Lecture 13: Thermodynamics Chapter 15 - PowerPoint PPT Presentation

1 / 18
About This Presentation
Title:

Lecture 13: Thermodynamics Chapter 15

Description:

... 15) Thermodynamic Systems. Thermodynamics: Fundamental laws that heat and work obey ... Diathermal Walls Allows heat to flow through ... – PowerPoint PPT presentation

Number of Views:149
Avg rating:3.0/5.0
Slides: 19
Provided by: vinod2
Category:

less

Transcript and Presenter's Notes

Title: Lecture 13: Thermodynamics Chapter 15


1
Lecture 13 Thermodynamics (Chapter 15)
2
Thermodynamic Systems
Thermodynamics Fundamental laws that heat and
work obey System Collection of objects on which
the attention is being paid Surrounding
Everything else around System can be separated
from surrounding by Diathermal Walls Allows
heat to flow through Adiabatic Walls -
Perfectly insulating walls that do not allow flow
of heat State of a system the physical
condition can be defined using various
parameters such as volume, pressure, temperature
etc.
3
Zeroth Law of Thermodynamics
  • Deals with Thermal equilibrium
  • Two systems are said to be in thermal
    equilibrium if there is no net flow of heat
    between them when they are brought into thermal
    contact.
  • Temperature is the indicator of thermal
    equilibrium
  • Two systems individually in thermal equilibrium
    with a third system are in thermal equilibrium
    with each other.

4
First Law of Thermodynamics
  • When a substance involves in a process involving
    energy in the form of work and heat, the internal
    energy of the substance can change.
  • First Law Relationship between work, heat and
    change in the internal energy
  • Internal energy changes when heat is imparted
    ?U Q
  • Internal energy changes when work is done on the
    system or by the system ?U - W
  • Work is ve when it is done by the system and
  • Work is ve when it is done on the system.
  • Thus system can lose or gain energy through heat
  • or work
  • ?U Uf -Ui Q - W

5
Thermal Processes
A System can interact with the surrounding in
several ways but has to obey the first law of
thermodynamics. Isobaric process Constant
pressure Work done is ve when it expands (Vf
gt Vi) Isobaric compression work done is -ve
6
Thermal Processes
Isochoric Process Constant volume process Area
under the P V graph 0 gt No work is done So
the heat given is only used to change the
internal energy. ?U Q W Q Isothermal
Process Constant temperature process. Adiabatic
Process Occurs without the transfer of heat gt Q
0 gt ?U W When work is done by a system
adiabatically, W is ve and when work is done on
the system adiabatically, W is ve
7
Thermal Processes
Area under a P-V graph is the work for any kind
of thermal process
8
Thermal Processes using an Ideal Gas
Ideal Gas A gas for which the potential energy
of interaction between the molecules is
independent of their separation and hence is
independent of the gas volume. The internal
energy of such a gas depends on the
temperature. Isothermal Compression or
Expansion When a system performs work
isothermally, the temperature stays constant.
9
Thermal Processes using an Ideal Gas
  • What is the origin of energy for this work?
  • - Internal energy of an ideal gas is proportional
    to its Kelvin temperature, U 3/2 (n RT)
  • Internal energy remains constant throughout an
    isothermal process and so the change in internal
    energy 0
  • (ie) ?U Q-W 0 gt Q W
  • Energy for the work originates from the heat
    provided.
  • Expansion heat flows from the hot water to the
    gas
  • Compression heat flows from the gas into the
    water

10
Thermal Processes using an Ideal Gas
Adiabatic Compression or Expansion When a system
performs work adiabatically, NO heat flows into
or out of the system.
Expands adiabatically gt ve work gt Ti gt
Tf Compress adiabatically gt -ve work gt
TiltTf Ti PiVi/(nR) and Tf PfVf/(nR) PiVi?
PfVf? ? Is the ratio of specific heat
capacities at constant pressure and constant vol
11
Specific Heat Capacities
Q c m ?T (last chapter) Q C n ?T molar
specific heat capacity in units of J/(molK) n
number of moles ?T Tf Ti
12
Specific Heat Capacities
The molar specific heat capacities can now be
determined
The ratio ? of the specific heats is
13
Second Law of Thermodynamic
THE SECOND LAW OF THERMODYNAMICS THE HEAT FLOW
STATEMENT Heat flows spontaneously from a
substance at a higher temperature to a substance
at a lower temperature and does not flow
spontaneously in the reverse direction.
14
Heat Engines
  • Uses heat to perform work
  • - Hot reservoir provides heat
  • Part of the heat is used to do work
  • remaining heat is rejected to a cold reservoir

Efficiency Work done / input heat W /
QH Should obey the principle of conservation of
energy QH W QC E (QH QC) / QH
15
Carnots Principle and Carnots Engine
Sadi Carnot (1796-1832) A reversible process is
one in which both the system and its environment
can be returned to exactly the states they were
in before the process occurred.
No irreversible engine operating between two
reservoirs at constant temperatures can have a
greater efficiency than a reversible engine
operating between the same temperatures.
Furthermore, all reversible engines operating
between the same temperatures have the same
efficiency.
16
Refrigerators, Air conditioners and Heat Pumps
17
Entropy
In general, irreversible processes cause us to
lose some, but not necessarily all, of the
ability to perform work. This partial loss can be
expressed in terms of a concept called
entropy. To introduce the idea of entropy we
recall the relation QC/QH TC/TH that applies to
a Carnot engine. This equation can be rearranged
as QC/TC QH/TH, which focuses attention on the
heat Q divided by the Kelvin temperature T. The
quantity Q/T is called the change in the entropy
?S
Change in entropy of Carnots engine
Reversible process does not alter the total
entropy of the universe (2nd Law in terms of
entropy). ?Suniverse 0 Entropy Degree of
disorder Entropy and Arrow of time
18
Third Law of Thermodynamics
THE THIRD LAW OF THERMODYNAMICSIt is not possible
to lower the temperature of any system to
absolute zero (T 0 K) in a finite number of
steps.
Write a Comment
User Comments (0)
About PowerShow.com