NCERT Solution (Class 11) > Physics Chapter 13 Kinetic Theory
NCERT Solution Class 11 Physics Chapter 13 Kinetic Theory: National Council of Educational Research and Training Class 11 Physics Chapter 13 Solution – Kinetic Theory. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
13 |
Chapter Name |
Kinetic Theory |
Topic |
Exercise Solution |
Kinetic Theory Chapter all Questions and Numericals Solution
Chapter 13 – Kinetic Theory
13.1)
13.3) Figure 13.8 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 H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)
ANSWER-
(a) The dotted plot is horizontal line which shows that PV/T is constant. This relation is ideal behavior of gas.
(b) The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. The dotted plot represents an ideal gas. We know that at higher temperature and lower pressure real gas behaves as ideal gas. Means plot with temperature T1 is near to ideal plot hence it is higher than t2 plot. Therefore, T1 > T2.
(c) From the ideal gas equation we have:
(d) Because molecular mass of hydrogen is different than oxygen. If we obtain similar plots for
1.00×10–3kg of hydrogen then we will not get the same value of PV/T at the point where the curves meet the y – axis.
We have:
13.4)
13.7)
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 vrms the largest?
ANSWER:
According to Avogadro’s law, gases with equal volume contain equal number of molecules hence the three vessels will contain an equal number of the molecules also has equal pressure and temperature.
The root mean square speed of molecules (vRMS) is given by;
VRMS = √(3kT/m)
k = Boltzmann constant
T = temperature of gas
m = mass of gas
for constant temperature,
VRMS α √(1/m)
here root mean square speed is inversely related to mass of gas and among given gases neon is monoatomic hence has lowest mass hence highest speed.
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13.9)
13.10)
13.12)
Additional Exercises
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 = n1 exp [ -mg (h2 – h1 )/ kB T] where n2 , n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [ -mg NA (ρ – ρ′ ) (h2 –h1 )/ (ρ RT)]
Where ρ is the density of the suspended particle, and ρ′ that of surrounding medium. [NA is Avogadro’s number, and R the 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 = n1 exp [ -mg (h2 – h1 )/ kB T] ….. 1
Where,
n1 is the number density at height .
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, apparent weight of the suspended particle is given by
Weight of the medium displaced -Weight of the suspended particle = mg – m’g
13.14) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms
Answer:
To find the size of the atom means to find radius of the atom.
We know that , volume of 1 mole of gas is given by,
volume = mass/density
For carbon –
M = 12.01 × 10–3 kg
ρ = 2.22 × 103 kg m–3
putting these values in equation 3 we get,
r = 1.59 Å
for gold –
M = 197.00 × 10–3 kg
ρ = 19.32 × 103 kg m–3
putting these values in equation 3 we get
r = 1.29 Å
for nitrogen –
M = 14.01 × 10–3 kg
ρ = 1 × 103 kg m–3
putting these values in equation 3 we get
r = 1.77 Å
for lithium –
M = 6.94 × 10–3 kg
ρ = 0.53 × 103 kg m–3
putting these values in equation 3 we get,
r = 1.73 Å
for fluorine –
M = 19 × 10–3 kg
ρ = 1.14 × 103 kg m–3
putting these values in equation 3 we get,
r = 1.88 Å
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