Download Physics 52 - Heat and Optics Dr. Joseph F. Becker Physics

Physics 52 - Heat and Optics
Dr. Joseph F. Becker
Physics Department
San Jose State University
© 2005 J. F. Becker
Chapter 19
The First Law of
Thermodynamics
Ch. 19 First Law of Thermodynamics
1. Thermodynamic systems
2. Work done during volume changes
3. Paths between thermodyn. states
4. Internal energy and the First Law
5. Kinds of thermodynamic processes
6. Internal energy of an ideal gas
7. Heat capacities of an ideal gas
8. Adiabatic processes for ideal gas
Sign conventions:
a) Q is positive
b) Q is negative
c) W is positive
d) W is negative
e) In a steam
engine the heat
in and the work
out (done) are
both defined to
be POSITIVE.
(a) Molecule hits a piston
moving away from it; speed
and KE of molecule
decrease. (expan. -> cooling)
(b) Molecule hits a piston
moving toward it; speed
and KE of molecule
increase. (comp. -> heating)
WORK-ENERGY THEOREM
When a molecule hits a wall moving away from
it, the molecule does work on the wall; the
molecule’s speed and kinetic energy decrease.
The gas does positive work (on the piston);
and the internal energy of the gas decreases.
Molecule hits a wall moving toward it, the wall
does work on the molecule; the molecule’s
speed and kinetic energy increase.
The gas does negative work (on the piston);
and the internal energy of the gas increases.
The infinitesimal work done by the
system (gas) during the small expansion
dx is dW = p A dx = F dx
so W =  p dV
The work done equals the area under
the curve on a p-V diagram.
(a) Volume increases: work and area are positive.
(b) Volume decreases: work and area are negative.
(c) A constant-pressure: Volume increases, W > 0.
(a) Three different paths
between state 1 and state 2.
(b)-(d) The work done by the system during a
transition between two states depends on the
PATH (or PROCESSES) chosen.
The final states are the same. The intermediate
states (p and V) during the transition from state 1 to
2 are entirely different. Heat, like work, depends on
the initial and final states AND the path between the
states. (a) Heat is added slowly so as to keep the
temperature constant. (b) Rapid, free expansion does
no work and there is no heat transfer.
HEAT ADDED
The First Law of
Thermodynamics
DU = Q - W
In a thermodynamic
process, the internal
energy of a system may
increase, decrease, or
remain the same:
(a) DU > 0
(b) DU < 0
(c) DU = 0
Daily thermodynamic process of your body
(a thermodynamic system)
The net work done by
the system in the
CYCLIC PROCESS
aba is
–500 J.
A p-V
diagram
of a
cyclic
process.
<
Q=?
KINDS OF THERMODYNAMIC PROCESSES
ADIABATIC – NO HEAT TRANSFER; Q=0; DU=-W
ISOCHORIC – CONSTANT VOLUME; W=0; DU = Q
ISOBARIC – CONSTANT PRESSURE; W=p (V2 – V1)
ISOTHERMAL – CONSTANT TEMPERATURE;
only for ideal gas DU = 0, Q = W
The First Law of Thermodynamics
DU = Q - W or Q = DU + W
W=0
Processes for an
ideal gas:
adiabatic Q=0
isochoric W=0
isothermal DU=0
DU = 0
Q=0
The internal energy of an ideal gas
depends only on its temperature,
not on its pressure or volume.
Raising the temperature of an
ideal gas by various processes:
Q = DU + W and DU = f(T)
Isochoric W = 0 so Q = DU
Isobaric Q = DU + p(V2-V1)
A p-V diagram of an
adiabatic process
for an ideal gas
from state a to state b.
g
pV = constant (adiabat)
pV = constant (isotherm)
ADIABATIC PROCESSES for IDEAL GAS
Q=0
so DU = Q – W = 0 – p DV
and DU = n Cv dT for an ideal gas
DU = n Cv dT = – p DV = -(n R T/V) dV
dT/T + (R/ Cv) dV/V = 0
dT/T + (g - 1) dV/V = 0 where g – 1 = (Cp/Cv ) –1
 dT/T + (g - 1) dV/V = 0 ; lnT+(g-1) lnV = const
ln (TV g-1 ) = constant and T1V1
And since pV = nRT, (pV/nR) V
(p/nR) V g = constant’
g-1
g-1
= T2V2
g-1
= constant’
p1V1 g = p2V2
g
Adiabatic
compression
of air in a
cylinder of a
diesel
engine.
p
Carnot Cycle for an
ideal gas
HOT
COLD
The Carnot
cycle for an
ideal gas.
Isotherms in
light-blue.
Adiabats in
dark-blue.
Along which path is a) work done and b)
heat transferred by the system
greatest?
p
Q = DU + W
The absolute value of heat transfer
during one cycle is 7200 J.
Assigned for HW
A quantity of air is taken from state
a to state b along the straight line.
Is Tb > Ta?
Given values for all
p’s and V’s, calculate
the work.
Q = 90 J
W = 60 J
Q=?
|W| = 35 J
Q=?
W = 15 J
Ub = 240 J.
Qbc = ?
Uc = 680 J.
Wabc =
450 J
Qab = ?
Qdc = ?
Wadc =
120 J
Ua = 150 J.
Qad = ?
Ud = 330 J.
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