Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Dr Mohamed Fahmy Hassan
Chapter 3
Part 2
Heat transfer techniques and thermodynamics
Energy transfer mechanisms:
Energy may be transferred by conduction, convection, or radiation.
Thermal Conduction:
In conductors, conduction takes place by means of vibration of atoms.
Conduction occurs only if there is a difference in temperature between two parts
of the conducting medium. Consider a slab of material of thickness x and crosssectional area A. One face of the slab is at a temperature Tc, and the other face is
at a temperature Th as shown in figure 1.
The rate (
dQ
) at which energy transfer (from the hotter side to the colder
dt
one) occurs is found to be proportional to the cross-sectional area (A) and the
temperature difference T= Th – Tc, and inversely proportional to the thickness
(x):
Figure 1. Energy transfer between the two sides of a slab.
dQ
T
A
dt
x
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
dQ
T
 kA
dt
x
Where the proportionality constant k is the thermal conductivity of the material.
Substances that are good thermal conductors have large thermal conductivity
values, whereas good thermal insulators have low thermal conductivity values. If
the slab is of infinitesimal thickness, then:

dQ
dT
 kA
dt
dx
Suppose that a long, uniform rod of length L is thermally insulated so that
energy cannot escape by heat from its surface except at the ends, as shown in
figure 2. One end is in thermal contact with an energy reservoir at temperature Tc
, and the other end is in thermal contact with a reservoir at temperature Th>Tc .
When a steady state has been reached, the temperature at each point along the rod
is constant, so:
dT Th  Tc

dx
L

dQ
 T  Tc 
 kA h

dt
L


Figure 2. Energy transfer between the two sides of an insulated rod.
For a compound slab containing several materials of thicknesses L1, L2,...
and thermal conductivities k1, k2,…, as shown in figure 3, the rate of energy
transfer through the slab at steady state is:

ATh  Tc 
dQ

Li
dt

i ki
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Dr Mohamed Fahmy Hassan
where Tc and Th are the temperatures of the outer surfaces (which are held
constant) and the summation is over all slabs.
Figure 3. Energy transfer by conduction through compound slab containing several
materials in thermal contact with each other. At steady state, the rate of energy
transfer through all slabs is the same.
Example:
Two slabs of area=1m2 and thicknesses, L1=10 cm and L2=5 cm, and
thermal conductivities k1 =397 W/m C, and k2 =238 W/m C are in thermal
contact with each other, as shown in figure. In the steady-state condition, the
temperatures of their outer surfaces are Tc=20C and Th=50C, respectively.
Determine the rate of energy transfer by conduction through the slabs and the
temperature at the interface in the steady-state condition.
Solution
The rate of energy transfer by conduction through the slabs is:
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Dr Mohamed Fahmy Hassan
dQ AT  T 

L
dt

k
h
c
i
i
i
dQ AT  T 

L L
dt

k k
h
c
1
1
1
1
dQ 1  50  20 

 64.93  10 W
0.1 0.05
dt

397 238
3
To get the temperature at the interface in the steady-state condition:
Notice the phrase “in the steady-state condition.” means that energy
transfers through the compound slab at the same rate at all points.
The rate at which energy is transferred through slab 1= the rate at which energy is
transferred through slab 2:
𝑘1 𝐴 (
𝑇−𝑇𝑐
𝐿1
) = 𝑘2 𝐴 (
𝑇ℎ −𝑇
𝐿2
) = 64.93 ×103
For slab 2
𝑑𝑄
𝑇ℎ − 𝑇
= 𝑘2 × 𝐴 × (
)
𝑑𝑡
𝐿2
 50  T 
64.93  10 3  238  1  

 0.05 
T  36.4  C
4
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Dr Mohamed Fahmy Hassan
Example
Two rods of the same length (20×10-2 m) and same Area (5×10-2 m2), but they are
formed from different materials. They can be connected in series as in Figure (a).
Th= 90oC and Tc= 30oC. When the energy transfer reaches steady state, calculate;
(a) The rate of energy transfer by conduction when connected in series.
(b) The temperature at the junction.
(c) The total rate of energy transfer by conduction when connected in parallel as
in fig (b). The thermal conductivity of rods 1 and 2 are 400 Wm-1K-1,and 200
Wm-1K-1
Solution
The rate of energy transfer by conduction when connected in series
𝑑𝑄 𝐴 × 𝛥𝑇
=
𝐿
𝑑𝑡
∑𝑖 𝑖
𝐾𝑖
5 × 10−2 (90 − 30)
=
= 2000𝑊
20 × 10−2 20 × 10−2
+
( 400
200 )
The temperature at the junction
400 × 5 × 10−2 (90 − 𝑇)
2000 =
= 70 𝑜𝐶
−
20 × 10
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The total rate of energy transfer by conduction when connected in parallel;
For rod 1
𝑑𝑄 𝐾 × 𝐴 × 𝛥𝑇 400 × 5 × 10−2 × 60
=
=
= 6000𝑊
𝑑𝑡
𝐿
20 × 10−2
For rod 2
𝑑𝑄 𝐾 × 𝐴 × 𝛥𝑇 200 × 5 × 10−2 × 60
=
=
= 3000𝑊
𝑑𝑡
𝐿
20 × 10−2
Then
𝑡𝑜𝑡𝑎𝑙 𝑑𝑄/𝑑𝑡 = 6000 + 3000 = 9000𝑊
Table of Thermal Conductivities
Convection:
At one time or another, you probably have warmed your hands by holding
them over an open flame. In this situation, the air directly above the flame is
heated and expands. As a result, the density of this air decreases and the air rises.
This hot air warms your hands as it flows by.
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Dr Mohamed Fahmy Hassan
Energy transferred by the movement of a warm substance is said to have been
transferred by convection. If it were not for convection currents, it would be very
difficult to boil water. As water is heated in a tea kettle, the lower layers are
warmed first. This water expands and rises to the top because its density is
lowered. At the same time, the denser, cool water at the surface sinks to the
bottom of the kettle and is heated. The same process occurs when a room is
heated by a radiator. The hot radiator warms the air in the lower regions of the
room. The warm air expands and rises to the ceiling because of its lower density.
The denser, cooler air from above sinks, and is wormed.
Radiation:
The third means of energy transfer is radiation. All objects radiate energy
continuously in the form of electromagnetic waves produced by thermal
vibrations of the molecules. The rate at which an object radiates energy is
proportional to the fourth power of its absolute temperature. This is known as
Stefan’s law and is expressed in equation form as:
dQ
 AeT 4
dt
Where
dQ
is the power in watts radiated from the surface of the object,  is a
dt
constant equal to 5.6696 10-8 W/m2 K4, A is the surface area of the object in
square meters, e is the emissivity, and T is the surface temperature in kelvins. The
value of e can vary between zero and unity, depending on the properties of the
surface of the object.
Object radiates energy at a rate given by
dQ
 AeT 4 , it also absorbs
dt
electromagnetic radiation. If the latter process did not occur, an object would
eventually radiate all its energy, and its temperature would reach absolute zero.
The energy an object absorbs comes from its surroundings, which consist of other
objects that radiate energy. If an object is at a temperature T and its surroundings
are at an average temperature T0, then the net rate of energy gained or lost by the
object as a result of radiation is:
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
dQ
  Ae T 4  To4
dt

When an object is in equilibrium with its surroundings, it radiates and absorbs
energy at the same rate, and its temperature remains constant. When an object is
hotter than its surroundings, it radiates more energy than it absorbs, and its
temperature decreases.
An ideal absorber is defined as an object that absorbs all the energy incident on it,
and for such an object, e =1. An object for which e =1 is often referred to as a
black body. An ideal absorber is also an ideal radiator of energy. In contrast, an
object for which e = 0 absorbs none of the energy incident on it. Such an object
reflects all the incident energy, and thus is an ideal reflector.
Example:
If the skin temperature of someone is 35°C, what is the net energy loss
from his body in 10 min by radiation? The surroundings are at 20°C. Assume that
the emissivity of skin is 0.9, the surface area of him is 1.50 m2 and
 = 5.6696 10-8 W/m2 K4.
Solution

dQ
  Ae T 4  To4
dt


dQ
4
4
 5.6696  10 8  1.5  0.9  35  273  20  273
dt
dQ
 125 W
dt
The total energy lost in 10 min is:
 dQ 
Q     t  125  600  75  10 3 J
 dt 
8
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Faculty of Engineering
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Dr Mohamed Fahmy Hassan
Ideal Gas Law
 The equation of state for an ideal gas is,
PV = n R T
 This is known as the ideal gas law
 R is a constant, called the Universal Gas Constant
 R = 8.314 J/ mol K
 From this, you can determine that 1 mole of any gas at
atmospheric pressure and at 0o C is 22.4 L
The internal energy of Monatomic Gas
 For a monatomic gas, translational kinetic energy is the only type
of energy the particles of the gas can have
 Therefore, the total energy is the internal energy:
Eint 
3
n RT
2
Thermodynamics
State Variables
 State variables describe the state of a system
 In the macroscopic approach to thermodynamics, variables are used to
describe the state of the system
 Pressure, temperature, volume, internal energy
 These are examples of state variables
 The macroscopic state of an isolated system can be specified only if the
system is in internal thermal equilibrium
Transfer Variables
 Transfer variables have values only when a process occurs in which energy is
transferred across the boundary of a system
 Transfer variables are not associated with any given state of the system, only
with changes in the state
 Heat and work are transfer variables
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 Example of heat: we can only assign a value of the heat if energy
crosses the boundary by heat
Work in Thermodynamics
 work can be done on a deformable
system, such as a gas
 Consider a cylinder with a moveable
piston
 A force is applied to slowly
compress the gas
 The compression is slow enough for
all the system to remain essentially
in thermal equilibrium
 This is said to occur in a quasistatic process
The piston is pushed downward by a force through a displacement:
⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗
𝑑𝑊 = 𝐹
𝑒𝑥𝑡 . 𝑑𝑟 = −𝐹𝑒𝑥𝑡 . 𝑑𝑦𝑗̂ = −𝑃𝐴𝑑𝑦
𝐴. 𝑑𝑦 is the change in volume of the gas, 𝑑𝑉
Therefore, the work done on the gas is
𝑑𝑊 = −𝑃 𝑑𝑉
 Interpreting 𝒅𝑾 = − 𝑷 𝒅𝑽
 If the gas is compressed, dV is negative and the work done on the
gas is positive
 If the gas expands, dV is positive and the work done on the gas is
negative
 If the volume remains constant, the work done is zero
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Dr Mohamed Fahmy Hassan
The total work done is
Vf
W    P dV
V
i
PV Diagrams
 Used when the pressure and volume
are known at each step of the process
 The state of the gas at each step can
be plotted on a graph called a PV
diagram
 This allows us to visualize the
process through which the gas
is progressing
 The curve is called the path
 The work done on a gas in a quasi-static process that takes the gas from an
initial state to a final state is the negative of the area under the curve on the PV
diagram, evaluated between the initial and final states
 This is true whether or not the pressure stays constant
 The work done does depend on the path taken
Work Done By Various Paths
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Dr Mohamed Fahmy Hassan
 Each of these processes have the same initial and final states
 The work done differs in each process
 The work done depends on the path
Example
A sample of ideal gas is expanded
to twice its original volume of
1.00 m3 in a quasi-static process
for which 𝑃 = 𝛼𝑉 2 , with, 𝛼 =
5.00 𝑎𝑡𝑚/𝑚6 , as shown in
Figure. How much work is done
on the expanding gas?
Solution
The First Law of Thermodynamics
The First Law of Thermodynamics is a special case of the Law of Conservation
of Energy
 It takes into account changes in internal energy and energy transfers by heat
and work
The First Law of Thermodynamics states that
∆𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊
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All quantities must have the same units of measure of energy
For infinitesimal changes in a system
𝑑𝐸 𝑖𝑛𝑡 = 𝑑𝑄 + 𝑑𝑊
No practical distinction exists between the results of heat and work on a
microscopic scale
 Each can produce a change in the internal energy of the system
 Once a process or path is defined, Q and W can be calculated or
measured
Sign rule
Quantity
Q
Q
W
W
sign
+ve
-ve
+ve
-ve
Meaning
Heat gained by the system
Heat lost by the system
Work done on the system
Work done by the system
Adiabatic Process
 An adiabatic process is one during which no energy enters or leaves the
system by heat
𝑄 = 0
This is achieved by
 Thermally insulating the walls of the system
 Having the process proceed so quickly that no heat can be
exchanged
 Since 𝑄 = 0, ∆𝐸𝑖𝑛𝑡 = 𝑊
 If the gas is compressed adiabatically, W is positive so ∆𝐸𝑖𝑛𝑡 is
positive and the temperature of the gas increases, Work is done on
the gas
 If the gas is expands adiabatically
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 W is negative so ∆𝐸𝑖𝑛𝑡 is negative and the temperature of the gas
decreases, Work is done by the gas
𝑄 = 0,
∆𝐸𝑖𝑛𝑡 = − 𝑊
−∆𝐸𝑖𝑛𝑡 = 𝑊
Adiabatic Processes, Examples
 Some important examples of adiabatic processes related to engineering are
 The expansion of hot gases in an internal combustion engine
 The liquefaction of gases in a cooling system
 The compression stroke in a diesel engine
Isobaric Process
 An isobaric process is one that occurs at a constant pressure
 The values of the heat and the work are generally both nonzero
 The work done is W = P (Vf – Vi) where P is the constant pressure
Isovolumic Process
 An isovolumic process is one in which there is no change in the volume
 Since the volume does not change, W = 0
 From the First Law, ∆𝐸𝑖𝑛𝑡 = Q
 If energy is added by heat to a system kept at constant volume, all of the
transferred energy remains in the system as an increase in its internal
energy
Isothermal Process
 An isothermal process is one that occurs at a constant temperature
 Since there is no change in temperature, ∆𝐸𝑖𝑛𝑡 = 0
 Therefore, 𝑄 = − 𝑊
 Any energy that enters the system by heat must leave the system by
work
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Dr Mohamed Fahmy Hassan
 This is a PV diagram of an
isothermal expansion
 The curve is a hyperbola
 The curve is called an isotherm
Special Processes, Summary
 Adiabatic
 No heat exchanged
 Q = 0 and ∆𝐸𝑖𝑛𝑡 = W
 Isobaric
 Constant pressure
 W = P (Vf – Vi) and ∆𝐸𝑖𝑛𝑡 = Q + W
 Isothermal
 Constant temperature
 ∆𝐸𝑖𝑛𝑡 = 0 and Q = - W
Cyclic Processes
 A cyclic process is one that
originates and ends at the same state
 This process would not be
isolated
 On a PV diagram, a cyclic
process appears as a
closed curve
 The internal energy must be zero
since it is a state variable
 If ∆𝐸𝑖𝑛𝑡 = 0, Q = -W
In a cyclic process, the net work done on the system per cycle equals the
area enclosed by the path representing the process on a PV diagram
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Dr Mohamed Fahmy Hassan
Example Heating a Solid
A 1.0-kg bar of copper is heated at atmospheric pressure, (1atm=1.013×105 Pa),
so that its temperature increases from 20°C to 50°C.
(A)What is the work done on the copper bar by the surrounding atmosphere?
(B) How much energy is transferred to the copper bar by heat?
(C) What is the increase in internal energy of the copper bar? Given that;
Linear expansion coefficient of copper 𝛼 = 1.7 × 10−5 oC-1
Density of copper ⍴ = 8.92 × 103 Kg /m3
The specific heat of copper c =387 J/Kg oC
Solution
the work done on the copper bar
Substitute for the volume in terms of
the mass and density of the copper:
Heat energy transferred to the copper
bar
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Use the first law of thermodynamics:
Example
Suppose 1.00 g of water vaporizes isobarically at atmospheric pressure (1.013 ×
105 Pa). Its volume in the liquid state is 𝑉𝑖 = 𝑉𝑙𝑖𝑞𝑢𝑖𝑑 = 1.00 𝑐𝑚3 , and its volume
in the vapor state is 𝑉𝑓 = 𝑉𝑣𝑎𝑝𝑜𝑢𝑟 = 1671 𝑐𝑚3. Find the work done in the
expansion and the change in internal energy of the system. Ignore any mixing of
the steam and the surrounding air; imagine that the steam simply pushes the
surrounding air out of the way. Specific latent heat of vaporization =2.26×106
J/Kg.
Solution
the work done on the system
as the air is pushed out of the
way:
the
latent
heat
of
vaporization for water to
find the energy transferred
into the system by heat:
Use the first law to find the
change in internal energy of
the system:
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Example
A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in
Fig. (a).
Fig. (a)
(i) State the change in internal energy of the gas during one complete cycle
PQRP.
(ii) Calculate the work done on the gas during the change from P to Q.
(iii) Some energy changes during the cycle PQRP are shown in Fig. (b).
Fig. (b)
Complete Fig. (b) to show all of the energy changes.
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Molar Specific Heat of an Ideal Gas
Consider an ideal gas undergoing several processes such that the change in
temperature is ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 for all processes. The temperature change can be
achieved by taking a variety of paths from one isotherm to another as shown in
Figure.
Figure: an ideal gas is taken from one isotherm at temperature T to another at
temperature 𝑇 + ∆𝑇 along three different paths.
Because ∆𝑇 is the same for each path, the change in internal energy ∆𝐸𝑖𝑛𝑡 is the
same for all paths. The work W done on the gas (the negative of the area under
the curves) is different for each path. Therefore, from the first law of
thermodynamics, the heat associated with a given change in temperature does
not have a unique value. We can address this difficulty by defining specific
heats for two special processes isovolumetric and isobaric. Because the number
of moles n is a convenient measure of the amount of gas, we define the molar
specific heats associated with these processes as follows:
𝑄 = 𝑛𝐶𝑉 ∆𝑇
(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒)
(1)
𝑄 = 𝑛𝐶𝑃 ∆𝑇
(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒)
(2)
Where CV is the molar specific heat at constant volume and C P is the molar
specific heat at constant pressure.
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Relation between Cv and R
The internal energy 𝐸𝑖𝑛𝑡 of N molecules (or n mol) of an ideal monatomic gas is
3
𝐸𝑖𝑛𝑡 = 𝑛𝑅𝑇
2
(3)
If energy is transferred by heat to a system at constant volume, no work is done
on the system. That is, 𝑊 = − ∫ 𝑃𝑑𝑉 = 0 for a constant-volume process. Hence,
from the first law of thermodynamics,
𝑄 = ∆𝐸𝑖𝑛𝑡
(4)
A constant-volume process from i to f
for an ideal gas is described in Figure,
where ∆T is the temperature
difference between the two isotherms.
Substituting the expression for Q
given by Equation 1 into Equation 4,
we obtain
∆𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑉 ∆𝑇
(5)
1 𝑑𝐸𝑖𝑛𝑡
𝐶𝑉 = 𝑛
𝑑𝑇
(6)
Substituting the internal energy from
Equation 3 into Equation (6) gives
3
𝐶𝑉 = 𝑅
2
Fig HMF
(7)
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Relation between CP and R
Now suppose the gas is taken along the constant-pressure path i →f’ S shown in
Figure (HMF). Along this path, the temperature again increases by ∆T. The
energy that must be transferred by heat to the gas in this process is 𝑄 = 𝑛𝐶𝑃 ∆T.
Because the volume changes in this process, the work done on the gas is
𝑊 = −𝑃∆𝑉, where P is the constant pressure at which the process occurs.
Applying the first law of thermodynamics to this process, we have
∆𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊 = 𝑛𝐶𝑃 ∆𝑇 + (−𝑃∆𝑉 )
(8)
In this case, the energy added to the gas by heat is channeled as follows. Part of it
leaves the system by work (that is, the gas moves a piston through a
displacement), and the remainder appears as an increase in the internal energy of
the gas. The change in internal energy for the process i →f’, however, is equal to
that for the process i →f because ∆𝐸𝑖𝑛𝑡 depends only on temperature for an ideal
gas and ∆T is the same for both processes.
In addition, because 𝑃𝑉 = 𝑛𝑅𝑇, note that for a constant pressure process,
𝑃∆𝑉 = 𝑛𝑅∆𝑇. Substituting this value for 𝑃∆𝑉 into Equation (8) with
∆𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑉 ∆T (Eq. 5) gives
𝑛𝐶𝑉 ∆𝑇 = 𝑛𝐶𝑃 ∆𝑇 − 𝑛𝑅∆T
𝐶𝑃 − 𝐶𝑉 = 𝑅
(9)
(10)
3
Because 𝐶𝑉 = 𝑅 for a monatomic ideal gas, Equation (10) predicts a
5
2
value 𝐶𝑃 = 𝑅 = 20.8 J/mol K for the molar specific heat of a monatomic gas at
2
constant pressure. The ratio of these molar specific heats is a dimensionless
quantity γ:
γ=
CP
CV
=
5R⁄
2
3R⁄
2
5
= = 1.67
3
21
(11)
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Dr Mohamed Fahmy Hassan
Table -Molar Specific Heats of Various Gases
Adiabatic Process for an Ideal Gas
 At any time during the process,
PV = nRT is valid
 None of the variables alone are
constant
 Combinations of the variables
may be constant
 The pressure and volume of an ideal
gas at any time during an adiabatic
process are related by
𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(12)
𝛾
𝛾
𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓
( 13)
 All three variables in the ideal gas
law (P, V, T) can change during an
adiabatic process.
22
The PV diagram for an adiabatic
expansion of an ideal gas
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Dr Mohamed Fahmy Hassan
Example
Air at 20.0oC in the cylinder of a diesel engine is compressed from an initial
pressure of 1.00 atm and volume of 800.0 cm3 to a volume of 60.0 cm3. Assume
air behaves as an ideal gas with γ = 1.40 and the compression is adiabatic. Find
the final pressure and temperature of the air.
Solution
Use Equation (13) to find the final
pressure:
Use the ideal gas law to find the final
temperature:
23