By | September 22, 2020

CBSE Class 11 Physics

NCERT Solutions
Chapter 13
KINETIC THEORY

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.

Ans.Given that,

1. Diameter of an oxygen molecule, d

r = 1.5  =

Now we know that,actual volume occupied by 1 mole of oxygen gas at STP = 22400

Molecular volume of oxygen gas,

Where, N is Avogadro’s number = 6.023  1023 molecules/mole

Ratio of the molecular volume to the actual volume of oxygen =

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 the measured volume at STP is 22.4 litres.

Ans. The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:

PVnRt

Where,

is the universal gas constant =

n= Number of moles = 1(Taken here)

T= Standard temperature = 273 K(Since given condition is STP)

P= Standard pressure = 1 atm =

= 0.0224

= 22.4 liters

Hence, the molar volume of a gas at STP is 22.4 liters.

3. Figure 13.8 shows the plot of PV/T versus P for kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true:?

(c) What is the value of PV/T where the curves meet on the y-axis?

(d) If similar plots are obtained for kg of hydrogen, should 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)? (Given, Molecular mass of u,  O2= 32.0 u, R =.)

Ans. (a) The dotted plot in the graph signifies the ideal behavior of the gas.We know fromthe ideal gas law that

=nR                                                                                (1)

where n=no. of moles

R= Universal Gas constant

Since both n and R are constants, hence the RHS is also constant.

From eq(1) hence we conclude that the LHS is also a constant. Thus PV/T is a constant quality and is independant to any change in pressure.

(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature is closer to the dotted plot than the curve of the gas at temperature. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, is true for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is nR. This is because the ideal gas equation is given as:

PV = nRT

PV/T=nR

Where,

P is the pressure

T is the temperature

V is the volume

n is the number of moles

R is the universal constant

Molecular mass of oxygen = 32.0 g

Mass of oxygen =  kg = 1 g

R = 8.314 J

= 0.26 J

Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is 0.26 J .

(d) If we obtain similar plots for  kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).

We have the value obtained from the last problem for oxygen that,

Now we know that, R = 8.314 J

Molecular mass (M) of = 2.02 u

Now, PV/T=nR   (at constant temperature)

Where, n=m/M

m = Mass of

Hence, kg of  will yield the same value of PV/T.

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, molecular mass of = 32 u).

Ans. Volume of oxygen, = 30 litres =

Gauge pressure, = 15 atm =

Temperature, = 27°C = 300 K

Universal gas constant, R= 8.314 J

Let the initial number of moles of oxygen gas in the cylinder be.

Now,the ideal gas equation is given as:

= 18.276

But,

Where,

= Initial mass of oxygen

M= Molecular mass of oxygen = 32 g

M = 18.276  32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, = 30 litres =

Gauge pressure, = 11 atm =

Temperature, = 17°C = 290 K

Let be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

= 13.86

But,

Where,

is the mass of oxygen remaining in the cylinder

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

= 584.84 g– 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

5. An air bubble of volume 1.0 cmrises 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. Volume of the air bubble,

Given that bubble rises to height, d = 40 m

Temperature at a depth of 40 m, = 12°C =(273+12)K= 285 K

Temperature at the surface of the lake, = 35°C=(273+35)K = 308 K

The pressure on the surface of the lake:

= 1 atm =

The pressure at the depth of 40 m:

= 1 atm + dg

Where,

is the density of water =

g is the acceleration due to gravity =

=

= 493300 Pa

Now we have from the ideal gas law that,

Where, is the volume of the air bubble when it reaches the surface

Therefore,

Therefore, when the air bubble reaches the surface, its volume becomes 5.263.

6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 at a temperature of 27 °C and 1 atm pressure.

Ans. Volume of the room, V= 25.0

Temperature of the room, T= 27°C = 300 K

Pressure in the room, P= 1 atm =

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:

PV =

Where,

is Boltzmann constant =

is the number of air molecules in the room

molecules

Therefore, the total number of air molecules in the given room is.

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).

Ans. (i) At room temperature, T= 27°C = (273+27)K=300 K

We know that, the average thermal energy

Where k is the Boltzmann constant =

Hence, the average thermal energy of a helium atom at room temperature (27°C) is.

(ii) On the surface of the sun, T= 6000 K

Average thermal energy

Hence, the average thermal energy of a helium atom on the surface of the sun is.

(iii) At temperature, T= 107K

Average thermal energy

Hence, the average thermal energy of a helium atom at the core of a star is.

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 iit is the largest?

Ans. Yes. All the vessels contain the same number of the respective molecules..

Since the three vessels have the same capacity, they have the same volume.

Also it is given that each gas has the same pressure and temperature.

According to the Avogadro’s law, equal volumes of different gases under same temperature and pressure will contain equal number of molecules. This number is equal to Avogadro’s number, N=.

Now ,we know that the root mean square speed () of a gas of mass m, at a  temperature T, is given by the relation:

Where, k is Boltzmann constant

Since the temperature for all the three gases is same , the vrms only depends on mass such that,

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.

9. At what temperature is the root mean square speed of an atom of an argon gas 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. Given,

Temperature of the helium atom, = –20°C= 253 K

Atomic mass of argon, = 39.9 u

Atomic mass of helium, = 4.0 u

Now, let, be the rms speed of argon.

Let be the rms speed of helium.

The rms speed of argon is given by:

… (i)

Where,

is the universal gas constant

is temperature of argon gas

And, the rms speed of helium is given by:

… (ii)

It is given that:

or,

or,

Therefore,

= 2523.675 =

Therefore, the temperature of the argon atom is .

10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at a pressure 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0. Compare the collision time with the time during which the molecule moves freely between two successive collisions (Molecular mass of).

Ans. Mean free path =

Collision frequency =

Successive collision time 500  (Collision time)

Pressure inside the cylinder containing nitrogen, P= 2.0 atm =

Temperature inside the cylinder, T17°C =290 K

Radius of a nitrogen molecule, r= 1.0 =

Diameter, d =

Molecular mass of nitrogen, M= 28.0 g =

The root mean square speed of nitrogen is given by the relation:

Where,

is the universal gas constant = 8.314 J

= 508.26 m/s

The mean free path (l) is given by the relation:

Where,

is the Boltzmann constant =

Collision frequency

Collision time is given as:

Time taken between successive collisions:

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

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?

Ans. 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, = 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

Let h cm of mercury flow out as a result of atmospheric pressure.

∴Length of the air column in the bore= 24 + cm

And, length of the mercury column = 76 – cm

Initial pressure, = 76 cm of mercury

Initial volume, = 15

Final pressure, = 76 –(76– h) = h cm of mercury

Final volume, = (24 + h

Temperature remains constant throughout the process.

Hence from Boyle’s law we have,

76  15 = h (24 + h)

+ 24h –1140 = 0

= 23.8 cm or – 47.8 cm

Since,height cannot be negative hence -47.8cm is an invalid answer.. 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.

12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3. The diffusion of another gas under the same conditions is measured to have an average rate of. Identify the gas.

[Hint: Use Graham’s law of diffusion:, where are diffusion rates of gases 1 and 2, andtheir respective molecular masses. The law is a simple consequence of kinetic theory.]

Ans. Rate of diffusion of hydrogen,

Rate of diffusion of another gas,

According to Graham’s Law of diffusion, we have:

Where,

is the molecular mass of hydrogen = 2.020 g

is the molecular mass of the unknown gas

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

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 atmosphere

Where refer to number density at heights respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

Where  is the density of the suspended particle, and ‘ that of surrounding medium. [is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent

Ans. According to the law of atmospheres, we have:

… (i)

Where,

is the number density at height is the number density at height

mg is the weight of the particle suspended in the gas column

Density of the medium =

Density of the suspended particle =

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– mg

………..(ii)

Gas constant, R =

… (iii)

Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:

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 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].

Ans.

Atomic mass of a substance = M

Density of the substance =

Volume of each atom

Volume of number of molecules N … (i)

Volume of one mole of a substance = … (ii)

N =

For carbon:

Hence, the radius of a carbon atom is 1.29 .

For gold:

M

Hence, the radius of a gold atom is 1.59 .

For liquid nitrogen:

Hence, the radius of a liquid nitrogen atom is 1.77 .

For lithium:

Mkg

p =

Hence, the radius of a lithium atom is 1.73 .

For liquid fluorine:

m-3

$r={\left(\frac{3×19.00×{10}^{-3}}{4\pi ×1.14×{10}^{3}×6.23×{10}^{23}}\right)}^{\frac{1}{3}}$=18.55