For Cs2013 and Sr2013

1. Thermodynamics

2. THERMODYNAMICS
Thermodynamics is
the study of energy
relationships that
involve heat,
mechanical work,
and other aspects of
energy and heat
transfer.
Central Heating

3. A THERMODYNAMIC SYSTEM
• A system is a closed environment in which
heat transfer can take place. (For example,
the gas, walls, and cylinder of an
automobile engine.)
Work done on
gas or work
done by gas

4. INTERNAL ENERGY OF SYSTEM
• The internal energy U of a system is the
total of all kinds of energy possessed by
the particles that make up the system.
Usually the internal energy consists
of the sum of the potential and
kinetic energies of the working gas
molecules.

5. TWO WAYS TO INCREASE THE INTERNAL
ENERGY, U.
+ U
WORK DONE HEAT PUT INTO
ON A GAS A SYSTEM
(Positive) (Positive)

6. TWO WAYS TO DECREASE THE INTERNAL
ENERGY, U.
Wout
Qout
- U
Decrease
hot hot
WORK DONE BY HEAT LEAVES A
EXPANDING GAS: SYSTEM
W is positive Q is negative

7. THERMODYNAMIC STATE
The STATE of a thermodynamic
system is determined by four
factors:
• Absolute Pressure P in
Pascals
• Temperature T in Kelvins
• Volume V in cubic meters
• Number of moles, n, of working gas

8. THE FIRST LAW OF
THERMODYAMICS:
• The net heat put into a system is equal to
the change in internal energy of the
system plus the work done BY the system.
Q= U+W final - initial)
• Conversely, the work done ON a system is
equal to the change in internal energy plus
the heat lost in the process.

9. SIGN CONVENTIONS FOR
FIRST LAW +Wout
+Qin
• Heat Q input is positive U
• Work BY a gas is positive -Win
• Work ON a gas is negative U
• Heat OUT is negative -Qout
Q= U+W final - initial)

10. APPLICATION OF FIRST
LAW OF THERMODYNAMICS
Example 1: In the figure, the
gas absorbs 400 J of heat and Wout =120 J
at the same time does 120 J
of work on the piston. What
is the change in internal
energy of the system? Qin
400 J
Apply First Law:
Q= U+W

11. Example 1 (Cont.): Apply First Law
Q is positive: +400 J (Heat IN) Wout =120 J
W is positive: +120 J (Work OUT)
Qin
Q= U+W
400 J
U=Q-W
U=Q-W
= (+400 J) - (+120 J) U = +280 J
= +280 J

12. FOUR THERMODYNAMIC
PROCESSES:
• Isochoric Process: V = 0, W = 0
• Isobaric Process: P=0
• Isothermal Process: T = 0, U = 0
• Adiabatic Process: Q=0
Q= U+W

13. HEAT ENGINES
A heat engine is any
Hot Res. TH device which through
Qhot Wout a cyclic process:
Engine • Absorbs heat Qhot
Qcold • Performs work Wout
Cold Res. TC • Rejects heat Qcold

14. THE SECOND LAW OF
THERMODYNAMICS
Hot Res. TH It is impossible to construct an
Qhot Wout
engine that, operating in a
cycle, produces no effect other
Engine than the extraction of heat
from a reservoir and the
Qcold performance of an equivalent
amount of work.
Cold Res. TC
Not only can you not win (1st law);
you can’t even break even (2nd law)!

15. THE SECOND LAW OF
THERMODYNAMICS
Hot Res. TH Hot Res. TH
400 J 100 J 400 J
400 J
Engine Engine
300 J
Cold Res. TC Cold Res. TC
• A possible engine. • An IMPOSSIBLE
engine.

16. EFFICIENCY OF AN ENGINE
The efficiency of a heat engine
Hot Res. TH is the ratio of the net work
QH W done W to the heat input QH.
Engine W QH- QC
e= =
QC QH QH
Cold Res. TC QC
e=1-
QH

17. EFFICIENCY EXAMPLE
An engine absorbs 800 J and
Hot Res. TH wastes 600 J every cycle. What
800 J W is the efficiency?
QC
Engine e=1-
600 J QH
Cold Res. TC 600 J
e=1- e = 25%
800 J
Question: How many joules of work is done?

18. REFRIGERATORS
A refrigerator is an engine
operating in reverse:
Hot Res. TH
Work is done on gas
Qhot Win extracting heat from cold
reservoir and depositing
Engine heat into hot reservoir.
Qcold Win + Qcold = Qhot
Cold Res. TC
WIN = Qhot - Qcold

19. THE SECOND LAW FOR
REFRIGERATORS
It is impossible to construct a
Hot Res. TH refrigerator that absorbs heat
Qhot from a cold reservoir and
deposits equal heat to a hot
Engine
reservoir with W = 0.
Qcold
Cold Res. TC If this were possible, we could
establish perpetual motion!