The
molecules of a gas move with high speeds at a given temperature but even then a
molecule of the gas takes a very long time to go from one point to another
point in the container of the gas. This is due to the fact that a gas molecule suffers
a number of collisions with other gas molecules surrounding it. As a result of
these collisions, the path followed by a gas molecule in the container of the
gas is zig - zag as shown in the figure. During two successive collisions, a
molecule of a gas moves in a straight line with constant velocity and the
distance travelled by a gas molecule between two successive collisions is known
as free path.
The
distance travelled by a gas molecule between two successive collisions is not
constant and hence the average distance travelled by a molecule during all
collisions is to be calculated. This average distance travelled by a gas
molecule is known as mean free path.
Let
λ₁, λ₂, λ₃, …λn be the
distance travelled by a gas molecule during n
collisions respectively, then the mean free path of a gas molecule is given by λ
= (λ₁ + λ₂ + λ₃ + … + λn)/ n
(1)
λ= 1/ (√2πnd)²; where d = Diameter of
the molecule, n = Number of molecules
per unit volume
(2)
As PV = m RT = m NkT Þ N/ V = P/ kT = n = Number of molecule per unit volume S λ = (1/√2)
(kT/ πd²P)
(3)
From λ= 1/ (√2πnd)² = m/ √2π (mn) d² = m/ √2πd²ρ [As mn = Mass per unit volume = Density =r]
(4)
If average speed of molecule is v
then λ = v x t/ N = v x T [As N =
Number of collision in time t, T = time interval between two
collisions]
(5)
Every gas consists of extremely small particles known as molecules. The
molecules of a given gas are all identical but are different than those of
another gas.
(6)
The molecules of a gas are identical, spherical, rigid and perfectly elastic
point masses.
(7)
Their size is negligible in comparison to intermolecular distance (10–9
m)
(8)
The gas molecules keep on colliding among themselves as well as with the walls
of containing vessel. These collisions are perfectly elastic. (i.e. the total energy before collision =
total energy after the collision).
(9)
Molecules move in a straight line with constant speeds during successive
collisions.
(10)
The distance covered by the molecules between two successive collisions is
known as free path and mean of all free paths is known as mean free path.
(11)
No attractive or repulsive force acts between gas molecules.
(12)
Gravitational attraction among the molecules is ineffective due to extremely
small masses and very high speed of molecules.
(13)
Molecules constantly collide with the walls of container due to which their
momentum changes.
The change in momentum is transferred to the walls of the
container. Consequently pressure is exerted by gas molecules on the walls of
container.
(1) Root mean
square speed: It is defined as the square root of mean of squares of
the speed of different molecules i.e.
vrms = √(v²₁ + v²₂ + v²₃ + v²₄ + …)/N
(i)
From the expression for pressure of ideal gas P = 1/3 mN/V v²rms
vrms
= √ [3PV/ mN] = √ [3PV/ Mass of gas] = √ [3P/ ρ] [As ρ = Mass of gas/ V]
(ii)
vrms = √ [3PV/ Mass of gas] = √ [3μRT/ μM] = √ [3RT/ M]
[As
if M is the molecular weight of gas PV = μRT and Mass of gas = μM]
(iii) vrms = √ [3RT/ M] = √ [3 NAkT/
NAM] = √ [3kT/ m] [As M = NAm and R = NAk]
∴ Root mean square velocity vrms
= √ [3P/ ρ] = √ [3 RT/ M] = √ [3kT/ m]
(i) With rise
in temperature rms speed of gas molecules
increases as vrms α √T.
(ii) With
increase in molecular weight rms
speed of gas molecule decreases as vrms α 1/ √M.
e.g., rms speed of hydrogen molecules is four times that of oxygen
molecules at the same temperature.
(iii)
rms speed of gas molecules is of the
order of km/s
e.g., At NTP for hydrogen gas (vrms)
= √ [3 RT/ M] = √ [3 x 8.31 x 273/ 2 x 103] = 1840 m/s.
(iv)
rms speed of gas molecules is √3/γ
times that of speed of sound in gas
As
vrms = √ [3 RT/ M] and vs = √ [γRT/ M] ∴ As vrms
= √3/γ (vs)
(v)
rms speed of gas molecules does not
depends on the pressure of gas (if temperature remains constant) because P µ r (Boyle’s
law) if pressure is increased n times
then density will also increases by n
times but vrms remains
constant.
(vi)
Moon has no atmosphere because vrms
of gas molecules is more than escape velocity (ve).
A
planet or satellite will have atmosphere only and only if vrms < ve
(vii)
At T = 0; vrms = 0 i.e.
the rms speed of molecules of a gas
is zero at 0 K. This temperature is
called absolute zero.
(2) Most
probable speed: The particles of a gas have a range of speeds. This
is defined as the speed which is possessed by maximum fraction of total number
of molecules of the gas. e.g., if
speeds of 10 molecules of a gas are 1, 2, 2, 3, 3, 3, 4, 5, 6, 6 km/s,
then the most probable speed is 3 km/s, as maximum fraction of total
molecules possess this speed.
Most probable
speed vmp
= √2P/ρ =
√2RT/M =
√2kT/m
(3) Average speed: It
is the arithmetic mean of the speeds of molecules in a gas at given
temperature.
vav
= (v₁ + v₂ + v₃ + v₄ + …)/ N
And according
to kinetic theory of gases
Average speed
vav = √ (8/π) (P/ρ) = √(8/π) (RT/M) = √(8/π) (kT/m)
vrms > vav > vmp (order remembering trick) (RAM)
vrms : vav : vmp = √3 : √8/π :√2
= √3 : √2.5 : √2
(i) P = 1/3 mN/V (v²rms) Or P α (mN)
T/ V [Asv²rms α T]
(a)
If volume and temperature of a gas are constant P µ mN i.e. Pressure µ (Mass of
gas).
i.e. if mass of gas is increased,
number of molecules and hence number of collision per second increases i.e. pressure will increase.
(b)
If mass and temperature of a gas are constant. P µ (1/V), i.e., if volume decreases, number of
collisions per second will increase due to lesser effective distance between
the walls resulting in greater pressure.
(c)
If mass and volume of gas are constant, P α (vrms)² α T
i.e., if temperature increases, the
mean square speed of gas molecules will increase and as gas molecules are
moving faster, they will collide with the walls more often with greater
momentum resulting in greater pressure.
(ii)
P = 1/3 mN/V (v²rms) = 1/3 M/V (v²rms) [As M = mN = Total mass of the gas]
∴ P = 1/3 ρv2rms
[As ρ = M/V]
(iii)
Relation between pressure and kinetic energy
Kinetic
energy = ½ Mv²rms
∴ Kinetic energy per unit volume (E)
= ½ (M/V) v2rms = ½ ρv2rms … (i)
And
we know P = 1/3 ρv2rms … (ii)
From
(i) and (ii), we get P = 2/3 E
i.e. the pressure exerted by an ideal
gas is numerically equal to the two third of the mean kinetic energy of
translation per unit volume of the gas.
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