Solutions for Class 11CBSE Physics Chapter 13 Kinetic Theory (Updated)
Kinetic Theory : NCERT Solutions – Class 11 Physics
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 3Å.
Answer
Diameter of an oxygen molecule,d= 3Å
Radius,r=d/2 = 3/2 = 1.5 Å = 1.5 × 10–8cm
Actual
Molecular volume of oxygen gas,V= (4/3)πr3N
Where,Nis Avogadro’s number = 6.023 × 1023molecules/mole
∴V= (4/3) × 3.14 × (1.5 × 10-8)3× 6.023 × 1023= 8.51 cm3
Ratio of the molecular volume to the actual volume of oxygen = 8.51 / 22400
= 3.8 × 10-4.
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 °C). Show that it is 22.4 litres.
Answer
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:
PV=nRT
Where,
Ris the universal gas constant = 8.314 J mol–1K–1
n= Number of moles = 1
T= Standard temperature = 273 K
P= Standard pressure = 1 atm = 1.013 × 105Nm–2
∴V=nRT/P
= 1 × 8.314 × 273 / (1.013 × 105)
= 0.0224 m3
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.
13.3. Figure 13.8 shows plot ofPV/Tversus P for 1.00×10–3kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true:T1>T2orT1<T2?
(c) What is the value ofPV/Twhere the curves meet on they-axis?
(d) If we obtained similar plots for 1.00 ×10–3kg of hydrogen, would we get the same value ofPV/Tat the point where the curves meet on they-axis? If not, what mass of hydrogen yields the same value ofPV/T(for low pressure high temperature region of the plot)? (Molecular mass of H2= 2.02μ, of O2= 32.0μ,R= 8.31 J mo1–1K–1.)
Answer
(a)The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratioPV/Tis equal.
μR(μis 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.
(b)The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperatureT1is closer to the dotted plot than the curve of the gas at temperatureT2. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore,T1>T2is true for the given plot.
(c) The value of the ratioPV/T, where the two curves meet, isμR. This is because the ideal gas equation is given as:
PV=μRT
PV/T =μR
Where,
Pis the pressure
Tis the temperature
Vis the volume
μ is the number of moles
Ris the universal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen = 1 × 10–3kg = 1 g
R= 8.314 J mole–1K–1
∴PV/T= (1/32) × 8.314
= 0.26 J K-1
Therefore, the value of the ratioPV/T, where the curves meet on they-axis, is
0.26 J K–1.
(d)If we obtain similar plots for 1.00 × 10–3kg of hydrogen, then we will not get the same value ofPV/Tat the point where the curves meet they-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have:
PV/T= 0.26 J K-1
R= 8.314 J mole–1K–1
Molecular mass (M) of H2= 2.02 u
PV/T= μRat constant temperature
Where, μ =m/M
m= Mass of H2
∴m= (PV/T) × (M/R)
= 0.26 × 2.02 / 8.31
= 6.3 × 10–2g = 6.3 × 10–5kg
Hence, 6.3 × 10–5kg of H2will yield the same value ofPV/T.
Page No: 334
13.4. An oxygen cylinder of volume 30 litres 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 J mol–1K–1, molecular mass of O2= 32μ).
Answer
Volume of oxygen,V1= 30 litres = 30 × 10–3m3
Gauge pressure,P1= 15 atm = 15 × 1.013 × 105Pa
Temperature,T1= 27°C = 300 K
Universal gas constant,R= 8.314 J mole–1K–1
Let the initial number of moles of oxygen gas in the cylinder ben1.
The gas equation is given as:
P1V1=n1RT1
∴n1=P1V1/RT1
= (15.195 × 105× 30 × 10-3) / (8.314 × 300) = 18.276
Butn1=m1/M
Where,
m1= Initial mass of oxygen
M= Molecular mass of oxygen = 32 g
∴m1=n1M= 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume,V2= 30 litres = 30 × 10–3m3
Gauge pressure,P2= 11 atm = 11 × 1.013 × 105Pa
Temperature,T2= 17°C = 290 K
Letn2be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P2V2=n2RT2
∴n2=P2V2/RT2
= (11.143 × 105× 30 × 10-3) / (8.314 × 290) = 13.86
Butn2=m2/M
Where,
m2is the mass of oxygen remaining in the cylinder
∴m2=n2M= 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
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
13.5. An air bubble of volume 1.0 cm3rises 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?
Answer
Volume of the air bubble,V1= 1.0 cm3= 1.0 × 10–6m3
Bubble rises to height,d= 40 m
Temperature at a depth of 40 m,T1= 12°C = 285 K
Temperature at the surface of the lake,T2= 35°C = 308 K
The pressure on the surface of the lake:
P2= 1 atm = 1 ×1.013 × 105Pa
The pressure at the depth of 40 m:
P1= 1 atm +dρg
Where,
ρis the density of water = 103kg/m3
gis the acceleration due to gravity = 9.8 m/s2
∴P1= 1.013 × 105+ 40 × 103× 9.8 = 493300 Pa
We haveP1V1/T1=P2V2/T2
Where,V2is the volume of the air bubble when it reaches the surface
V2=P1V1T2/T1P2
= 493300 × 1 × 10-6× 308 / (285 × 1.013 × 105)
= 5.263 × 10–6m3or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
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 m3at a temperature of 27 °C and 1 atm pressure.
Answer
Volume of the room,V= 25.0 m3
Temperature of the room,T= 27°C = 300 K
Pressure in the room,P= 1 atm = 1 × 1.013 × 105Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:
PV=kBNT
Where,
KBis Boltzmann constant = 1.38 × 10–23m2kg s–2K–1
Nis the number of air molecules in the room
∴N=PV/kBT
= 1.013 × 105× 25 / (1.38 × 10-23× 300)
= 6.11 × 1026molecules
Therefore, the total number of air molecules in the given room is 6.11 × 1026.
13.7. 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).Answer
(i) At room temperature,T= 27°C = 300 K
Average thermal energy = (3/2)kT
Wherekis Boltzmann constant = 1.38 × 10–23m2kg s–2K–1
∴ (3/2)kT= (3/2) × 1.38 × 10-38× 300
= 6.21 × 10–21J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21J.
(ii) On the surface of the sun,T= 6000 K
Average thermal energy = (3/2)kT
= (3/2) × 1.38 × 10-38× 6000
= 1.241 × 10-19J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19J.
(iii) At temperature,T= 107K
Average thermal energy = (3/2)kT
= (3/2) × 1.38 × 10-23× 107
= 2.07 × 10-16J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16J.
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 isvrmsthe largest?
Answer
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’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 × 1023.The root mean square speed (vrms) of a gas of massm, and temperatureT, is given by the relation:vrms= (3kT/m)1/2
where,
kis Boltzmann constant
For the given gases,kandTare constants.
Hencevrmsdepends only on the mass of the atoms, i.e.,
vrms∝ (1/m)1/2
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.
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 – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer
Temperature of the helium atom,THe= –20°C= 253 K
Atomic mass of argon,MAr= 39.9 u
Atomic mass of helium,MHe= 4.0 u
Let, (vrms)Arbe the rms speed of argon.
Let (vrms)Hebe the rms speed of helium.
The rms speed of argon is given by:
= 2523.675 = 2.52 × 103K
Therefore, the temperature of the argon atom is 2.52 × 103K.
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 °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 N2= 28.0 u).
Answer
Mean free path = 1.11 × 10–7m
Collision frequency = 4.58 × 109s–1
Successive collision time ≈ 500 × (Collision time)
Pressure inside the cylinder containing nitrogen,P= 2.0 atm = 2.026 × 105Pa
Temperature inside the cylinder,T= 17°C =290 K
Radius of a nitrogen molecule,r= 1.0 Å = 1 × 1010m
Diameter,d= 2 × 1 × 1010= 2 × 1010m
Molecular mass of nitrogen,M= 28.0 g = 28 × 10–3kg
The root mean square speed of nitrogen is given by the relation:
Collision frequency =vrms/ l
= 508.26 / (1.11 × 10-7) = 4.58 × 109s-1
Collision time is given as:
T=d/vrms
= 2 × 10-10/ 508.26 = 3.93 × 10-13s
Time taken between successive collisions:
T‘ = l /vrms
= 1.11 × 10-7/ 508.26 = 2.18 × 10-10s
∴T‘ /T= 2.18 × 10-10/ (3.93 × 10-13) = 500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.
Page No: 335
Additional Exercises
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?
Answer
Length of the narrow bore,L= 1 m = 100 cm
Length of the mercury thread,l= 76 cm
Length of the air column between mercury and the closed end,la= 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 – (76 + 15) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Lethcm of mercury flow out as a result of atmospheric pressure.
∴Length of the air column in the bore = 24 +hcm
And, length of the mercury column = 76 –hcm
Initial pressure,P1= 76 cm of mercury
Initial volume,V1= 15 cm3
Final pressure,P2= 76 – (76 –h) =hcm of mercury
Final volume,V2= (24 +h) cm3
Temperature remains constant throughout the process.
∴P1V1=P2V2
76 × 15 =h(24 +h)
h2+ 24h– 1140 = 0
= 23.8 cm or –47.8 cm
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.
13.12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3s–1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R1/R2= (M2/M1)1/2, where R1, R2are diffusion rates of gases 1 and 2, and M1and M2their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer
Rate of diffusion of hydrogen,R1= 28.7 cm3s–1
Rate of diffusion of another gas,R2= 7.2 cm3s–1
According to Graham’s Law of diffusion, we have:
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.
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
n2=n1exp [-mg(h2–h1)/kBT]
Wheren2,n1refer to number density at heightsh2andh1respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2=n1exp [-mg NA(ρ– P′) (h2–h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NAis Avogadro’s number, andRthe universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer
According to the law of atmospheres, we have:
n2=n1exp [-mg(h2–h1) /kBT] …(i)
where,
n1is the number density at heighth1, andn2is the number density at heighth2
mg is the weight of the particle suspended in the gas column
Density of the medium =ρ‘
Density of the suspended particle =ρ
Mass of one suspended particle =m‘
Mass of the medium displaced =m
Volume of a suspended particle =V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:
Weight of the medium displaced – Weight of the suspended particle
=mg–m‘g
=mg–Vρ’g =mg– (m/ρ)ρ‘g
=mg(1 – (ρ‘/ρ) ) ….(ii)
Gas constant,R=kBN
kB=R/N ….(iii)
Substituting equation(ii)in place ofmg in equation(i)and then using equation(iii), we get:
n2=n1exp [-mg(h2–h1) /kBT]
=n1exp [-mg(1 – (ρ’/ρ) )(h2–h1)(N/RT) ]
=n1exp [-mg(ρ – ρ’)(h2–h1)(N/RTρ) ]
13.14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s 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 Å].AnswerIf r is the radius of the atom, then volume of each atom = 4/3 π r3Volume of all atoms in one mole of substance = 4/3 π r3× N = M/ρ
∴ r = [ 3M / 4πρN]1/3
For Carbon,
M= 12.01 × 10-3Kg
ρ= 2.22 × 103Kg m-3
Similarly,for gold,r= 1.59 Åfor liquid nitrogen,r= 1.77 Åfor lithium, r = 1.73 Åfor liquid fluorine, r = 1.88 Å
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