{"id":201102,"date":"2021-03-03T09:28:16","date_gmt":"2021-03-03T09:28:16","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=201102"},"modified":"2023-09-20T08:31:49","modified_gmt":"2023-09-20T08:31:49","slug":"ncert-solutions-for-11th-class-physics-chapter-13-kinetic-theory","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-11th-class-physics-chapter-13-kinetic-theory\/","title":{"rendered":"NCERT Solutions for 11th Class Physics: Chapter 13-Kinetic Theory"},"content":{"rendered":"\n

Class 11: Physics Chapter 13 solutions. Complete Class 11 Physics Chapter 13 Notes.<\/p>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 13-Kinetic Theory<\/strong><\/h2>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

Page No: 333<\/p>\n\n\n\n

Excercises<\/h4>\n\n\n\n

13.1. 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\u00c5.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Diameter of an oxygen molecule, d<\/em> = 3\u00c5
Radius, r<\/em> = d<\/em>\/2 = 3\/2 = 1.5 \u00c5 = 1.5 \u00d7 10\u20138<\/sup> cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3<\/sup>
Molecular volume of oxygen gas, V<\/em> = (4\/3)\u03c0r3<\/sup>N
Where, N<\/em> is Avogadro\u2019s number = 6.023 \u00d7 1023 <\/sup>molecules\/mole
\u2234 V<\/em> = (4\/3) \u00d7 3.14 \u00d7 (1.5 \u00d7 10-8<\/sup>)3<\/sup> \u00d7 6.023 \u00d7 1023<\/sup> = 8.51 cm3<\/sup>
Ratio of the molecular volume to the actual volume of oxygen = 8.51 \/ 22400
= 3.8 \u00d7 10-4<\/sup>.<\/p>\n\n\n\n

13.2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 \u00b0C). Show that it is 22.4 litres.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The ideal gas equation relating pressure (P<\/em>), volume (V<\/em>), and absolute temperature (T<\/em>) is given as:
PV<\/em> = nRT<\/em>
Where,
R<\/em> is the universal gas constant = 8.314 J mol\u20131<\/sup> K\u20131<\/sup>
n<\/em> = Number of moles = 1
T<\/em> = Standard temperature = 273 K
P<\/em> = Standard pressure = 1 atm = 1.013 \u00d7 105<\/sup> Nm\u20132<\/sup>
\u2234 V<\/em> = nRT<\/em> \/ P<\/em>
= 1 \u00d7 8.314 \u00d7 273 \/ (1.013 \u00d7 105<\/sup>)
= 0.0224 m3<\/sup>
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.<\/p>\n\n\n\n

13.3. Figure 13.8 shows plot of PV<\/em>\/T <\/em>versus P for 1.00\u00d710\u20133<\/sup> kg of oxygen gas at two different temperatures.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

(a) What does the dotted plot signify?
(b) Which is true: T<\/em>1 <\/sub>> T<\/em>2<\/sub> or T<\/em>1<\/sub> < T<\/em>2<\/sub>?
(c) What is the value of PV<\/em>\/T <\/em>where the curves meet on the y-<\/em>axis?
(d) If we obtained similar plots for 1.00 \u00d710\u20133<\/sup> kg of hydrogen, would we get the same value of PV<\/em>\/T <\/em>at the point where the curves meet on the y<\/em>-axis? If not, what mass of hydrogen yields the same value of PV<\/em>\/T <\/em>(for low pressure high temperature region of the plot)? (Molecular mass of H2 <\/sub>= 2.02\u03bc, of O2<\/sub> = 32.0\u03bc, R <\/em>= 8.31 J mo1\u20131<\/sup> K\u20131<\/sup>.)<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a)The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV<\/em>\/T<\/em> is equal.
\u03bc<\/em>R<\/em> (\u03bc<\/em> is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.<\/p>\n\n\n\n

(b)The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T<\/em>1<\/sub> is closer to the dotted plot than the curve of the gas at temperature T<\/em>2<\/sub>. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T<\/em>1<\/sub> > T<\/em>2<\/sub> is true for the given plot.<\/p>\n\n\n\n

(c) The value of the ratio PV<\/em>\/T<\/em>, where the two curves meet, is \u03bc<\/em>R<\/em>. This is because the ideal gas equation is given as:
PV<\/em> = \u03bc<\/em>RT<\/em>
PV\/T = <\/em>\u03bcR
Where,
P<\/em> is the pressure
T<\/em> is the temperature
V <\/em>is the volume
\u03bc is the number of moles
R<\/em> is the universal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen = 1 \u00d7 10\u20133<\/sup> kg = 1 g
R<\/em> = 8.314 J mole\u20131<\/sup> K\u20131<\/sup>
\u2234 PV<\/em>\/T<\/em> = (1\/32) \u00d7 8.314
= 0.26 J K-1<\/sup>
Therefore, the value of the ratio PV<\/em>\/T<\/em>, where the curves meet on the y<\/em>-axis, is
0.26 J K\u20131<\/sup>.<\/p>\n\n\n\n

(d)If we obtain similar plots for 1.00 \u00d7 10\u20133<\/sup> kg of hydrogen, then we will not get the same value of PV<\/em>\/T<\/em> at the point where the curves meet the y<\/em>-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have:
PV<\/em>\/T<\/em> = 0.26 J K-1<\/sup>
R<\/em> = 8.314 J mole\u20131<\/sup> K\u20131<\/sup>
Molecular mass (M<\/em>) of H2<\/sub> = 2.02 u
PV<\/em>\/T<\/em> = \u03bcR<\/em> at constant temperature
Where, \u03bc = m<\/em>\/M<\/em>
m<\/em> = Mass of H2<\/sub>
\u2234 m<\/em> = (PV<\/em>\/T<\/em>) \u00d7 (M<\/em>\/R<\/em>)
= 0.26 \u00d7 2.02 \/ 8.31
= 6.3 \u00d7 10\u20132<\/sup> g = 6.3 \u00d7 10\u20135<\/sup> kg
Hence, 6.3 \u00d7 10\u20135<\/sup> kg of H2<\/sub> will yield the same value of PV<\/em>\/T<\/em>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

Page No: 334<\/p>\n\n\n\n

13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 \u00b0C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 \u00b0C. Estimate the mass of oxygen taken out of the cylinder (R <\/em>= 8.31 J mol\u20131<\/sup> K\u20131<\/sup>, molecular mass of O2<\/sub> = 32\u03bc).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Volume of oxygen, V<\/em>1<\/sub> = 30 litres = 30 \u00d7 10\u20133<\/sup> m3<\/sup>
Gauge pressure, P<\/em>1<\/sub> = 15 atm = 15 \u00d7 1.013 \u00d7 105<\/sup> Pa
Temperature, T<\/em>1<\/sub> = 27\u00b0C = 300 K
Universal gas constant, R<\/em> = 8.314 J mole\u20131<\/sup> K\u20131<\/sup>
Let the initial number of moles of oxygen gas in the cylinder be n<\/em>1<\/sub>.
The gas equation is given as:
P<\/em>1<\/sub>V<\/em>1<\/sub> = n<\/em>1<\/sub>RT<\/em>1<\/sub>
\u2234 n<\/em>1<\/sub> = P<\/em>1<\/sub>V<\/em>1<\/sub>\/ RT<\/em>1<\/sub>
= (15.195 \u00d7 105<\/sup> \u00d7 30 \u00d7 10-3<\/sup>) \/ (8.314 \u00d7 300)  =  18.276
But n<\/em>1<\/sub> = m<\/em>1<\/sub> \/ M<\/em>
Where,
m<\/em>1<\/sub> = Initial mass of oxygen
M<\/em> = Molecular mass of oxygen = 32 g
\u2234m<\/em>1<\/sub> = n<\/em>1<\/sub>M <\/em>= 18.276 \u00d7 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V<\/em>2<\/sub> = 30 litres = 30 \u00d7 10\u20133<\/sup> m3<\/sup>
Gauge pressure, P<\/em>2<\/sub> = 11 atm = 11 \u00d7 1.013 \u00d7 105<\/sup> Pa
Temperature, T<\/em>2<\/sub> = 17\u00b0C = 290 K
Let n<\/em>2<\/sub> be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P<\/em>2<\/sub>V<\/em>2<\/sub> = n<\/em>2<\/sub>RT<\/em>2<\/sub>
\u2234 n<\/em>2<\/sub> = P<\/em>2<\/sub>V<\/em>2<\/sub>\/ RT<\/em>2<\/sub>
= (11.143 \u00d7 105<\/sup> \u00d7 30 \u00d7 10-3<\/sup>) \/ (8.314 \u00d7 290)  =  13.86
But n<\/em>2<\/sub> = m<\/em>2<\/sub> \/ M<\/em>
Where,
m<\/em>2<\/sub> is the mass of oxygen remaining in the cylinder
\u2234 m<\/em>2<\/sub> = n<\/em>2<\/sub>M <\/em>= 13.86 \u00d7 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 \u2013 Final mass of oxygen in the cylinder
= m<\/em>1<\/sub> \u2013 m<\/em>2<\/sub>
= 584.84 g \u2013 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.<\/p>\n\n\n\n

13.5. An air bubble of volume 1.0 cm3<\/sup> rises from the bottom of a lake 40 m deep at a temperature of 12 \u00b0C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 \u00b0C?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Volume of the air bubble, V<\/em>1<\/sub> = 1.0 cm3<\/sup> = 1.0 \u00d7 10\u20136<\/sup> m3<\/sup>
Bubble rises to height, d<\/em> = 40 m
Temperature at a depth of 40 m, T<\/em>1<\/sub> = 12\u00b0C = 285 K
Temperature at the surface of the lake, T<\/em>2<\/sub> = 35\u00b0C = 308 K
The pressure on the surface of the lake:
P<\/em>2<\/sub> = 1 atm = 1 \u00d71.013 \u00d7 105<\/sup> Pa
The pressure at the depth of 40 m:
P<\/em>1<\/sub> = 1 atm + d\u03c1<\/em>g
Where,
\u03c1<\/em> is the density of water = 103<\/sup> kg\/m3<\/sup>
g<\/em> is the acceleration due to gravity = 9.8 m\/s2<\/sup>
\u2234 P<\/em>1<\/sub> = 1.013 \u00d7 105<\/sup> + 40 \u00d7 103<\/sup> \u00d7 9.8 = 493300 Pa
We have P<\/em>1<\/sub>V<\/em>1<\/sub> \/ T<\/em>1<\/sub> = P<\/em>2<\/sub>V<\/em>2<\/sub> \/ T<\/em>2<\/sub>
Where, V<\/em>2<\/sub> is the volume of the air bubble when it reaches the surface
V<\/em>2<\/sub> = P<\/em>1<\/sub>V<\/em>1<\/sub>T<\/em>2<\/sub> \/ T<\/em>1<\/sub>P<\/em>2<\/sub>
= 493300 \u00d7 1 \u00d7 10-6<\/sup> \u00d7 308 \/ (285 \u00d7 1.013 \u00d7 105<\/sup>)
= 5.263 \u00d7 10\u20136<\/sup> m3<\/sup> or 5.263 cm3<\/sup>
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3<\/sup>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

13.6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3<\/sup> at a temperature of 27 \u00b0C and 1 atm pressure.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Volume of the room, V<\/em> = 25.0 m3<\/sup>
Temperature of the room, T<\/em> = 27\u00b0C = 300 K
Pressure in the room, P<\/em> = 1 atm = 1 \u00d7 1.013 \u00d7 105<\/sup> Pa
The ideal gas equation relating pressure (P<\/em>), Volume (V<\/em>), and absolute temperature (T<\/em>) can be written as:
PV <\/em>= k<\/em>B<\/em><\/sub>NT<\/em>
Where,
K<\/em>B<\/em><\/sub> is Boltzmann constant = 1.38 \u00d7 10\u201323<\/sup> m2<\/sup> kg s\u20132<\/sup> K\u20131<\/sup>
N<\/em> is the number of air molecules in the room
\u2234 N<\/em> = PV<\/em> \/ kB<\/sub>T<\/em>
= 1.013 \u00d7 105<\/sup> \u00d7 25 \/ (1.38 \u00d7 10-23<\/sup> \u00d7 300)
= 6.11 \u00d7 1026<\/sup> molecules
Therefore, the total number of air molecules in the given room is 6.11 \u00d7 1026<\/sup>.<\/p>\n\n\n\n

13.7. Estimate the average thermal energy of a helium atom at (i) room temperature (27 \u00b0C), (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).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) At room temperature, T<\/em> = 27\u00b0C = 300 K
Average thermal energy = (3\/2)kT<\/em>
Where k <\/em>is Boltzmann constant = 1.38 \u00d7 10\u201323<\/sup> m2<\/sup> kg s\u20132<\/sup> K\u20131<\/sup>
\u2234 (3\/2)kT<\/em> = (3\/2) \u00d7 1.38 \u00d7 10-38<\/sup> \u00d7 300
= 6.21 \u00d7 10\u201321<\/sup>J
Hence, the average thermal energy of a helium atom at room temperature (27\u00b0C) is 6.21 \u00d7 10\u201321<\/sup> J.<\/p>\n\n\n\n

(ii) On the surface of the sun, T<\/em> = 6000 K
Average thermal energy = (3\/2)kT<\/em>
= (3\/2) \u00d7 1.38 \u00d7 10-38<\/sup> \u00d7 6000
= 1.241 \u00d7 10-19<\/sup> J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 \u00d7 10\u201319<\/sup> J.<\/p>\n\n\n\n

(iii) At temperature, T<\/em> = 107<\/sup> K
Average thermal energy = (3\/2)kT<\/em>
= (3\/2) \u00d7 1.38 \u00d7 10-23<\/sup> \u00d7 107<\/sup>
= 2.07 \u00d7 10-16<\/sup> J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 \u00d7 10\u201316<\/sup> J.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

13.8. 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 v<\/em>rms<\/sub> the largest?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

All the three vessels have the same capacity, they have the same volume.
Hence, each gas has the same pressure, volume, and temperature.
According to Avogadro\u2019s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro\u2019s number, N<\/em> = 6.023 \u00d7 1023<\/sup>.
The root mean square speed (v<\/em>rms<\/sub>) of a gas of mass m<\/em>, and temperature T<\/em>, is given by the relation:
v<\/em>rms<\/sub> = (3kT<\/em>\/m<\/em>)1\/2<\/sup>
where,
k<\/em> is Boltzmann constant<\/p>\n\n\n\n

For the given gases, k<\/em> and T<\/em> are constants.
Hence v<\/em>rms<\/sub> depends only on the mass of the atoms, i.e.,
v<\/em>rms<\/sub> \u221d (1\/m<\/em>)1\/2<\/sup>
Therefore, 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. Hence, neon has the largest root mean square speed among the given gases.<\/p>\n\n\n\n

13.9. 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 \u2013 20 \u00b0C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Temperature of the helium atom, T<\/em>He<\/sub> = \u201320\u00b0C= 253 K
Atomic mass of argon, M<\/em>Ar <\/sub>= 39.9 u
Atomic mass of helium, M<\/em>He<\/sub> = 4.0 u
Let, (v<\/em>rms<\/sub>)Ar<\/sub> be the rms speed of argon.
Let (v<\/em>rms<\/sub>)He<\/sub> be the rms speed of helium.
The rms speed of argon is given by:<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

= 2523.675 = 2.52 \u00d7 103<\/sup> K
Therefore, the temperature of the argon atom is 2.52 \u00d7 103<\/sup> K.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

13.10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 \u00b0C. Take the radius of a nitrogen molecule to be roughly 1.0 \u00c5. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2<\/sub> = 28.0 u).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Mean free path = 1.11 \u00d7 10\u20137<\/sup> m
Collision frequency = 4.58 \u00d7 109<\/sup> s\u20131<\/sup>
Successive collision time \u2248 500 \u00d7 (Collision time)
Pressure inside the cylinder containing nitrogen, P<\/em> = 2.0 atm = 2.026 \u00d7 105<\/sup> Pa
Temperature inside the cylinder, T<\/em> = 17\u00b0C =290 K
Radius of a nitrogen molecule, r<\/em> = 1.0 \u00c5 = 1 \u00d7 1010<\/sup> m
Diameter, d<\/em> = 2 \u00d7 1 \u00d7 1010<\/sup> = 2 \u00d7 1010<\/sup> m
Molecular mass of nitrogen, M<\/em> = 28.0 g = 28 \u00d7 10\u20133<\/sup> kg
The root mean square speed of nitrogen is given by the relation:<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Collision frequency = v<\/em>rms<\/sub> \/ l
= 508.26 \/ (1.11 \u00d7 10-7<\/sup>)  =  4.58 \u00d7 109<\/sup> s-1<\/sup>
Collision time is given as:
T<\/em> = d<\/em> \/ v<\/em>rms<\/sub>
= 2 \u00d7 10-10<\/sup> \/ 508.26  =  3.93 \u00d7 10-13<\/sup> s<\/p>\n\n\n\n

Time taken between successive collisions:
T<\/em> ‘ = l \/ v<\/em>rms<\/sub>
= 1.11 \u00d7 10-7<\/sup> \/ 508.26  =  2.18 \u00d7 10-10<\/sup> s
\u2234 T<\/em> ‘ \/ T<\/em> = 2.18 \u00d7 10-10<\/sup> \/ (3.93 \u00d7 10-13<\/sup>)  =  500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

Page No: 335

Additional Exercises<\/strong><\/p>\n\n\n\n

13.11. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Length of the narrow bore, L<\/em> = 1 m = 100 cm
Length of the mercury thread, l<\/em> = 76 cm
Length of the air column between mercury and the closed end, l<\/em>a<\/em><\/sub> = 15 cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 \u2013 (76 + 15) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Let h<\/em> cm of mercury flow out as a result of atmospheric pressure.
\u2234Length of the air column in the bore = 24 + h<\/em> cm
And, length of the mercury column = 76 \u2013 h<\/em> cm
Initial pressure, P<\/em>1<\/sub> = 76 cm of mercury
Initial volume, V<\/em>1<\/sub> = 15 cm3<\/sup>
Final pressure, P<\/em>2<\/sub> = 76 \u2013 (76 \u2013 h<\/em>) = h<\/em> cm of mercury
Final volume, V<\/em>2<\/sub> = (24 + h<\/em>) cm3<\/sup>
Temperature remains constant throughout the process.
\u2234 P<\/em>1<\/sub>V<\/em>1<\/sub> = P<\/em>2<\/sub>V<\/em>2<\/sub>
76 \u00d7 15 = h<\/em> (24 + h<\/em>)
h<\/em>2<\/sup> + 24h<\/em> \u2013 1140 = 0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

= 23.8 cm or \u201347.8 cm<\/p>\n\n\n\n

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.<\/p>\n\n\n\n

13.12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3<\/sup> s\u20131<\/sup>. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3<\/sup> s\u20131<\/sup>. Identify the gas.
[Hint: Use Graham\u2019s law of diffusion: R1<\/sub>\/R2<\/sub> = (M2<\/sub>\/M1<\/sub>)1\/2<\/sup>, where R1<\/sub>, R2<\/sub> are diffusion rates of gases 1 and 2, and M1<\/sub> and M2<\/sub> their respective molecular masses. The law is a simple consequence of kinetic theory.]<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Rate of diffusion of hydrogen, R<\/em>1 <\/sub>= 28.7 cm3<\/sup> s\u20131<\/sup>
Rate of diffusion of another gas, R<\/em>2<\/sub> = 7.2 cm3<\/sup> s\u20131<\/sup>
According to Graham\u2019s Law of diffusion, we have:<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.<\/p>\n\n\n\n

13.13. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n<\/em>2<\/sub> = n<\/em>1<\/sub> exp [-mg <\/em>(h<\/em>2 <\/sub>\u2013 h<\/em>1<\/sub>)\/ k<\/em>B<\/em><\/sub>T<\/em>]
Where n<\/em>2<\/sub>, n<\/em>1<\/sub> refer to number density at heights h<\/em>2<\/sub> and h<\/em>1<\/sub> respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n<\/em>2<\/sub> = n<\/em>1<\/sub> exp [-mg N<\/em>A<\/em><\/sub>(\u03c1<\/em> – P<\/em>\u2032) (h<\/em>2<\/sub> \u2013h<\/em>1<\/sub>)\/ (\u03c1RT<\/em>)]
Where \u03c1 is the density of the suspended particle, and \u03c1\u2019 that of surrounding medium. [N<\/em>A<\/sub> is Avogadro\u2019s number, and R <\/em>the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

According to the law of atmospheres, we have:
n<\/em>2<\/sub> = n<\/em>1<\/sub> exp [-mg <\/em>(h<\/em>2 <\/sub>\u2013 h<\/em>1<\/sub>) \/ kBT<\/em>] \u2026 (i)<\/strong>
where,
n<\/em>1 <\/sub>is the number density at height h<\/em>1<\/sub>, and n<\/em>2<\/sub> is the number density at height h<\/em>2<\/sub>
m<\/em>g is the weight of the particle suspended in the gas column
Density of the medium = \u03c1<\/em>‘
Density of the suspended particle = \u03c1<\/em>
Mass of one suspended particle = m<\/em>‘
Mass of the medium displaced = m<\/em>
Volume of a suspended particle = V<\/em>
According to Archimedes\u2019 principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:<\/p>\n\n\n\n

Weight of the medium displaced \u2013 Weight of the suspended particle
= mg<\/em> \u2013 m<\/em>‘g<\/em>
=<\/em> mg<\/em> – V<\/em> \u03c1’<\/em> g<\/em>  =  mg<\/em> – (m<\/em>\/\u03c1<\/em>)\u03c1<\/em>‘g<\/em>
= mg<\/em>(1 – (\u03c1<\/em>‘\/\u03c1<\/em>) )   ….(ii)<\/strong>
Gas constant, R<\/em> = k<\/em>B<\/em><\/sub>N<\/em>
kB<\/sub> = R<\/em> \/ N<\/em>     ….(iii)<\/strong>
Substituting equation (ii)<\/strong> in place of m<\/em>g in equation (i)<\/strong> and then using equation (iii)<\/strong><\/em>, we get:
n<\/em>2<\/sub> = n<\/em>1<\/sub> exp [-mg <\/em>(h<\/em>2 <\/sub>\u2013 h<\/em>1<\/sub>) \/ k<\/em>B<\/em><\/sub>T<\/em>]
= n<\/em>1<\/sub> exp [-mg<\/em> (1 – (\u03c1’\/\u03c1) )(h<\/em>2 <\/sub>\u2013 h<\/em>1<\/sub>)(N<\/em>\/RT<\/em>) ]
= n<\/em>1<\/sub> exp [-mg<\/em> (\u03c1 – \u03c1’)(h<\/em>2 <\/sub>\u2013 h<\/em>1<\/sub>)(N<\/em>\/R<\/em>T\u03c1<\/em>) ]<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

13.14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

[Hint: Assume the atoms to be \u2018tightly packed\u2019 in a solid or liquid phase, and use the known value of Avogadro\u2019s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few \u00c5].<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

If r is the radius of the atom, then volume of each atom = 4\/3 \u03c0 r3<\/sup><\/p>\n\n\n\n

Volume of all atoms in one mole of substance = 4\/3 \u03c0 r3<\/sup> \u00d7 N = M\/\u03c1<\/em><\/p>\n\n\n\n

\u2234 r = [ 3M \/ 4\u03c0\u03c1N<\/em>]1\/3<\/sup>

For Carbon, 
M<\/em> = 12.01 \u00d7 10-3<\/sup> Kg
\u03c1 <\/em>= 2.22 \u00d7 103<\/sup> Kg m-3<\/sup><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Similarly,<\/p>\n\n\n\n

for gold, r<\/em> = 1.59 \u00c5<\/p>\n\n\n\n

for liquid nitrogen, r <\/em>= 1.77 \u00c5<\/p>\n\n\n\n

for lithium, r = 1.73 \u00c5<\/p>\n\n\n\n

for liquid fluorine, r = 1.88 \u00c5<\/p>\n\n\n\n

NCERT 11th Physics Chapter 13, class 11 Physics chapter 13 solutions<\/p>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 13: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 13-Kinetic Theory<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 11th Class Physics: Chapter 13-Kinetic Theory PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 11 Physics<\/strong>:<\/h4>\n\n\n\n