NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Param Himalaya
12.1 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 3 Å.
Solution :
Diameter of an oxygen molecule, d = 3 Å
Radius, r = d / 2
r = 3 / 2 = 1.5 Å = 1.5 × $10^{-8}$ cm
The actual volume occupied by 1 mole of oxygen gas at STP = $22400 cm^{3}$
The molecular volume of oxygen gas, V = $4 / 3 πr^{3}$. N
Where, N is Avogadro’s number = 6.023 × $10^{23}$ molecules/ mole
Hence,
V = 4 / 3 × 3.14 × (1.5 x $10^{-8})^{3}$ × $6.023 \times 10^{23}$
We get,
$V = 8.51 cm^{3}$
Therefore, the ratio of the molecular volume to the actual volume of oxygen
= 8.51/ 22400 = 3. 8 × $10^{-4}$
12 .2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Solution:
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as PV = nRT
Where R is the universal gas constant = $8.314 J mol^{-1}K^{-1}$
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 × $10^{5}Nm^{-2}$
Hence, V = nRT / P = 1 × 8.314 × 273 / 1.013 × $10^{5}$= 0.0224 $m^{3}$= 22.4 litres
12.3 Figure 12.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two
different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00×$10^{–3}$ kg of hydrogen, would we get the same
value of PV/T at the point where the curves meet on the y-axis? If not, what mass
of hydrogen yields the same value of PV/T (for low pressure high temperature
region of the plot) ? (Molecular mass of $H_{2}$ = 2.02 u, of $O_{2}$ = 32.0 u,R = 8.31 J $mo1^{–1} K^{–1}$)
Solution:
(a) The dotted plot is parallel to X-axis, signifying that nR [PV/T = nR] is independent of P. Thus, it represents ideal gas behaviour.
(b) The graphs at temperature T1 are closer to ideal behaviour (because they are closer to the dotted line) hence, T1 > T2 (the higher the temperature, the ideal behaviour is higher)
(c) Use PV = nRT
PV/ T = nR
Mass of the gas = 1 x 10-3 kg = 1 g
Molecular mass of O2 = 32g/ mol
Hence, Number of mole = Given weight / Molecular weight = 1/ 32
So, nR = 1/ 32 × 8.314 = 0.26 J/ K
Hence, Value of PV / T = 0.26 J/ K
(d) 1 g of $H_{2}$ doesn’t represent the same number of mole
E.g., Molecular mass of $H_{2}$ = 2 g/mol
Hence, the number of moles of H2 requires is 1/32 (as per the question).
Therefore, Mass of $H_{2}$ required = No. of mole of $H_{2}$ × Molecular mass of $H_{2}$= 1/ 32 × 2 = 1 / 16 g = 0.0625 g = 6.3 × $10^{-5}$ kg
Hence, 6.3 × $10^{-5}$ kg of $H_{2}$ would yield the same value.
12.4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder ($R = 8.31 Jmol^{–1} K^{–1}$, molecular mass of $O_{2}$= 32 u).
Solution:
Volume of gas, $V_{1}$ = 30 litres = 30 × $10^{-3} m^{3}$
Gauge pressure, $P_{1}$ = 15 atm = 15 × 1.013 × $10^{5}$ Pa
Temperature, $T_{1}= 27^{0}C = 300 K$
Universal gas constant, R = 8.314 J $mol^{-1} K^{-1}$
Let the initial number of moles of oxygen gas in the cylinder be n1
The gas equation is given as follows:
$P_{1}V_{1} = n_{1}RT_{1}$
Hence,
$n_{1} = P_{1}V_{1} / RT_{1}$ = (15.195 × 105 × 30 × $10^{-3}$) / (8.314 × 300) = 18.276
But $n_{1} = m_{1} / M$
Where,
$m_{1}$ = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
Thus, $m_{1} = N_{1}M$ = 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.
Volume, $V_{2}$ = 30 litres = 30 × $10^{-3} m^{3}$
Gauge pressure, $P_{2}$ = 11 atm = 11 × 1.013 × $10^{5}$ Pa
Temperature, T2 = $17^{0}C$ = 290 K
Let n2 be the number of moles of oxygen left in the cylinder.
The gas equation is given as $P_{2}V_{2}= n_{2}RT_{2}$
Hence, $n_{2} = P_{2}V_{2} / RT_{2}$
= (11.143 × $10^{5}$ × 30 × $10^{-30}$) / (8.314 × 290) = 13.86
But $n_{2} = m_{2} / M$
Where, $m_{2}$ is the mass of oxygen remaining in the cylinder.
Therefore, $m_{2}$ = $n_{2}$ × M = 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation,
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= m1 – m2 = 584.84 g – 453.1 g
We get, = 131.74 g = 0.131 kg
Hence, 0.131 kg of oxygen is taken out of the cylinder.
12.5 An air bubble of volume $1.0 cm^{3}$ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Solution :
The volume of the air bubble,
V1 = $1.0 cm^{3}$ = 1.0 × $10^{-6} m^{3}$
Air bubble rises to height, d = 40 m
The temperature at a depth of 40 m,
T1 = 12 °C = 285 K
The temperature at the surface of the lake, T2 = 35 °C = 308 K
The pressure on the surface of the lake $P_{2}$ = 1 atm = 1 × 1.013 × $10^{5}$ Pa
The pressure at a depth of 40 m $P_{1}$ = 1 atm + dρg
Where, ρ is the density of water = $10^{3} kg / m^{3}$
g is the acceleration due to gravity = $9.8 m/s^{2}$
Hence, $P_{1}$ = 1.013 × $10^{5}$ + 40 × $10^{3}$ × 9.8
We get, = 493300 Pa
We have $P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}$
Where, $V_{2}$ is the volume of the air bubble
when it reaches the surface $V_{2} = P_{1}V_{1}T_{2}/ T_{1}P_{2}$
= 493300 × 1 × $10^{-6}$ × 308 / (285 × 1.013 × $10^{5}$)
We get, = 5.263 × $10^{-6} m^{3}$ or 5.263 $cm^{3}$
Hence, when the air bubble reaches the surface, its volume becomes $5.263 cm^{3}$
12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Solution:
The volume of the room, V = $25.0 m^{3}$
The temperature of the room, T = 270 C = 300 K
The pressure in the room, P = 1 atm = 1 × 1.013 × $10^{5}$ Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as
$PV = (k_{B}NT)$
Where, $K_{B}$ is Boltzmann constant = (1.38 ×$10^{-23}) m^{2} kg s^{-2} K^{-1}$
N is the number of air molecules in the room.
Therefore, N = (PV / $k_{B}$T) = (1.013 × $10^{5}$× 25) / (1.38 × $10^{-23}$ × 300)
We get, = 6.11 × $10^{26}$ molecules
Hence, the total number of air molecules in the given room is 6.11 × $10^{26}$
12.7 Estimate the average thermal energy of a helium atom at
(i) room temperature (27 °C),
(ii) the temperature on the surface of the Sun (6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Solution :
(i) At room temperature, T = 27 °C = 300 K
Average thermal energy = (3 / 2) kT
Where, k is the Boltzmann constant = 1.38 × $10^{-23} m^{2} kg s^{-2} K^{-1}$
Hence, (3 / 2) kT = (3 / 2) × 1.38 × $10^{-23}$ × 300
On calculation, we get = 6.21 × $10^{-21}$ J
Therefore, the average thermal energy of a helium atom at a room temperature of 27 °C is 6.21 × $10^{-21}$ J
12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ?
Solution :
All three vessels have the same capacity, they have the same volume. So, each gas has the same pressure, volume and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.
This number is equal to Avogadro’s number, N = 6.023 × $10^{23}$
The root mean square speed (Vrms) of a gas of mass m and temperature T is given by the relation
$V_{rms}$ = √3kT / m
Where, k is the Boltzmann constant.
For the given gases, k and T are constants.
Therefore, Vrms depends only on the mass of the atoms, i.e.,
$V_{rms}$ ∝ $(1/m)^{1/2}$
Hence, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine and uranium hexafluoride, the mass of neon is the smallest. Therefore, neon has the largest root mean square speed among the given gases.
12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar= 39.9 u, of He = 4.0 u).
Given :
The temperature of the helium atom,
$T_{He}$ = -20°C = 253 K
The atomic mass of argon, $M_{Ar} = 39.9 u$
The atomic mass of helium, $M_{He} = 4.0 u$
Let $(V_{rms})_{Ar}$ be the rms speed of argon and
Let $(V_{rms})_{He}$ be the rms speed of helium
The rms speed of argon is given by $(V_{rms})_{Ar}$ = √3R$T_{Ar}$ / $M_{Ar}$ ………… (i)
Where, R is the universal gas constant.
$T_{Ar}$ is the temperature of argon gas.
The rms speed of helium is given by $(V_{rms})_{He}$ = √3R$T_{He}$ / $M_{He}$ … (ii)
Given that $(V_{rms})_{Ar}$ = $(V_{rms})_He$
√3$RT_{Ar} / M_{Ar}$= √3$RT_{He} / M_{He}$
$T_{Ar} / M_{Ar} = T_{He} / M_{He}$
$T_{Ar} = T_{He} / M_{He}$ × $M_{Ar}$ = (253 / 4) × 39.9
We get $= 2523.675 = 2.52 × 10^{3} K$
Hence, the temperature of the argon atom is $2.52 × 10^{3}K$
12.10 Estimate the mean free path and collision frequency f a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions Molecular mass of $N_{2}$ = 28.0 u.
Solution :
Mean free path = 1.11 × $10^{-7}$ m
Collision frequency = 4.58 × $10^{9} s^{-1}$
Successive collision time ≅ 500 × (Collision time)
The pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × $10^{5}$ Pa
The temperature inside the cylinder, T = 17°C = 290 K
The radius of a nitrogen molecule, r = 1.0 Å = 1 × $10^{10}$ m
Diameter, d = 2 × 1 × $10^{10}$ = 2 × $10^{10}$ m
The molecular mass of nitrogen, M = 28.0 g = 28 × $10^{-3}$ kg
The root mean square speed of nitrogen is given by the relation $V_{rms}$ = √3RT / M
Where, R is the universal gas constant = 8.314 J $mol^{-1} K^{-1}$
Hence, $V_{rms}$ = 3 × 8.314 × 290 / 28 × $10^{-3}$
On calculation,
we get = 508.26 m/s
The mean free path (l) is given by the relation l = KT / √2 × π × $d^{2}$ × P
Where, k is the Boltzmann constant = 1.38 × $10^{-23} kg m^{2} s^{-2} K^{-1}$
Hence, l = (1.38 × $10^{-23}$ × 290) / (√2 × 3.14 × (2 × $10^{-10})^{2}$ × 2.026 × $10^{5}$
We get, = 1.11 × $10^{-7}$ m
Collision frequency = $V_{rms}$ / l = 508.26 / 1.11 × $10^{-7}$
On calculation, we get = 4.58 × $10^{9} s^{-1}$
The collision time is given as
T = d / Vrms = 2 × $10^{-10}$ / 508.26
On further calculation, we get = 3.93 × $10^{-13}$ s
Time taken between successive collisions T’ = l / $V_{rms}$ = 1.11 × $10^{-7}$ / 508.26
We get, = 2.18 × $10^{-10}$
Hence,
T’ / T = 2.18 × $10^{-10}$ / 3.93 × $10^{-13}$
On calculation, we get
= 500
Therefore, the time taken between successive collisions is 500 times the time taken for a collision.