V1
W i  f  Nk B T ln
Vi
Vf
Wi-f > 0 if Vi >Vf (compression)
Wi-f < 0 if Vi <Vf (expansion)
V2

V1
Q1  2   W 1  2
dV
V
  Nk B T ln
V2
V1
Adiabatic Process in an Ideal Gas
Q1 2  0
adiabatic (thermally isolated system)
dU  W1 2
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
V2
W 1 2    P (V , T ) dV
P
V1
2
PV= NkBT2
PV= NkBT1
1
V2
PV  Nk B T
V1

to calculate W1-2 , we need to know P (V,T)
for an adiabatic process
f
U 
2
Nk B T
PdV  VdP  Nk B dT
V
ln 
 V1
f
dU 
2
Nk B dT   PdV
( f – the # of “unfrozen” degrees of freedom )
V
dV 
2  dP
0
 1   
V 
f 
P


,
 1
PdV  VdP  
2
f
V
2 Adiabatic

f exponent

dV
V1


P 
  ln  1 

 P 


 PV
PdV
PV


 P1V1  const
V
P


P1
dP
P
0
Adiabatic Process in an Ideal Gas (cont.)
PV
P


 P1V1  const
2
V2
1
PV= NkBT2
PV= NkBT1
V1
V
V2
An adiabata is “steeper” than an isotherma:
in an adiabatic process, the work flowing
out of the gas comes at the expense of its
thermal energy  its temperature will
decrease.
V2
W 1  2    P (V , T ) dV   
V1

 P1V1

V1

P1V1
V


dV   P1V 1
1
  1
V2
V
  1
V1
 1
1 
  1   1 
  1  V2
V1 
1
 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
Prove
W 1 2 
f
2
D  PV


f
2
NkB DT  DU
Summary of quasi-static processes of ideal gas
DU  U
Quasi-Static
process
isobaric
(DP=0)
isochoric
(DV=0)
isothermal
(DT=0)
adiabatic
(Q=0)
f
Ui
DU
f
DU 
2
DU 
f
2
Q
f
NkB DT 
NkB DT 
f 2
P DV
2
f
2
 D P V
DU 
2
NkB DT 
2
D  PV

Vi
 PDV
Ti
f
Pi
 D P V
W
f
W
2
2
0
f
PDV
Ideal gas
law
0
0
 N k B T ln
DU

Tf

Ti
Vf
Vf
Pf
Tf
PV
 Pf V f
i i
Vi


PV
 Pf V f
i i
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