First Law of Thermodynamics
Eint  Q  W  Qin  Won
Heat ADDED
to system:
Heat REMOVED
from system:
 Qin  0
 Qin  0
Qout  Qin 
Qin
Eint
Work done ON system BY environment:
 Won  0
Work done BY system ON environment:
 Won  0
Won
Wby  Won 
ADD heat to system or
do work ON system:
REMOVE heat from system or
work done BY system:
 INCREASE internal energy
 DECREASE internal energy
 Eint  0
 Eint  0
Work
 
W   F  ds
W 0
to
W 0
Force PARALLEL to
displacement:
Force OPPOSITE
displacement:
Wby

ds

Fby gas


  Fby  ds
  P  Ads
 Ads  dV

Wby
ds

Fby gas
Volume INCREASE
Volume DECREASE
Wby  0
Wby  0
Won  0
Won  0
  PdV
Won  Wby
   PdV
(based on 20.27) The figure shows two
processes, A and B, carried out on an ideal
gas. Consider the following choices:
1. A
2. B
3. Same
4. Not enough information
a)
In which process was the amount
of work done by the gas greater?
b) In which process was the change
in energy of the gas greater?
c)
In which process was the energy
transferred thermally to the gas
greater?
Using a P-V diagram
P, kPA
Consider a monatomic ideal gas, that begins at
an initial pressure and volume of 200 kPa and
0.01 m3 . The pressure and the volume of the
gas are reduced as shown in the P-V diagram, to
a final pressure and volume of 100 kPa and
0.005 m3.
What are Won, Qin, and Eint for this process?
Wby   PdV
Won   Wby
200
100
.005
0.01
V, m3
 Area under curve = Area of rectangle + Area of Triangle
= –0.75 kJ
= +0.75 kJ
Eint  Eint, f  Eint,i
 32 NkT f  32 NkTi
 23 Pf V f  23 PiVi
Don’t forget the First Law!
Eint  Qin  Won
Qin  Eint  Won
= –2.25 kJ – (+0.75 kJ) = –3.0 kJ
= –2.25 kJ
Was heat added or removed during this process?
(monatomic)
Constant Volume Process
Ideal Gas Processes –
Special Case
PV  NkT
P
Eint  32 NkT f  32 NkTi
 32 ( Pf V f  PiVi )
V
 32 ( Pf  Pi )V
P increase  Pf  Pi  T f  Ti
 Eint increase  Eint  0
P decrease  Pf  Pi  T f  Ti
 Eint decrease  Eint  0
Wby   PdV
 Area under curve
Won
0
 Wby  0
Constant Volume Process
 NO WORK DONE
Eint  Qin  Won
Qin  Eint  Won
Qin  Eint
Constant Pressure Process
(monatomic)
PV  NkT
P
Eint  32 NkT f  32 NkTi
 32 ( Pf V f  PiVi )
V
 32 P (V f  Vi )
Ideal Gas Processes –
Special Case
V increase  V f  Vi  T f  Ti
 Eint increase  Eint  0
V decrease  V f  Vi  T f  Ti
 Eint decrease  Eint  0
Wby   PdV
Vf
 P  dV  P(V f  Vi )
Vi
Won   Wby
Eint  Qin  Won
Qin  Eint  Won
Constant Temperature Process
Iso
(monatomic)
Thermal
P
Vf
Ideal Gas Processes –
Special Case
Vf
NkT
Wby   PdV  
dV
V
Vi
Vi
Vf
dV
 NkT 
 NkT ln
V
 Vi
Vi
Vf
V
PV  NkT
NkT cons.
P

V
V
Eint  32 NkT f  32 NkTi
0
Isothermal Process 
NO CHANGE IN INTERNAL ENERGY
Vf
 PiVi ln
 Vi




Vf 
  Pf V f ln 

 Vi 
Won   Wby
Eint  Qin  Won
Qin  Won
Isentropic (Adiabatic) Process
PV  NkT
(monatomic)
Ideal Gas Processes –
Special Case
 
P V   cons.
monatomic

cons.
P 
V
Pi Vi  Pf V f
 
Qin  0
 
5
3
Consider adiabatic expansion of gas:
V increase  Wby  0  Won  0
Won  0  Eint  0  T decrease
P
Isotherm
(high T)
Eint  NkT f  NkTi
3
2
3
2
 32 ( Pf V f  PiVi )
Eint  Qin  Won
Won  Eint
Isotherm
(low T)
Adiabat
Vi
Vf
Adiabat (Isentrope) STEEPER than
Isotherm (through same point)
V
(based on 20.58) A sample of diatomic gas (d = 5,  = 5/3) begins in the state 1 shown
on the PV diagram and ends in the state 3. Determine everything you can about states,
changes in states, and processes.