9 Pius Augustine Gases

1
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Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
Gases
we will
overcome

Gases
No shape and size and can be
contained in a vessel of any
size or shape.
Expands indefinitely and
uniformly to fill the available
space.
Exerts pressure on its
surroundings.
Dr. Pius Augustine, S H College, Kochi

Kinetic theory explains
macroscopic properties of
gases, such as pressure,
temperature, or volume, by
considering their molecular
composition and motion

Brownian motion or Pedesis (Greek word
meaning leaping)
random movement of particles suspended in a
fluid . molecules are so large that they are not
driven by their own thermal energy but by
collisions with solvent particles.
Advection, is a transport mechanism of a
substance, by a fluid, due to the fluid's bulk
motion in a particular direction.
movements in share prices may arise due to
unforeseen events which do not repeat
themselves

Diffusion
Do not require a concentration gradient to spread
out and it is a thermal motion of all (liquid and
gas) molecules at temp above absolute zero.
Molecular diffusion is relevant only on length
scales between nanometer and millimeter.
larger length scales, transport in liquids and
gases is normally due to another transport
phenomenon, convection.Dr. Pius Augustine, S H College, Kochi

Perfect or ideal gas
Hypothetical gas in which
molecules are point masses
and donot attract or repel
each other.
An actual gas behaves like
ideal gas most closely at low
pressure and high
temperature.

Permanent gases
Gas which cannot be
liquefied easily.
Eg. O2, H2, He etc

Gas laws
Assuming permanent gases to be
ideal, it was established that
gases irrespective of their
nature obey the following gas
laws.
i. Boyle’s law
ii. Charles’ law
iii. Gay Lussac’s or pressure law.

i. Boyle’s law
For a given mass of a gas, the
volume of a gas at costant
temperature (isothermal
process) is inversely
proportional to its pressure.
V α 1/P or PV = constant .
PiVi = PfVf.

P – V graph in an isothermal
process is a rectangular hyperbola
or PV vs V (or P) graph is a straight
line parallel to V (or P) axis.
PV
V or P

PV has the dimension of energy.
PV = constant
ie. Energy of a gas at a constant
temperature has a constant value.
Doesnot increase or decrease with
expansion or contraction of the gas so
long as its temperature remains
constant.
No gas obey Boyle’s law fully. A gas
obey Boyle’s law is ideal gas.

Alternative form of Boyle’s law.
PV = constant = Pm/ρ
m – is a constant.
P/ρ = constant.
“At constant temperature ,
density of a gas increases
directly with its pressure.”

VELOCITY OF SOUND AT
CONSTANT
TEMPERATURE IS
INDEPENDENT OF
CHANGES IN PRESSURE

A faulty barometer tube is 90cm long
and it contains some air above
mercury. The reading is 74.5cm when
the true atmospheric pressure is 76cm.
What will be the true atmospheric
pressure if the reading on this
barometer is 74cm?

P1V1 = P2V2
1.5 x A x 15.5 = (P – 74) x A x 16
Ans : P = 75.45 cm of Hg

ii. Charles law
For a given mass of a gas
,the volume of a gas at
constant pressure (isobaric
process) is directly
proportional to absolute
temperature
V α T or V /T = constant
V/T = V /T .

V – T graph in isobaric process is a
straight line passing through origin.
Or V/ T vs V (or T) graph is a straight
line parallel to V or T axis
V/T
V or T

Charle’s law states that the volume of a
gas changes by γp of its volume at 0oC
for each 1oC temperature change,
provided pressure remains constant.
V = V0 ( 1 + γp ∆T)
All gases have nearly the same
value for γp (= 1/273 oC-1)

vol
temp
P1
P2
V-T curves for a certain mass of a gas at
two pressures P1 and P2. Which is higher P1
or P2?
Hint : V = ( nR/P) T , slope = ( nR/P)

iii. Gay Lussac’s law or Regnault’s law
or Pressure law
For a given mass of a gas, the pressure
of a gas at constant volume ( isochoric
process ) is directly proportional to its
absolute temperature T.
ie . P α T or P/T = constant
P vs T graph and P/T vs P or T graph
can be drawn as before.

Gay Lussac’s law states that the Pressure
of a gas changes by γv of its pressure at
0oC for each 1oC temperature change ,
provided volume remains constant.
P = P0 ( 1 + γv ∆T)
All gases have nearly the same
value for γv = γp(= 1/273 oC-1)

Avogadro’s law
Equal volumes of all gases
under same condition of
temperature and pressure,
contain equal no. of
molecules.Dr. Pius Augustine, S H College, Kochi

Gas equation
PV / T = a constant.
Value of constant depends on the mass of
the gas taken.
For 1 mole of gas constant is written as R.
Since one mole of any gas at STP occupies
the same volume (22.4litres) , R is same
for all gases. Hence it is a universal
constant.
R = PV/T = (1.0135 x 22.4-3) / 273.15
= 8.31 J/mol/K

An air bubble of volume 1cc rises from the bottom of a
lake 40m deep at a temperature of 12oC. To what
volume does it grow when it reaches the surface, which
is at a temperature of 35oC?
P1V1 = P2V2.
T1 T2.
V2. = 5.3 x 10-6 m3.

Other forms of gas equation
i. PV = nRT , n – no. of moles of gas
ii. PV = (m/M) RT
m- mass of gas and M – molar mass.
iii. P/ρT = R/M Using Boyle’s law 2nd form.
iv. P/ρT = r or PV’ = rT ( V’ is volume of unit mass )
r = R/M is called specific gas constant and is
not a universal constant, but has different
values for different gases.
v. PVm = RT or NkT for one mole of gas.(N –
Avogadro’s number and k – Boltzmann const.
vi. P = nkT. n = N/Vm. - no. of molecules /unit volume.

r = 4.163 J/kg/K for hydrogen
and 2.083 J/kg/K for helium.

4 gas laws can be written in one single
equation known as ideal gas equation
PV = nRT = (m/M) RT
m- total mass of the gas
M – molecular mass of the gas
n = m/M – no. of moles of the gas .
R – universal gas constant = 8.31 J/mol K = 2.0
cal / molKDr. Pius Augustine, S H College, Kochi
Ideal gas equation

Note : 4 gas laws can be derived from
this equation.
For a given mass of gas, m – constant
i. PV = constant at constant T
(Boyle’s law)
ii. P/T = constant at constant V
iii. V/T = constant at constant P
iv. If P, V , T are constant , n – constant
for all gases
Ideal gas equation

Kinetic theory
The science of understanding the
thermal properties from molecular
mechanics is referred to as
Kinetic theory.
Applicable for all 3 states of
matter.

Kinetic theory of gas R.Claussius & J.C.Maxwell
Postulates (applicable in ideal gas)
i. A gas consists of particles called
molecules which are identical in all
respect and different from molecules of
other gases.
ii. Molecules of gas are in incessant
random motion colliding with one
another and also with the walls of the
container. (obey Newton’s laws of
motion)

iii. Dimensions of molecule is negligible
compared with the average distance
b/w the molecules. (volume of the
molecules is negligibly small fraction
of the volume occupied by the gas)
iv. No intermolecular attraction or
repulsion among the molecules or
b/w the molecules and the walls.
v. Collisions are instantaneous and
elastic.

Mean free path
Between two collisions molecule
travel straight line distance
called free path.(Average
distance b/w successive
collision)

Expression for mean free path
Mean free path = 1/πnd2.
n- no of molecules / unit
volume
Since all the molecules are
moving modified equation is
1/√2 times (1/πnd2)

Assumptions
i. The molecules of a gas are considered as hard
spheres of diameter ‘a’.
ii. Collisions b/w molecules are elastic.
iii. All molecules of a gas except the molecule under
consideration are at rest.
iv. A molecule under consideration collides with all
those molecules whose centre are at a distance
‘d’(diameter of molecule) from the centre of the
molecule.

Root Mean Square velocity (Crms)
Root of the mean of the squares of
velocities of gas molecules.
Crms = v1
2 + v2
2 + …+ vn
2.
n
= ∑vi
2
n

Expression for pressure
Molecular collision on the walls (elastic).
Billions of molecules strike against the walls
in 1sec, exert a sizeable force and hence
pressure.
Let n molecules in cubical vessel of side
1m, and v – velocity of molecule in x-
direction (parallel to one side of cube)
After elastic collision on EFGH bounces
back with –v and strike ABCD.

P (one molecule) before collision = mv
P after collision = - mv
∆P of molecule = -2mv
∆P imparted to wall = 2mv
Distance traveled b/w two successive
collision on EFGH = 2metre.
Distance traveled in 1sec = v metre.
No. of collisions on EFGH in 1sec =
Distance traveled in 1sec ÷ distance
traveled for one collision = v/2.

∆P imparted to wall EFGH in 1sec due
to collision of one molecule
= (v/2) x 2mv = mv2.
Of the n molecules , n/3 move along X-
axis , n/3 along Y and n/3 along Z.
Total change in momentum of EFGH in
1sec = nmC2.
3
Force on the wall EFGH = nmCrms
2
3

Pressure = F/A = mnC2. A = 1m2
3
m is mass of molecule and n is no. of
molecules in unit volume . (cube of
side 1m)
ie. nm is mass / unit vol = density ρ
Pressure = ρ C2.
3

Pressure = F/A = mnC2. A = 1m2
3
Pressure P = ρ C2.
3
C = 3P/ρ - equation for rms velocity.
P = ρ C2. 2 = 2 (½ ρ C2 )
3 2 3
= 2 (½ M/V C2 ) =2 KE/V = 2 E
3 3 3
E – kinetic energy / unit volume .

State the main assumptions of
kinetic theory of gases and
prove that the pressure exerted
by a perfect gas is two – thirds
of it kinetic energy per unit
volume?

In the upper atmosphere , the kinetic
temperature is of the order of 1000K, even
then it is quite cold there . Explain.
Heat sensation depends on the
quantity of heat energy transferred
to the skin.
Since no of molecules in the upper
atmosphere is very small , KE /
unit volume of the air is not large
in the upper atmosphere.

KE and Temperature
Pressure P = ρ C2. = MC2.
3 3 V
PV = MC2. = RT C = 3RT
3 M
MC2. = 3RT = K.E of one mole of gas
2 2
Mean kinetic energy is
proportional to TDr. Pius Augustine, S H College, Kochi

At what temperature the rms
speed of hydrogen molecules
would be equal to the escape
velocity from the surface of
the earth.
Ans : 104K.

Estimate the temperature at which the
oxygen molecules will have the same
rms velocity as the hydrogen molecules
at 150oC. Molecular wt of oxygen is 32
and that of hydrogen is 2.
3RT1/M1 = 3RT2/M2.
T1 = 423 K
Ans . T2 = 6768 K

What is the temperature
at which gas loses all its
energy ?
At zero Kelvin.
Absolute zero

Expression for mean KE of a molecule
Molecular wt M = Nm
m – mass of molecule
MC2. = 3RT = NmC2.
2 2 2
3RT = mC2. = 3 kT
2N 2 2
k = R/N = 1.38 x 10-23 J/K is
Boltzmann constant is gas constant
for one molecule.

ie. C2 α T
Or RMS velocity is independent
of pressure, volume and nature
of the ideal gas but depends on
absolute temperature.

Find the temperature at which the
rms velocity of a molecule would
become twice of its value at 0oC.
C1
2 = T1
C2
2 T2.
T2 = 273 x 4 = 1092K

Absolute temperature of the gas is
increased 3 times. What will be the
increase in rms velocity of the gas
molecules ?
C α√T
Rms velocity becomes √3 times.
Increase in velocity = 0.732 C

What is the physical significance of
molar gas constant R ?
It represents the work
done in increasing the
temperature of 1mole of
gas through 1K

Mention two conditions
when real gases obey the
ideal gas equation
pV = RT.
Low pressure and
high temperature.

Degree of freedom (f)
Refers to the no. of possible independent
ways in which system can have energy.
Or. It is the number of independent ways in
which a mechanical system can move
without violating any constraint which may
be imposed.

Degree of freedom of gas molecules
i. Monoatomic gas eg . He.
It can have translational
motion in any direction in space.
So , 3 degrees of freedom ( all
translational )
f = 3

ii. Diatomic gas and linear polyatomic gas eg .
O2,H2,CO2 etc.
At room temperature
f = 5 ( 3 translnal + 2 rotnal)
( about axis along the diatomic molecule is neglected )
At higher temperature
f = 7 (3 trnl+ 2 rtnl+ 2 vibrnl)
Vibrational KE + Vibrational PE

iii. Non linear poly atomic gas
eg . NH3.
At room temperature f = 6
(3 translational + 3 rotational)
At high temperature f = 8
( 3 + 3 + 2 )

Note :
• There is no clear cut demarcation
above which vibrational energy
become significant. ( if it is not
mentioned , take the values for room
temperature )
• When a diatomic or poly atomic gas
dissociates into atoms , it behaves
as a monoatomic gas and degree of
freedom changes accordingly .

Law of equipartition of energy
It states that, if a system is in
equilibrium, at absolute temperature
T, the total energy is distributed
equally in different energy modes of
absorption.
(energy of each degree of freedom for
1mole of an ideal gas = ½ RT)

What is the value of specific heat of
a gas in an adiabatic process ?
Zero
C = ∆Q / m∆T
∆Q = 0 in adiabatic process.

What is specific heat of gas in
isothermal process?
Infinite.
C = ∆Q / m∆T
∆T = 0 for isothermal
process

What is SI and CGS units of heat .
How are they related?
joule and calorie.
1cal = 4.18 J

What is the value of specific heat of
water in SI unit? Does it vary with
temperature?
Specific heat of water is 4180
J/kg/K .
It vary with temperature
(variation is very small)

What is meant by Boltzmann constant?
Calculate its value in SI units?
k = R/N = 8.31 Jmole-1K-1.
6.02323mole-1.
= 1.38-23J/K

Obtain dimensional formula for R
R = PV / T
= [ ML2T-2K-1]

Can the temperature of a gas
be increased keeping its
pressure and volume
constant ?
No.
Temperature cannot be
change without changing
either V or P. Dr. Pius Augustine, S H College, Kochi

In given samples of 1cc of H
and 1cc of oxygen at NTP,
which sample has larger
number of molecules ?
Both the sample contain the same
number of molecules, in
accordance with Avogadro’s law.

H and O2 are at the same
temperature T. What is the ratio
of KE of O and H molecules?
(O2 is 16 times heavier than H)
One.
KE / molecule of the gas
depends only upon the
temperature.

Are the average KE of
molecules of different
gases at same
temperature equal ?
Yes = 3kT
2Dr. Pius Augustine, S H College, Kochi

The density of a gas is doubled,
keeping all other factors
unchanged. What will be the
effect on the pressure of the gas
Doubled

A container contains a monatomic
gas.What would be the effect on the
average kinetic energy of the atoms of
the gas if the temperature of the gas is
increased from 200K to 400K ?
doubled

A vessel is filled with a mixture of
two different gases. Will the
mean kinetic energies per
molecule of both the gases be
equal ?
Yes.
Depends only on temperature.

Write two important properties of
gases.
i. Most of the energy of the
gas molecules is the kinetic
energy .
ii. Interaction between gas
molecules is negligible.

Give equation of state for ideal gas
and real gas ?
Ideal gas PV = RT
Real gas (P+a/v2) (V-b) = RT

Two different gases have exactly the
same temperature. Does this mean that
their molecules have the same rms
speed?
Average KE / molecule (3/2kT)
for each gas is same.
But different gases may have
molecules of different mases,
hence, Crms of different gases
shall be different.

Velocity of air molecules is about
0.5km/s. But smell of the scent spreads
at a much slower rate. Why?
Because molecules donot travel
uninterrupted. They have random
motion . Scent vapour molecules
undergo a number of collisions and
trace a zig – zag path. Hence
effective displacement per unit time is
low and spreading is at a much slower

Molecular speeds are
comparable with those of rifle
bullet, yet a gas with a strong
odour takes a few minutes to
spread throughout a room.
Why ? Dr. Pius Augustine, S H College, Kochi

The volume of vessel A is twice the volume of
another vessel B and both of them are filled with
the same gas. If the gas in A is at twice the
temperature and twice the pressure in
comparison to the gas in B, what is the ratio of
gas molecules in A and B ?
PAVA = nARTA.
PBVB = nBRTB.
Taking ratio , nA / nB = 2.

When a gas filled in a closed vessel is
heated through 10C , its pressure increases
by 0.4%. What is the initial temperature of
the gas ?
Pf = 1.004Pi. Tf = (Ti+1)
P/T = constant.
Pi/Ti = Pf/Tf.
Ans: 250K

What is an ideal gas? Explain its
main characteristics .
One which obey Boyle’s law ,Charle’s law, Gay-
Lussac’s law etc.
Characteristics of an ideal gas
i. Size of the molecules of a gas
is zero ie . Each molecules is
a point mass with no
dimensions.
ii. No intermolecular force.

Under what conditions do the real
gases obey ideal gas equation?
i. Low pressure. – volume will be
large. Molecular size is negligible
compared to total volume of gas.
ii. High temperature: molecules
have large KE, so that
intermolecular force becomes
negligible.

On reducing the volume of the gas at
constant temperature, the pressure of
the gas increases. Explain on the basis
of kinetic theory.
On reducing volume space decreases ,
and no. of molecules / unit volume
increases.
More molecules collide with the walls of
the vessel per second and hence large
momentum is transferred to the walls
per second and pressure increases

On the basis of what
happens to individual
molecules, explain why
the temperature of a
gas rises when the gas
is compressed ?

Explain the concept of absolute zero of temperature on
the basis of kinetic theory of gases.
C α √T
Square root of absolute temperature of an
ideal gas is directly proportional to root
mean square velocity of its molecules.
When T = 0, C= 0.ie. absolute zero is temp
at which C = 0. ie. molecular motion
ceases.
Note: applicable only for ideal gases. Real
gases show deviation particularly at low
temperature .

Why is cooling caused by evaporation ?
Evaporation occurs on account
of faster molecules escaping
from the surface of the liquid.
Liquid is left with molecules
having lower speed.
Average speed decreases –
gives cooling.

Air pressure in a car tyre increased
during driving. Why ?
Temperature of air inside the
tyre increases due to
motion. Also heat is
produced due to friction.
According to Charle’s law,
P α T.

Show that molar volume is 22.4litres.
Molar volume is the volume
occupied by 1mol of ideal gas at
STP.
V = RT / P
= ( 8.31 x 273 ) / 1.0135
= 0.0224 m3
= 0.2246 cm3
= 22400 ml. ( 1cc = 1ml )Dr. Pius Augustine, S H College, Kochi

Estimate the fraction of molecular volume to
the actual volume occupied by Oxygen gas
at STP . Take the diameter of an oxygen
molecule to be 3Å.
Molar volume = 22400 cc.
Molecular volume of N molecules
= 4/3 π (D/2)3 N = 8.52 cc.
(N = 6.02323.)
Fraction = 8.52 / 22400
= 3.8 x 10-4.

At what temperature is the Crms of an atom
in an argon gas cylinder equal to the Crms of
a helium gas atom at -20oC? ( Atomic mass
of Ar = 39.9u and He = 4u. )
1u = 1.67-27kg.
Crms = 3kBT/m
3kT = 3kT’
m m’
T = T’m/m’ = 2.522 x 103 K.

Two perfect gases at absolute
temperatures T1 and T2 are mixed.
There is no loss of energy. Find the
temperature of the mixture if the
masses of the molecules are m1 and
m2 and the number of the molecules in
the gases are n1and n2 respectively.

Kinetic theory -Average KE/ molecule
= 3/2 (kBT)
Before mixing
Average KE of all the molecules of two gases
= 3/2 (n1kBT1) + 3/2 (n2kBT2)
After mixing
Average KE of all the molecules of two gases
= 3/2 (n1 + n2) kBT
Since there is no loss of energy ,
3/2 (n1kBT1) + 3/2 (n2kBT2) = 3/2 (n1 + n2) kBT
Solving , T = (n1T1 + n2T2)
(n1 + n2)

Estimate the average thermal energy of
a helium atom at i) room temperature
(270C), ii) the temperature on the
surface of the sun (6000K) , the
temperature of 10million K (typical core
temperature of a star).
Average thermal energy of an atom
= 3/2 kT
i) T = 300K Ans: 6.21-21 J
ii) T = 6000K Ans: 1.24-19 J
iii) T = 107K Ans : 2.1-16 JDr. Pius Augustine, S H College, Kochi

One mole : is defined as the amount of
substance which contains the same
number of particles (atoms , molecules ,
ions) as there are molecules (Avogadro
no) in 12g of carbon isotope C12.
Number of moles in a sample of any
substance containing N molecules is n =
N/NA.

Phenomena supporting Kinetic
theory of gas.
1.Diffusion
2.Expansion of gas
3.Evaporation
4.Brownian movemet.

Diffusion
Mixing of one gas into another
against gravity.
Eg. Cylinder of hydrogen is
placed over a cylinder of
carbon dioxide, the two gases
mix and a uniform mixture will
be formed .

Brownian movement
Irregular zig – zag motion of fine
suspended particles in liquid
due to the impact of the
surrounding molecules of the
liquid on the suspended
particles. Dr. Pius Augustine, S H College, Kochi

Very small smoke
particles in air seen by a
microscope move in zig-
zag path.

Graham’s law of Diffusion of Gases.
Rate of diffusion of a gas is inversely
proportional to the square root of the
density of the gas.
P = 1/3 (ρC2.)
C = ( 3P/ρ)1/2.
C α √1/ρ if P is constant.
Since rms velocity is proportional to the rate of
diffusion (r) of the gas,
r α √1/ρ
Denser the gas, slower is the rate of diffusion .

Dalton’s law of partial pressures
Resultant pressure exerted by a
mixture of non – interacting
gases is equal to the sum of
their individual pressures.
P = ⅓ ρC1
2 + ⅓ ρC2
2 + ⅓ ρC3
2 ….
= P1+P2+P3……..
Limitation : not applicable for
reacting gas. Dr. Pius Augustine, S H College, Kochi

Amagat’s law of partial volumes.
Partial volume of one of the components of
mixture of mixture of gases is the volume
which that constituent gas would have
occupied, if it had the same pressure and
temperature as that of the mixture.
Total volume of a mixture = sum of
partial volumes

Define average speed.
It is the arithmetic mean of
the speeds of the molecules
of a gas.

Most probable speed of of gas molecules.
It is the speed at which fraction of
molecules having speeds b/w v and
v+dv according to Maxwell –
Boltzmann speed distribution law is
maximum.
Vmp = (2RT/M)1/2.
Vrms > Vav > Vmp.

Critical temperature
It is a characteristic
temperature of a gas below
which it can be liquefied by
pressure and above which
cannot be liquefied by the
pressure however high it be.

Distinguish b/w vapor and gas?
Vapor can be liquefied by the
application of pressure
alone.
Gas can be liquefied by
pressure only if its
temperature is below critical
temperature.

At 0oC , CO2 can be liquefied
by applying pressure but O2
cannot be. Why?
Critical temperature for
CO2 is 31oC , while that
of O2 is below 0oC.

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Appeal: Please Contribute to Prime Minister’s or Chief
Minister’s fund in the fight against COVID-19
Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
we will
overcome
Thank You
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