Embed Size (px)
Citation preview
PHYSICS 2 (FLUID MECHANICS AND THERMAL PHYSICS)
02 credits (30 periods)
Chapter 1 Fluid Mechanics
Chapter 2 Heat, Temperature and the Zeroth
Law of Thermodynamics
Chapter 3 Heat, Work and the First Law of
Thermodynamics
Chapter 4 The Kinetic Theory of Gases
Chapter 5 Entropy and the Second Law of
Thermodynamics
CHAPTER 4
The Kinetic Theory of Gases
Ideal Gases, Experimental Laws and the Equation
of State
Molecular Model of an Ideal Gas The Equipartition
of Energy
The Boltzmann Distribution Law
The Distribution of Molecular Speeds
Mean Free Path
The Molar Specific Heats of an Ideal Gas
Adiabatic Expansion of an Ideal Gas
1. Ideal Gases, Experimental Laws and the Equation of State 1.1 Notions
► Properties of gases
A gas does not have a fixed volume or pressure
In a container, the gas expands to fill the container
► Ideal gas
Collection of atoms or molecules that move randomly
Molecules exert no long-range force on one another
Molecules occupy a negligible fraction of the volume of their container
► Most gases at room temperature
and pressure behave approximately
as an ideal gas
1.2 Moles
► It’s convenient to express the amount of gas in a given volume in terms of the number of moles, n
mass
nmolar mass
► One mole is the amount of the substance that contains as many particles as there are atoms in 12 g of carbon-12
1.3 Avogadro’s Hypothesis
“Equal volumes of gas at the same temperature and pressure contain the same numbers of molecules”
Corollary: At standard temperature and pressure, one mole quantities of all gases contain the same number of molecules
This number is called NA
Can also look at the total number of particles: AN nN
The number of particles in a mole is called
Avogadro’s Number
NA=6.02 x 1023 particles / mole
The mass of an individual atom :
atom
A
molar massm
N
The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al2O3). One carat is equivalent to a mass of 0.200 g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al2O3 molecules in the ruby Rosser Reeves.
SOLUTION
(44.8 )(0.200 / ) 8.90Hopem carats g carat g
8.900.741
12.011 /
Hope
Hope
m gn mol
mass per mole g mol
The mass of the Hope diamond : (a)
The number of moles in the Hope diamond :
(0.741 )H AN mol N
The number of carbon atoms in the Hope diamond :
PROBLEM 1
23 23(0.741 )(6.022 10 / ) 4.46 10mol atoms mol atoms
The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al2O3). One carat is equivalent to a mass of 0.200 g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al2O3 molecules in the ruby Rosser Reeves.
SOLUTION
(138 )(0.200 / ) 27.6Rm carats g carat g
27.60.271
101.9612 /R
R
m gn mol
mass per mole g mol
The mass of the Rosser Reeves : (b)
Molecular mass :
The number of moles in the Rosser Reeves :
PROBLEM 1
2(26.9815 ) 3(15.9994 ) 101.9612Rm u u u 101.9612 /g mol
(b)
The number of Al2O3 molecules in the Rosser Reeves :
(0.271 )R AN mol N
23 23(0.271 )(6.022 10 / ) 1.63 10mol atoms mol molecules
The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al2O3). One carat is equivalent to a mass of 0.200 g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al2O3 molecules in the ruby Rosser Reeves.
SOLUTION
PROBLEM 1
1.4 Experimental Laws
Boyle’s Law
Experiment :
Conclusion :
When the gas is kept at a constant temperature, its pressure is inversely proportional to its volume (Boyle’s law)
PV const
Charles’ Law
Experiment :
Conclusion :
At a constant pressure, the temperature is directly proportional to the volume (Charles’ law)
V CT
( C : constant )
Gay-Lussac’s Law
Experiment :
Conclusion :
At a constant volume, the temperature is directly proportional to the pressure (Gay-Lussac’ law)
T CP
( C : constant )
1.5 Equation of State for an Ideal Gas
Gay-Lussac’ law : T CPV = constant
T = const Boyle’s law : PV const
Charles’ law : V CTP = const
The number of moles n of a substance of mass m (g) :
mn
M (M : molar mass-g/mol)
Equation of state for an ideal gas :
PV = nRT (Ideal gas law)
T : absolute temperature in kelvins
R : a universal constant that is the same for all gases R =8.315 J/mol.K
Definition of an Ideal Gas : “An ideal gas is one for which PV/nT is constant at all pressures”
AN nN Total number of molecules :
PVR
nT
With Boltzmann’s constant :
A
NPV = RT
NA
R= nT
N
23
23 1
8.315 / .1.38 10 /
6.22 10
J mol Kk J K
mol
B
A
R= =
N
BPV = Nk T Ideal gas law :
PV = CTIdeal gas law for an quantity of gas:
Test
An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples? a. 1/9 b. 1/3 c. 3.0 d. 9.0
An ideal gas occupies a volume of 100cm3 at 20°C and 100 Pa.
(a)Find the number of moles of gas in the container (b) How many molecules are in the container?
SOLUTION
PVn
RT
The number of moles of gas : (a)
PROBLEM 2
4 36(100 )(10 )
4.10 10(8.315 / )(293 )
Pa mmol
J mol K
The number molecules in the container : 6(4.10 10 ) AN mol N 6 23
18
(4.10 10 )(6.022 10 / )
2.47 10
mol atoms mol
molecules
(b)
A certain scuba tank is designed to hold 66 ft3 of air when it is at atmospheric pressure at 22°C. When this
volume of air is compressed to an absolute pressure of 3 000 lb/in.2 and stored in a 10-L (0.35-ft3) tank, the air becomes so hot that the tank must be allowed to cool before it can be used. (a) If the air does not cool, what is its temperature? (Assume that the air behaves like an ideal gas.)
PROBLEM 3
SCUBA (Self-Contained Underwater Breathing Apparatus)
The number of moles n remains constant :
1 1 2 2
1 2
;PV PV
n RT T
2 3
2 3
(3000 / . )(0.35 )(295 ) 319
(14.7 / . )(66 )
lb in ftK K
lb in ft
2 22 1
1 1
PV
T TPV
(a)
A certain scuba tank is designed to hold 66 ft3 of air when it is at atmospheric pressure at 22°C. When this
volume of air is compressed to an absolute pressure of 3 000 lb/in.2 and stored in a 10-L (0.35-ft3) tank, the air becomes so hot that the tank must be allowed to cool before it can be used. (b) What is the air temperature in degrees Celsius and in degrees Fahrenheit?
PROBLEM 3
45.9°C; 115°F. (b)
A sculpa consists of a 0.0150 m3 tank filled with compressed air at a pressure of 2.02107 Pa. Assume that air is consumed at a rate of 0.0300 m3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under seawater at a depth of (a) 10.0 m and (b) 30.0 m The density of seawater is = 1025 kg/m3.
PROBLEM 4
SOLUTION
2 1P P gh
5 3 21.01 10 (1025 / )(9.80 / )(10.0 )Pa kg m m s m
1 12
2
PVV
P
(a)
52.01 10 Pa
31.51 m5 3
5
(2.02 10 )(0.0150 )
(1.01 10 )
Pa m
Pa
The volume available for breathing : 3 3 31.51 0.0150 1.50m m m
A sculpa consists of a 0.0150 m3 tank filled with compressed air at a pressure of 2.02107 Pa. Assume that air is consumed at a rate of 0.0300 m3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under seawater at a depth of (a) 10.0 m and (b) 30.0 m The density of seawater is = 1025 kg/m3.
PROBLEM 4
SOLUTION
(a)
3
3
1.5050.0 min
0.0300 /min
mt
m
The compressed air will last for :
(b) 24.6 mint
The deeper dive must have a shorter duration
A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125 cm3 is at 22°C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195°C, what is
the pressure inside the can? Assume any change in the volume of the can is negligible.
PROBLEM 5
The number of moles n remains constant :
1 1 2 2
1 2
PV PVn R
T T
(468 )(202 ) 320
(295 )
KkPa kPa
K 2
2 1
1
TP P
T
SOLUTION
Because the initial and final volumes of the gas are assumed to be equal :
1 2
1 2
;P P
T T
An ideal gas at 20.0OC at a pressure of 1.50 105 Pa when has a number of moles of 6.1610-2 mol.
SOLUTION
nRTV
P
The volume : (a)
PROBLEM 6
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
(a) Find the volume of the gas.
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
3 31.00 10 1.00m L
An ideal gas at 20.0OC at a pressure of 1.50 105 Pa when has a number of moles of 6.1610-2 mol. (b) The gas expands to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature.
SOLUTION
nRTV
P
The volume : (a)
PROBLEM 6
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
(b)
3 31.00 10 1.00m L
1 1 2 2
1 2
;PV PV
n RT T
5
5
(1.01 10 )(2.00 )(293 )
(1.50 10 )(1.00 )
Pa LK
Pa L
2 22 1
1 1
PVT T
PV 395 K
A beachcomber finds a corked bottle containing a message. The air in the bottle is at the atmospheric pressure and a temperature of 30.0OC. The cork has the cross-sectional area of 2.30 cm3. The beachcomber places the bottle over a fire, figuring the increased pressure will pushout the cork. At a temperature of 99oC the cork is ejected from the bottle (a) What was the the pressure in the bottle just before the cork left it ?
PROBLEM 7
1 1 2 1
1 2
;PV PV
n RT T
5 5(372 )(1.01 10 ) 1.24 10
(303 )
KPa Pa
K 2
2 1
1
TP P
T
(a) SOLUTION
Message in a bottle found 24 years later - Yahoo!7
A beachcomber finds a corked bottle containing a message. The air in the bottle is at the atmospheric pressure and a temperature of 30.0OC. The cork has the cross-sectional area of 2.30 cm3. The beachcomber places the bottle over a fire, figuring the increased pressure will push out the cork. At a temperature of 99oC the cork is ejected from the bottle. (b) What force of friction held the cork in place?
PROBLEM 7
0 ;F
5 5 4 2(1.24 10 1.01 10 )(2.30 10 )Pa Pa m
1 0in out fricP A P A F (b)
SOLUTION
5.29 N
( )fric in outF P P A
A room of volume 60.0 m3 contains air having an equivalent molar mass of 29.0 g/mol. If the temperature of the room is raised from 17.0°C to 37.0°C, what mass of air (in
kilograms) will leave the room? Assume that the air pressure in the room is maintained at 101 kPa.
PROBLEM 8
mPV n RT RT
3 5 3(29.0 10 / )(1.01 10 ) 60.0 1 1
(8.31 / . ) 290 310
kg mol Pa m
J mol K K K
1 2
1 2
1 1PVm m
R T T
SOLUTION
4.70 kg
2 Molecular Model of an Ideal Gas
2.1 Assumptions of the molecular model of an ideal gas
A container with volume V contains a very large number N of identical molecules, each with mass m.
The molecules behave as point particles; their size is small in comparison to the average distance between particles and to the dimensions of the container.
The molecules are in constant motion; they obey Newton's laws of motion. Each molecule collides occasionally with a wall of the container. These collisions are perfectly elastic.
The container walls are perfectly rigid and infinitely massive and do not move.
A particle having a brownian motion inside a polymer like network
Brownian motion
2.2 Collisions and Gas Pressure
Consider a cubical box with sides of length d containing an ideal gas. The molecule shown moves with velocity v.
Consider the collision of one molecule moving with a velocity v toward the right-hand face of the box
Elastic collision with the wall Its x component of momentum is reversed, while its y component remains unchanged :
The average force exerted on the molecule :
The average force exerted by the molecule on the wall :
( ) 2x x x xp mv mv mv
2
1
2 2
2 /x x x
x
mv mv mvF
t d v d
2 2
1x xmv mv
Fd d
The total force F exerted by all the molecules on the wall :
The average value of the square of the velocity in the x direction for N molecules :
The total pressure exerted on the wall:
2 2
1 2...x x
mF v v
d
2 2 22 1 2
...x x xNx
v v vv
N
2
x
NmF v
d
2 2 2 2 ;x y zv v v v 2 2 2 2 ;x y zv v v v 2 23 ;xv v
2
3
N mvF
d
2 2
2 3
1 1;
3 3
F F N NP mv mv
A Vd d
22 1
3 2
NP mv
V
The equation of state for an ideal gas :
Temperature is a direct measure of average molecular kinetic energy
The average translational kinetic energy per molecule is
Each degree of freedom contributes to the energy of a
system:
(the theorem of equipartition of energy)
22 1
3 2T mv
k
22 1
3 2
NP mv
V
PV NkT
21 3
2 2mv kT
3
2kT
2 21
3xv v 21 1
;2 2
xmv kT 21 1;
2 2ymv kT 21 1
2 2zmv kT
1
2kT
The total translational kinetic energy of N molecules of gas
: The number of moles of gas
: Boltzmann’s constant
21 3
2 2mv kT
21 3 3
2 2 2transE N mv NkT nRT
A
Nn
N
A
Rk
N
Assume: Ideal gas is a monatomic gas (which has individual atoms rather than molecules: helium, neon, or argon) and the internal energy Eint of ideal gas is simply the sum of the translational kinetic energies of its atoms
int
3 3
2 2transE E NkT nRT
The root-mean-square (rms) speed of the molecules :
2 21 1 3;
2 2 2rms Bmv mv k T
3 3rms
kT RTv
m M
M is the molar mass in kilograms per mole : M = mNA
2
rmsv v
Five gas molecules chosen at random are
found to have speeds of 500, 600,700, 800, and 900 m/s.
Find the rms speed. Is it the same as the average speed?
SOLUTION
PROBLEM 9
In general, vrms and vav are not the same.
A tank used for filling helium balloons has a
volume of 0.300 m3 and contains 2.00 mol of helium gas at
20.0°C. Assuming that the helium behaves like an ideal gas,
(a) what is the total translational kinetic energy of the
molecules of the gas?
SOLUTION
PROBLEM 10
(a)
A tank used for filling helium balloons has a
volume of 0.300 m3 and contains 2.00 mol of helium gas at
20.0°C. Assuming that the helium behaves like an ideal gas,
(b) What is the average kinetic energy per molecule?
(c) Using the fact that the molar mass of helium is 4.00103 kg/mol, determine the rms speed of the atoms at 20.0°C.
SOLUTION
PROBLEM 10
(b)
(c)
(a) What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 27°C ?
(b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas? (c) What is the root-mean-square speed of oxygen molecules at this temperature ?
SOLUTION
PROBLEM 11
(a)
(b)
(a) What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 27°C ?
(b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas? (c) What is the root-mean-square speed of oxygen molecules at this temperature ?
SOLUTION
PROBLEM 11
(c)
(a) A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million K. What is the rms speed of the deuterons? Is this a significant fraction of the speed of light (c = 3.0 x 108 m/s) ? (b) What would the temperature of the plasma be if the deuterons had an rms speed equal to 0.10c ?
SOLUTION
PROBLEM 12
2.3 The Boltzmann Distribution Law The Maxwell–Boltzmann distribution function
Consider the distribution of molecules in our atmosphere : Determine how the number of molecules per unit volume varies with altitude
V Vmgn V mgn Ady
VdP mgn dy
Consider an atmospheric layer of thickness dy and cross-sectional area A, having N particles. The air is in static equilibrium :
( )PA P dP A mgN
where nV is the number density.
;BPV Nk T
Law of Exponential Atmospheres
From the equation of state : ;V BP n k T B VdP k Tdn
;V B Vmgn dy k Tdn ;V
V B
dn mgdy
n k T
0 0
;Vn y
V
V Bn
dn mgdy
n k T
0 0
;Vn y
V
V Bn
dn mgdy
n k T 0 0
ln Vn y
V nB
mgn y
k T
0ln ln ;VB
mgn n y
k T
0
ln ;V
B
n mgy
n k T
0
expV
B
n mgy
n k T
0
ln ;V
B
n mgy
n k T
0
expV
B
n mgy
n k T
/0
Bmgy k TVn n e
/0
BU k TVn n e
The Boltzmann distribution law : the probability of finding the molecules in a particular energy state varies exponentially as the negative of the energy divided by kBT.
What is the number density of air at an
altitude of 11.0 km (the cruising altitude of a commercial
jetliner) compared with its number density at sea level?
Assume that the air temperature at this height is the same as
that at the ground, 20°C.
SOLUTION
PROBLEM 13
/0
Bmgy k TVn n e The Boltzmann distribution law :
Assume an average molecular mass of : 2628.9 4.80 10 u kg
Density of the number of molecules with speeds between v and dv :
The Maxwell–Boltzmann distribution function
2 2/2 /2 2( ) sinmv kT mv kT
VN v dV e dV e v dv d d
2
2/2 2
0 0
( ) sinmv kTVN v dv e v dv d d
2 /2 2( ) 4 mv kT
VN v dv e v dv
2 /2 2( ) 4 mv kT
VN v dv A e v dv
With : ( )VN v dv N
2 /2 24 mv kTA e v dv N
Poisson's Integral Formula:
2
0
1
2
axe dxa
Density of the number of molecules with speeds between v and dv is
23 / 2
2 / 242
mv kTV
mN N v e
kT
Density of the number of molecules with speeds between v and dv is
The rms speed :
The average speed:
The most probable speed:
23 /2
2 /242
mv kTV
mN N v e
kT
2 3 / 1.73 /rmsv v kT m kT m
8 / 1.60 /v kT m kT m
2 / 1.41 /mpv kT m kT m
rms mpv v v
Definition: The average value of v n : PROOF:
0
1n nvv v N dv
N
2
3/22 /2
0
14
2
mv kTmv v N v e dv
N kT
2
3/22 2 2 /2
0
14 3 /
2
mv kTmv v N v e dv kT m
N kT
The average speed:
8 / 1.60 /v kT m kT m
The mean square speed:
2 3 / 1.73 /rmsv v kT m kT m
The most probable speed:
2
3/2
2 /20 ; 4 0 ;2
mv kTvdN d mN v e
dv dv kT 2 /mpv kT m
For diatomic carbon dioxide gas ( CO2 , molar
mass 44.0 g/mol) at T = 300 K, calculate
(a) the most probable speed vmp;
(b) the average speed vav;
(c) the root-mean-square speed vrms.
SOLUTION
PROBLEM 14
The rms speed :
The average speed:
The most probable speed:
2 3 / 1.73 /rmsv v kT m kT m
8 / 1.60 /v kT m kT m
2 / 1.41 /mpv kT m kT m
At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at 20.00C?
SOLUTION
PROBLEM 15
The rms speed : 2 3 /rmsv v kT m
A N2 molecule has more mass so N2 gas must be at a higher temperature to have the same v rms .
2.4 The mean free path
A molecule moving through a gas collides with other molecules in a random fashion.
Notion of the mean free path
Between collisions, the molecules move with constant speed along straight lines. The average distance between collisions is called the mean free path.
The mean free path for a gas molecule
Consider N spherical molecules with radius r in a volume V. Suppose only one molecule is moving.
When it collides with another molecule, the distance between centers is 2r.
In a short time dt a molecule with speed v travels a distance vdt ; during this time it collides with any molecule that is in the cylindrical volume of radius 2r and length vdt.
The volume of the cylinder : 24 r vdt
The number of the molecules with centers in this cylinder :
The number of collisions per unit time :
2(4 )N
dN r vdtV
2(4 )dN N
r vdt V
When all the molecules move at once : 22(4 )dN N
r vdt V
The average time between collisions (the mean free time)
The mean free path (the average distance traveled between collisions) is
For the ideal-gas :
PV NkT
2
2
1
2(4 )2(4 )mean
Vt
N r v Nr vV
24 2
mean
Vvt
r N
24 2
mean
kTvt
r P
Approximate the air around you as a
collection of nitrogen molecules, each of which has a diameter
of 2.00 10-10 m.
How far does a typical molecule move before it collides with
another molecule?
SOLUTION
PROBLEM 16
Assume that the gas is ideal:
The mean free path:
A cubical cage 1.25 m on each side contains
2500 angry bees, each flying randomly at 1.10 m/s. We
can model these insects as spheres 1.50 cm in diameter. On
the average, (a) how far does a typical bee travel between
collisions,
(b) what is the average time between collisions,
and (c) how many collisions per second does a bee make?
SOLUTION
PROBLEM 17
3. The Molar Specific Heats of an ldeal Gas
Constant volume: VQ nC T
CV : the molar specific heat at constant volume
PQ nC T
Constant pressure:
CP : the molar specific heat at constant pressure
First law of thermodynamics:
int
30
2VE Q W nC T nR T
3
2VC R int VE nC T
C : molar specific heat of Various Gases
3
2VC R
Gas constant: R = 8.315 J/mol.K
5
2VC R
7
2VC R
C : molar specific heat of Various Gases
3
2VC R
5
2VC R
7
2VC R
monatomic molecules:
diatomic molecules: (not vibration)
polyatomic molecules:
f : degree of freedom (the number of independent coordinates to specify the motion of a molecule)
2
V
fC R
V = const dW = 0
PdQ nC dT
VdQ nC dT
If the heat capacity is measured under constant- volume conditions: the molar heat capacity CV at constant volume
First law dU = dQ = nCVdT
By definition :
dW PdV nRdT
(Ideal gas : PV = nRT)
First law : dQ = dU + dW PnC dT dU nRdT
VnC dT nRdT
P VC C R
Relating Cp and Cv for an Ideal Gas
The total work done by the gas as its volume changes from
V1 to Vf : f
i
V
V
W PdV
Ideal gas : PV nRT
f
i
V
V
nRTW dV
V
Isothermal process: T const
;f
i
V
V
dVW nRT
V ln f
i
VW nRT
V
Work done by an ideal gas at constant temperature
Also : i i f fPV PV
ln f
i
VW nRT
V
ln i
f
PW nRT
P
:f iV V 0W
When a system expands : work is positive.
When a system is compressed, its volume decreases and
it does negative work on its surroundings
Work done by an ideal gas at constant volume
0f
i
V
V
W PdV
Work done by an ideal gas at constant pressure
0 ( )f
i
V
f i
V
W PdV P V V P V
PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of 20.00C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion?
SOLUTION
PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of 20.00C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion? (b) What is the change in the internal energy of the helium during the temperature increase?
SOLUTION
PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of 20.00C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion? (b) What is the change in the internal energy of the helium during the temperature increase? (c) How much work is done by the helium as it expands against the pressure of the surrounding water during the temperature increase?
SOLUTION
For adiabatic process : no energy is transferred by heat between the gas and its surroundings: dQ = 0
dU = dQ – dW = -dW
P
V
C
C Definition of the Ratio of Heat Capacities :
The Ratio of Heat Capacities
4 Adiabatic Expansion of an Ideal Gas
;dU dQ dW dW
For ideal gas : ;PV nRT PdV VdP nRdT
From : R = CP - CV :
PV const
VnC dT PdV
V
RPdV VdP PdV
C
P V
V
C CPdV VdP PdV
C
Divide by PV :
P V
V
C CdV dP dV
V P C V (1 )
dV
V
0 ;dP dV
P V ln lnP V const
i i f fPV PV
For ideal gas : PV nRT
PV const i i f fPV PV
1nRTV nRTV const
V
1 1
i i f fTV T V 1TV const
PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion?
SOLUTION
PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion? b/ What would be the final temperature if the gas had expanded adiabatically to this same final volume? Oxygen (O2 is diatomic and here has rotation but not oscillation.)
SOLUTION
PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion? b/ What would be the final temperature if the gas had expanded adiabatically to this same final volume? Oxygen (O2 is diatomic and here has rotation but not oscillation.) c/ What would be the final temperature and pressure if, instead, the gas had expanded freely to the new volume, from an initial pressure of.2.0 Pa?
SOLUTION
The temperature does not change in a free expansion:
PROBLEM 20 Air at 20.0°C 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 that air behaves as an ideal gas with = 1.40 and that the compression is adiabatic. Find the final pressure and temperature of the air.
SOLUTION
PROBLEM 21 A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from 23.9°C to 11.6°C at a constant
pressure of 1.00 atm. Treat the air as an ideal gas with = 1.400.
SOLUTION
PROBLEM 22 The compression ratio of a diesel engine is 15 to 1; this means that air in the cylinders is compressed to 1/15 of its initial volume (Fig). If the initial pressure is 1.01 105 Pa and the initial temperature is 27°C (300 K), (a) find the final
pressure and the temperature after compression. Air is mostly a mixture of diatomic oxygen and nitrogen; treat it as an ideal gas with = 1.40.
SOLUTION (a)
The compression ratio of a diesel engine is 15 to 1; this means that air in the cylinders is compressed to 1/15 of its initial volume (Fig). If the initial pressure is 1.01 105 Pa and the initial temperature is 27°C (300 K),(b) how much work
does the gas do during the compression if the initial volume of the cylinder is 1.00 L? Assume that CV for air is 20.8 J/mol.K and = 1.40.
SOLUTION (b)
PROBLEM 22
Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to 1/3 this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain.
SOLUTION
PROBLEM 23
Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to 1/3 this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain.
SOLUTION
The internal energy increases because work is done on the gas (ΔU > 0) and Q = 0. The temperature increases because the internal energy has increased.
PROBLEM 23
On a warm summer day, a large mass of air (atmospheric pressure 1.01 105 Pa) is heated by the ground to a temperature of 26.0°C and then begins to rise through the
cooler surrounding air. (This can be treated as an adiabatic process). Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 105 Pa. Assume that air is an ideal gas, with = 1.40.
SOLUTION
PROBLEM 24
