Title: Chapter 12: Laws of Thermodynamics
1Chapter 12 Laws of Thermodynamics
Suggested homework assignment 4,21,33,39,48
- Work in Thermodynamics Processes
- Energy can be transferred to a system by
- heat and by work done on the system.
- In this chapter, all the systems we are
- concerned with are volumes of gas and
- they are in thermodynamic equilibrium
- every part of the gas is at the same
- temperature and pressure.
- Consider a gas contained by a cylinder
- with a movable piston and in equilibrium.
- The gas occupies a volume V and exerts
- a uniform pressure P on the cylinder wall.
2- Work in Thermodynamics Processes
- Work done on a gas (contd)
- The gas is compressed slowly enough so
- that the system remains essentially in
- thermodynamic equilibrium.
- As the piston is pushed downward by an
- external force F through a distance Dy, the
- work done on gas is
under constant pressure
- If the gas is compressed, DV is negative and the
work done on the - gas is positive. If the gas expands, the work
done on the gas is - negative.
3- Work in Thermodynamics Processes
isobaric process
- In general, the area under the graph
- in PV diagram is equal in magnitude
- to the work done on the gas, whether
- or not the process proceeds at
- constant pressure. Note that Pa x m3
- J.
4- First Law of Thermodynamics
- First law of thermodynamics
- The first law of thermodynamics is another
energy conservation law - that relates changes in internal energy the
energy associated with - the position and movement of all the molecules
of a system to - energy transfers due to heat and work.
- Two mechanisms to transfer energy between a
system and its - environment a macroscopic displacement of an
object by a force - and by heat which occurs through random
molecular collisions. - Both mechanisms result in a change in internal
energy DU.
First law of thermodynamics
Ui initial internal energy Uf final internal
energy Q energy transferred to the system by
heat W work done on the system
5- First Law of Thermodynamics
- First law of thermodynamics (contd)
- The internal energy of any isolated system must
remain constant.
- Molar specific heat at constant volume
- Molar specific heat at constant volume Cv is the
amount of - heat Q needed to change the temperature by DT
per mole - at constant volume and is defined as
n number of mole
- Internal energy of a monatomic ideal gas
- We learned that the internal energy of a
monatomic ideal gas is
- Therefore a change in U is proportional to a
change in T
6- First Law of Thermodynamics
- From the first law of thermodynamics
- At constant volume, W0. Therefore the first law
becomes
- However, for an ideal gas
- Therefore, whether its volume is constant or
not,
7- First Law of Thermodynamics
- Molar specific heat (contd)
- A gas with a larger specific heat requires more
energy to realize - a given temperature change. The size of molar
specific heat - depends on the structure of the gas molecule
and how many ways - it can store energy.
- A monatomic gas (He etc.) can store energy as
motion in three - different directions (3-dimension). Degrees of
freedom 3
- A diatomic gas (H2 etc.) can store energy as
motion in three - different directions (3-dimension), and can
tumble and rotate - in two different directions.
Degrees of freedom 5
- Other molecules can store energy as vibrations,
which add more - degrees of freedom.
Each degree of freedom contribute to the molar
specific heat by (1/2)R.
So, for example, the molar specific heat of a
diatomic gas such as oxygen O2 is 5 x (1/2)R
(5/2)R.
8- First Law of Thermodynamics
- Molar specific heat (contd)
- See Table of molar specific heats
- An isobaric process is a process
- where the pressure remains
- constant.
- Isobaric process and first law
9- First Law of Thermodynamics
- Ideal gas in isobaric process
For a monatomic ideal gas
For an isobaric process
Define molar specific heat at constant pressure
as
For an ideal gas in an isobaric process
This is valid for all ideal gasses
10- First Law of Thermodynamics
- An adiabatic process is a process where no
energy enters or leaves - the system by heat the system is thermally
isolated.
For an adiabatic process,
First law of thermodynamics
- It can also be shown that for an ideal gas in an
adiabatic process
11- First Law of Thermodynamics
- An isovolumetric process, also called an
isochoric process, is a - process where the volume is constant.
- Since the volume does not change, there is no
work done.
For an ideal gas
12- First Law of Thermodynamics
- An isothermal process is a process where the
temperature is constant.
- For an ideal gas, since the internal energy
depends only on the - temperature
First law of thermodynamics
- It can be shown that the work
- done on an ideal gas is given by
13- First Law of Thermodynamics
- Example 12.4 Expanding gas
Suppose a system of monatomic ideal gas at
2.00x105 Pa and an initial temperature of 293 K
slowly expands at constant pressure from a volume
of 1.00 L to 2.50 L. (a) Find the work done on
the environment.
(b) Find the change in the internal energy of the
gas.
14- First Law of Thermodynamics
- Example 12.4 Expanding gas (contd)
(c) Use the first law to obtain the energy
transferred by heat.
(d) Use the molar heat capacity at constant
pressure to obtain Q.
(e) How would the answers change for a diatomic
gas?
15- First Law of Thermodynamics
- Example 12.5 Work and engine cylinder
In a car engine operating at 1.80x103 rev/min,
the expansion of hot, high-pressure gas against a
piston occurs in about 10 ms. because energy
transfer by heat typically takes a time on the
order of minutes or hours, it is safe to assume
that little energy leaves the hot gas during
expansion. Estimate the work done by the gas on
the piston during this adiabatic expansion by
assuming the engine cylinder contains 0.100 moles
of an ideal monatomic gas which goes
from 1.20x103 K to 4.00x102 K typical engine
temperatures, during the expansion.
16- First Law of Thermodynamics
- Example 12.6 An adiabatic expansion
A monatomic ideal gas at a pressure 1.01x105 Pa
expands adiabatically from an initial volume of
1.50 m3, doubling its volume. (a) Find the new
pressure.
(b) Estimate the work done on the gas.
17- First Law of Thermodynamics
- Example 12.7 An isovolumetric process
How much thermal energy must be added to 5.00
moles of monatomic ideal gas at 3.00x102 K and
with a constant volume of 1.50 L in order to
raise the temperature of the gas by 3.80x102 K?
- Example 12.8 An isothermal expansion
A balloon contains 5.00 moles of monatomic ideal
gas. As energy is added to the system by heat,
the volume increases by 25 at a constant
temperature of 27.0oC. Find the work Wenv done by
the gas in expanding the balloon, the thermal
energy Q transferred to the gas, and the work W
done on the gas.
18- Second Law of Thermodynamics
- A heat engine takes in energy by heat and
partially converts it to - other forms, such as electrical and mechanical
energy.
- A heat engine, in general, carries some working
substance through - a cyclic process during which
- (1) energy is transferred by heat from a source
at a high temperature - (2) work is done by the engine
- (3) energy is expelled by the engine by heat to
a source at lower - temperature.
- A steam engine, the working substance is water.
The water in the - engine is carried through a cycle in which it
first evaporates into - steam in a boiler and then expands against a
piston. After the steam - is condensed with cooling water, it returns to
the boiler, and the - process is repeated.
19- Second Law of Thermodynamics
- The engine absorbs energy Qh from
- the hot reservoir, does work Weng,
- then gives up energy Qc to the cold
- reservoir. Because the working sub-
- stance goes through a cycle, always
- returning to its initial thermodynamic
- state the initial and final internal
- energy is the same, so DU0.
The work Wenv done by a heat engine equals the
net energy absorbed by the engine.
20- Second Law of Thermodynamics
- If the working substance is a gas,
- the work done by the engine for a
- cyclic process is the area enclosed
- by the curve representing the process
- on a PV diagram.
- The thermal efficiency of a heat
- engine is defined by
21- Second Law of Thermodynamics
- Example 12.10 Efficiency of an engine
During one cycle, an engine extracts 2.00x103 J
of energy from a hot reservoir and transfers
1.50x103 J to a cold reservoir. (a) Find the
thermal efficiency of the engine.
(b) How much work does this engine do in one
cycle?
(b) How much power does the engine generate if it
goes through four cycles in 2.50 s?
22- Second Law of Thermodynamics
- Example 12.11 Analyzing an engine cycle
- A heat engine contains an ideal gas
- confined to a cylinder by a movable
- piston. The gas starts at A where
- T3.00x102 K and B-gtC is an iso-
- thermal process.
- Find the number n of moles of
- the gas and the temperature at B.
23- Second Law of Thermodynamics
- Example 12.11 Analyzing an engine cycle
(contd)
(b) Find DU, Q, and W for iso- volumetric
process A-gtB.
24- Second Law of Thermodynamics
- Example 12.11 Analyzing an engine cycle
(contd)
(c) Find DU, Q, and W for iso- thermal
process B-gtC.
25- Second Law of Thermodynamics
- Example 12.11 Analyzing an engine cycle
(contd)
(d) Find DU, Q, and W for iso- baric
process C-gtA.
26- Second Law of Thermodynamics
- Example 12.11 Analyzing an engine cycle
(contd)
(e) Find the net change in internal energy
DUnet
(f) Find the energy input, Qh the energy
rejected, Qc the thermal efficiency and
the net work performed by the engine.
27- Second Law of Thermodynamics
- Refrigeration and heat pump
- A reversed heat engine is called refrigeration!
Energy is injected into the engine called heat
pump and that results in extraction of energy
from the cold reservoir to the hot reservoir.
Examples are refrigerator and air conditioner.
- Coefficient of performance
For a heat pump in the cooling mode
For a heat pump in the heating mode
28- Second Law of Thermodynamics
- Example 12.12 Cooling the leftovers
- 2.00 L of leftover soup at T323 K is placed in
a refrigerator. Assume - the specific heat of soup is the same as that
of water and the density - 1.25x103 kg/m3. The refrigerator cools the soup
to 283 K. - (a) If the COP of the refrigerator is 5.00,
find the energy needed, in the - form of work, to cool the soup.
(b) If the compressor has a power rating 0.250 hp
find the time needed to cool the food.
29- Second Law of Thermodynamics
- Second law of thermodynamics
- There are limits to the efficiency of heat
engines. - An ideal engine which would convert all the
input energy into work - does not exist.
- The Kelvin-Planck formulation of the second law
of thermodynamics
No heat engine operating in a cycle can absorb
energy from a reservoir and use it entirely for
the performance of an equal amount of work.
- This means that the efficiency eWeng/Qh of
engines must always - be less than one. Some energy Qc must always
be lost to the - environment.
- It is theoretically impossible to construct a
heat engine with an - efficiency 100.
We cannot get a greater amount of energy out of a
cyclic process that we put in.
30- Second Law of Thermodynamics
- Reversible and irreversible processes
- No engine can operate with 100 efficiency, but
different designs - yield different efficiencies.
- One design called Carnot cycle (engine) delivers
the maximum - possible efficiency.
- A reversible process is a process in which every
state along the same - path is an equilibrium state. In a reversible
process, the system can - return to its initial condition (state) by
going along the same path in - reverse direction.
- An irreversible process is a process which does
not satisfy the - condition for a reversible process.
- Most natural processes are irreversible, but
some of them are almost - reversible. If a real process occurs so slowly
that the system is virtually - always in equilibrium, the process can be
considered reversible.
31- Second Law of Thermodynamics
- Consider a heat engine operating in an ideal,
reversible cycle called - a Carnot cycle.
32- Second Law of Thermodynamics
- Consider a heat engine operating in an ideal,
reversible cycle called - a Carnot cycle.
A-gtB Isothermal expansion at Th. Qh
from hot reservoir. WAB done by gas.
B-gtC Adiabatic expansion Th-gtTc. No
heat goes out or comes in. WBC done
by gas.
C-gtD Isothermal compression at Tc.
Qc to cold reservoir. WCD done on gas.
D-gtA Adiabatic compression Tc-gtTh.
No heat goes out or comes in. WDA
done on gas.
33- Second Law of Thermodynamics
- Ratio of heat input to output vs. ratio of
temperatures
- Thermal efficiency of a Carnot enegine
All Carnot engines operating reversibly between
the same two temperatures have the same
efficiency.
- All real engines operate irreversibly, due to
friction and brevity of their - cycles, and are therefore less efficient that
the Carnot engine.
34- Second Law of Thermodynamics
- Example 12.13 Steam engine
- A steam engine has a boiler that operates at
5.00x102 K. The energy - from the boiler changes water to steam which
drives the piston. The - temperature of the exhaust is that of the
outside air, 300 K. - (a) What is the engines efficiency if it is an
ideal engine?
(b) If the 3.50x103 J of energy is supplied from
the boiler, find the work done by the engine
on its environment.
35- Let Qr be the energy absorbed or expelled during
a reversible, - constant temperature process between two
equilibrium states. - Then the change in entropy during any constant
temperature - process connecting the two equilibrium states
id defined by
SI unit joules/kelvin (J/K)
- A similar formula holds even when the
temperature is not constant. - Although calculation of DS during a transition
between two equilibrium - states requires finding a reversible path that
connects the states, the - entropy change calculated on that reversible
path is taken to be DS - for the actual path. This is valid logic as the
change in entropy DS - depends only on the initial and final states
and not on the path taken.
- The entropy of the Universe increases in all
natural processes.
36- In nature a disorderly arrangement is much more
probable than an - orderly one if the laws of nature are allowed
act without interference.
- Using statistical mechanics it can be concluded
that isolated systems - tend toward great disorder, and entropy is a
measure of that disorder.
kB Boltzman constant W a number proportional
to the probability that system has a
particular configuration.
- The second law of thermodynamics is really a
statement of what is - most probable rather than of what must be.
- The entropy of the Universe always increases
37- Example 12.14 Melting a piece of lead
- Find the change in entropy of 300 g of lead when
it melts at 327oC. - Lead has a latent heat of fusion of 2.45x104
J/kg.
(b) Suppose the same amount of energy is used to
melt part of a piece of silver, which is
already at its melting point of 961oC. Find the
change in the entropy of the sliver.
38- Example 12.15 Ice, steam, and the entropy of
the Universe
A block of ice at 273 K is put in thermal contact
with a container at 373 K, converting 25.0 g of
ice to water at 273 K while condensing some of
the steam to water at 373 K. (a) Find the change
in entropy of the ice.
(b) Find the change in entropy of the steam.
Thermal energy lost by the steam is equal to the
thermal energy gained by the ice.
(c) Find the change in entropy of the Universe.
39- Example 12.16 A falling boulder
A chunk of rock of mass 1.00x103 kg at 293 K
falls from a cliff of height 125 m into a large
lake, also at 293 K. Find the change in entropy
of the lake, assuming that all of the rocks
kinetic energy upon entering the lake converts to
thermal energy absorbed by the lake.
The rocks kinetic energy at the time of entrance
to the lake.
40- Application of thermodynamics to living organisms
- Animals do work and give off energy by heat, and
this lead us to - believe the first law of thermodynamics can be
applied to living - organisms.
- Lets apply the first law in terms of the time
rates of change of DU, - Q, and W.
On average, energy Q flows out of the body,
and work is done by the body on its
surroundings Q/Dt and W/Dt are negative.
DU/Dt is negative
- Without supply of energy, the internal energy
and the body - temperature would decrease. But in reality, all
animals acquire - internal energy (chemical potential energy) by
eating and breathing.
- Overall the energy from oxidation of food
ultimately supplies the work - done by the body and energy lost from the body
by heat. From this - point of view, DU/Dt is the rate at which
internal energy is added to - our bodies by food, which balances the rate of
energy loss by heat - and work.
41- Measuring the metabolic rate
- The metabolic rate DU/Dt is the rate at which
chemical potential - energy in food and oxygen are transformed into
internal energy to - balance the body losses of internal energy by
work and heat.
- The metabolic rate DU/Dt is directly
proportional to the rate of oxygen - consumption by volume.
For an average diet, the consumption of one liter
of oxygen releases 4.8 kcal or 20 kJ of energy.
L/s
42- Metabolic rate, activity, and weight gain
- Table below summarizes the measured rate of
oxygen consumption - in mL/(min kg) and the calculated metabolic
rate for 65-kg male - engaged in various activities.
43- Physical fitness and efficiency of the human
body as a - machine
- One measure of a persons physical fitness is
his or her maximum - capacity to use or consume oxygen. Table below
gives some idea - how well a person fit.
- The bodys efficiency e is defined as the
- ratio of the mechanical power supplied by
- a human to the metabolic rate
44- Example 12.17 Fighting fat
- In the course of 24 hours, a65-kg person spends
8 h at a desk - puttering around the house, 1h jogging 5 miles,
5 h in moderate - activity, and 8 h sleeping. What is the change
in her internal energy - during this period?