NCERT Class 11 Physics Chapter 12 Kinetic Theory Solutions, NCERT Class 11 Physics Chapter 12 Kinetic Theory Notes to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 11 Physics Chapter 12 Kinetic Theory Question Answer and select needs one.
NCERT Class 11 Physics Chapter 12 Kinetic Theory
Also, you can read the SCERT book online in these sections NCERT Class 11 Physics Chapter 12 Kinetic Theory Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given NCERT Class 11 Physics Chapter 12 Kinetic Theory Solutions for All Subjects, You can practice these here.
Kinetic Theory
Chapter: 12
| Part – II |
EXERCISE
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.
Ans: Volume of a single oxygen Molecule:
Diameter = 3 Å = 3 × 10-10 m
Radius = 1.5 × 10-10m
Volume = 4/3π (1.5 × 10-10)3
= 4.52 × 10-30 m3
1 mole of oxygen gas contains 6.022 × 1023 molecules.
Total volume = 4.52 × 10-30 × 6.022 × 1023 m3
Volume occupied by 1 mole of Oxygen gas at STP:
= 22.141 litres = 22.141 × 10-3 m3
Fraction:
= 2.72 × 10-6 / 22.414 × 10-3
= 1.23 × 10-4
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.
Ans: PV = nRT
P = 1 atm
T = 273.15 k
R = 0.0821 L
For 1 mole of gas (n = 1 mol)
V = nRT/P
V = 1 × 0.0821 × 273.15 / 1
V = 22.4 L.
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?
Ans: The dotted plot shows that pV/T is a constant quantity (=μR). This signifies the ideal gas behaviour.
(b) Which is true: T1 > T2 or T1 < T2?
Ans: T1 > T2.
(c) What is the value of PV/T where the curves meet on the y-axis?
Ans: At the point where the curve meets the y-axis, pV / T
= μR, Where μ is the number of moles of oxygen gas. Here,
mass of oxygen
m = 1.00 × 10-3 kg
Also, molecular mass,
M = 32 × 10-3 kg
No. of moles μ = m/M = 1.00 × 10-3 kg) / (32.0 × 1-3 kg/mol) = ⅓ mol.
pV / T = μR = ⅓ × 8.31
= 0.26 JK-1.
(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 H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)
Ans: Here: (mO / MO) × MH
Where:
mH is the mass of hydrogen
mO is the mass of oxygen (give as 1.00 × 10-3 kg)
MO is the molar mass of oxygen (32.0 g/mol)
MH is the molar mass of hydrogen (2.02 g/mol)
mH = (1.00 × 10-3 kg/ 32.0 g/mol) × 2.02 g/mol
= 6.3 × 10-5 kg.
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 J mol–1 K–1, molecular mass of O2 = 32 u).
Ans: Pressures to absolute pressures:
Initial = 15 atm + 1 atm = 16 atm
Final = 11 atm + 1 atm = 12 atm
Pressure to Pascals:
Initial = P1 = 16 atm × 101325 pa/atm = 1621200 pa
Final P2 = 12 atm × 101325 pa/atm = 1215900 pa
Here,
V = 30 L = 30 × 10-3 m3
Initial moles (n1) = P1V / RT1
= 1621200 pa × 30 × 10-3 m3 / 8.31 J × 300.15 K
= 48636 / 2496.66
= 19.48 mol
Final moles (n2) = P2V / RT2
= 1215900 Pa × 30 × 10-3 m3 / 8.31 j × 290.15 K
= 35477 / 2415.9
= 15.08 mol
Mass = (n1 – n2) × molecular mass
= 19.48 – 15.08
= 4.40 mol
Molecular mass of O2 = 32 g/mol
Mass = 4.40 mol × 32 g/mol
Mass = 140.8 g.
5. An air bubble of volume 1.0 cm3 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?
Ans: V1 = 1.0 cm3
T1 = 12 °C = 285.15 k
T2 = 35°C = 308.15 k
Depth of lake = 40m
Initial Pressure P1 = atmospheric pressure + water pressure
Water pressure = (40m × 1000 kg/m3 × 9.8 m/s2) / 101325 pa/atm
= 3.87 atm.
P1 = 1 atm + 3.87 atm = 4.87 atm
P2 = 1 atm
Combined gas law:
P1V1/T1 = P2V2 / T2
V2 = P1V1T2 / P2T1
= 4.87 atm × 1.0 cm3 × 308.15 K / 1 atm × 285.15 K
V2 = 1503.80 / 285.15
= 5.27 cm3.
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.
Ans: Given:
V = 25.0 m3
T = 27°C = 300 K
P = 1 atm = 101325 Pa
R = 8.31 J/mol K
NA = 6.022 × 1023 molecules/mol
The ideal gas law:
PV = nRT
n = PV / RT
n = (101325 Pa × 25.0m3 / 8.31 J) / mol K × 300.15 K
= n = 2533125 / 2493.24
= 1016.06 mol.
Total number of air molecules.
= n × NA
= 1016.66 mol × 6.022 × 1023 molecules/mol
= 6.12 × 1026 molecules.
7. Estimate the average thermal energy of a helium atom at:
(i) Room temperature (27 °C).
Ans: The average thermal energy of an atom is:
E = 3/2 KB T.
Where, KB = 1.38 × 10-23 JK-1
Is Boltzmann’s constant.
T = 27°C
= 27 + 273 = 300 k
E = 3/2 × 1.38 x 10-23 × 300
= 6.21 × 10-21J.
(ii) The temperature on the surface of the Sun (6000 K).
Ans: Average thermal energy = 3/2 × 1.38 × 10-23 J/K × 6000 K
= 3/2 × 8.28 × 10-20 J
= 1.24 × 10-19J.
(iii) The temperature of 10 million kelvin (the typical core temperature in the case of a star).
Ans: Average thermal energy = 3/2 × 1.38 × 10-23 J/K × 10,000,000 K
= 3/2 × 1.38 × 10-16 J
= 2.07 × 10-16 J.
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 an 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?
Ans: (i) Yes, because, according to Avogadro’s hypothesis, equal volume of all the gases contains equal number of molecules at the same temperature and pressure.
(ii) No, vr.m.s = √3 KBT / m
Because, m is different for different gases, therefore, v r.m.s. Is also different.
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).
Ans: TAr / mAr = THe / mHe
TAr = THe × mAr / mHe
THe = 20∘C + 273.15 = 253.15 K
TAr = 253.15 × 39.9 / 4.0
= 253.15 × 9.975
= 2523.17 K.
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 0C. 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).
Ans: λ = kT / (πd2P)
f = v / λ
Where:
k = 1.38 × 1023 J/K
T = 290 K
d = 2 × 1.0 Å = 2.0 × 10−10 m
P = 2.0 atm × 1.01 × 105 Pa/atm = 2.02 × 105 Pa
v = Average speed
Calculate average speed (v):
v = √(8RT/πM)
= √(8 × 8.31 J/mol·K × 290 K / (π × 28.0 u × 1.66 × 10−27 kg/u))
= 508.5 m/s
Calculate free path (λ):
λ = (1.38 × 10−23 J/K × 290 K) / (π × (2.0 × 10−10 m)2 × 2.02 × 105 Pa)
= 1.14 × 10−7 m
Calculate (f):
f = 508.5 m/s / 1.14 × 10−7 m
= 4.46 × 109 s−1
τ = d / v
= 2.0 × 10−10 m / 508.5 m/s
= 3.93 × 10−13 s
t = λ / v
= 1.14 × 10−7 m / 508.5 m/s
= 2.24 × 10−10 s.
