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Chapter 13- Kinetic Theory Interview Questions Answers

Question 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.

Answer 1 :

Diameter of an oxygen molecule, d = 3 Å

Radius, r = d / 2

r = 3 / 2 = 1.5 Å = 1.5 x 10 -8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3

Molecular volume of oxygen gas, V = 4 / 3 πr3. N

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

Hence,

V = 4 / 3 x 3.14 x (1.5 x 10-8)3 x 6.023 x 1023

We get,

V = 8.51 cm3

Therefore, ratio of the molecular volume to the actual volume ofoxygen = 8.51/ 22400 = 3. 8 x 10-4

Question 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 2 :

The ideal gas equation relating pressure (P), volume (V), andabsolute temperature (T) is given as:

PV = nRT

Where, R is 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 x 105Nm-2

Hence,

V = nRT / P

= 1 x 8.314 x 273 / 1.013 x 105

= 0.0224 m3

= 22.4 litres

Therefore, the molar volume of a gas at STP is 22.4 litres

Question 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 thedotted plot signify?

(b) Which is true: T1 > T2 or T1 < T2?

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

(d) If we obtained similar plotsfor 1.00×10-3 kg of hydrogen,would we get the same value of PV/T at the point where the curves meet on they-axis? If not, what mass of hydrogen yields the same value of PV/T (for lowpressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31J mo1–1 K–1.)

Answer 3 :

(a) dotted plot is parallel to X-axis, signifying that nR [PV/T= nR] is independent of P. Thus it is representing ideal gas behaviour

(b) the graph at temperature T1 is closer to ideal behaviour (because closer todotted line) hence, T1 >T2 (higher thetemperature, ideal behaviour is the higher)

(c) use PV = nRT

PV/ T = nR

Mass of the gas = 1 x 10-3 kg= 1 g

Molecular mass of O2 = 32g/ mol

Hence,

Number of mole = given weight / molecular weight

= 1/ 32

So, nR = 1/ 32 x 8.314 = 0.26 J/ K

Hence,

Value of PV / T = 0.26 J/ K

(d) 1 g of H2 doesn’trepresent the same number of mole

Eg. molecular mass of H2 = 2 g/mol

Hence, number of moles of H2 require is 1/32 (as per the question)

Therefore,

Mass of H2 required= no. of mole of H2 xmolecular mass of H2

= 1/ 32 x 2

= 1 / 16 g

= 0.0625 g

= 6.3 x 10-5 kg

Hence, 6.3 x 10-5 kgof H2 wouldyield the same value

Question 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-1 K-1, molecular mass of O2 = 32 u).

Answer 4 :

Volume of gas, V1 =30 litres = 30 x 10-3 m3

Gauge pressure, P1 =15 atm = 15 x 1.013 x 105 P a

Temperature, T1 =270 C = 300 K

Universal gas constant, R = 8.314 J mol-1 K-1

Let the initial number of moles of oxygen gas in the cylinder ben1

The gas equation is given as follows:

P1V1 = n1RT1

Hence,

n1 =P1V1 / RT1

= (15.195 x 105 x30 x 10-3) / (8.314 x 300)

= 18.276

But n1 =m1 / M

Where,

m1 =Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Thus,

m1 =N1M = 18.276 x 32 =584.84 g

After some oxygen is withdrawn from the cylinder, the pressureand temperature reduce.

Volume, V2 =30 litres = 30 x 10-3 m3

Gauge pressure, P2 =11 atm

= 11 x 1.013 x 10Pa

Temperature, T2 =170 C = 290 K

Let n2 bethe number of moles of oxygen left in the cylinder

The gas equation is given as:

P2V2 = n2RT2

Hence,

n2 =P2V2 / RT2

= (11.143 x 105 x30 x 10-30) / (8.314 x 290)

= 13.86

But

n2 =m2 / M

Where,

misthe mass of oxygen remaining in the cylinder

Therefore,

m2 =n2 x M = 13.86 x32 = 453.1 g

The mass of oxygen taken out of the cylinder is given by therelation:

Initial mass of oxygen in the cylinder – Final mass of oxygen inthe cylinder

= m1 –m2

= 584.84 g – 453.1 g

We get,

= 131.74 g

= 0.131 kg

Hence, 0.131 kg of oxygen is taken out of the cylinder

Question 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?

Answer 5 :

Volume of the air bubble, V1 = 1.0 cm3

= 1.0 x 10-6 m3

Air bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 120 C= 285 K

Temperature at the surface of the lake, T= 350 C= 308 K

The pressure on the surface of the lake:

P2 =1 atm = 1 x 1.013 x 105 Pa

The pressure at the depth of 40 m:

P1= 1atm + dρg

Where,

ρ is the density of water = 103 kg/ m3

g is the acceleration due to gravity = 9.8 m/s2

Hence,

P1 =1.013 x 105 + 40 x 103 x 9.8

We get,

= 493300 Pa

We have

P1V1 / T1 = P2V/ T2

Where, V2 isthe volume of the air bubble when it reaches the surface

V2 =P1V1T2 / T1P2

= 493300 x 1 x 10-6 x308 / (285 x 1.013 x 105)

We get,

= 5.263 x 10-6 m3 or 5.263 cm3

Hence, when the air bubble reaches the surface, its volumebecomes 5.263 cm3

Question 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.

Answer 6 :

Volume of the room, V = 25.0 m3

Temperature of the room, T = 270 C= 300 K

Pressure in the room, P = 1 atm = 1 x 1.013 x 105 Pa

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

PV = (kBNT)

Where,

KB isBoltzmann constant = (1.38 x 10-23) m2 kg s-2 K-1

N is the number of air molecules in the room

Therefore,

N = (PV / kBT)

= (1.013 x 105 x25) / (1.38 x 10-23 x 300)

We get,

= 6.11 x 1026 molecules

Hence, the total number of air molecules in the given room is6.11 x 1026

Question 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 7 :

(i) At room temperature, T = 270 C= 300 K

Average thermal energy = (3 / 2) kT

Where,

k is the Boltzmann constant = 1.38 x 10-23 m2 kg s-2 K-1

Hence,

(3 / 2) kT = (3 / 2) x 1.38 x 10-23 x300

On calculation, we get,

= 6.21 x 10-21 J

Therefore, the average thermal energy of a helium atom at roomtemperature of 270 C is 6.21 x 10-21 J

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

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10-23 x6000

We get,

= 1.241 x 10-19 J

Therefore, the average thermal energy of a helium atom on thesurface of the sun is 1.241 x 10-19 J

(iii) At temperature, T = 107 K

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10-23 x107

We get,

= 2.07 x 10-16 J

Therefore, the average thermal energy of a helium atom at thecore of a star is 2.07 x 10-16 J

Question 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 8 :

All the three vessels have the same capacity, they have the samevolume.

So, each gas has the same pressure, volume and temperature

According to Avogadro’s law, the three vessels will contain anequal number of the respective molecules.

This number is equal to Avogadro’s number, N = 6.023 x 1023.

The root mean square speed (Vrms) of a gas of mass m and temperature T is given bythe relation:

Vrms = 3kT / m

Where,

k is Boltzmann constant

For the given gases, k and T are constants

Therefore, Vrms dependsonly on the mass of the atoms, i.e., Vrms  (1/m)1/2

Hence, the root mean square speed of the molecules in the threecases is not the same.

Among neon, chlorine and uranium hexafluoride, the mass of neonis the smallest.

Therefore, neon has the largest root mean square speed among thegiven gases.

Question 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 9 :

Given

Temperature of the helium atom, THe = -200 C= 253 K

Atomic mass of argon, MAr = 39.9 u

Atomic mass of helium, MHe = 4.0 u

Let (Vrms)Ar be the rms speedof argon and

Let (Vrms)He be the rms speedof helium

The rms speed of argon is given by:

(Vrms)Ar = √3RTAr / MAr ………… (i)

Where,

R is the universal gas constant

TAr istemperature of argon gas

The rms speed of helium is given by:

(Vrms)He = √3RTHe / MHe ………… (ii)

Given that,

(Vrms)Ar = (Vrms)He

√3RTAr /MAr = √3RTHe / MHe

TAr /MAr = THe / MHe

TAr =THe / MHe x MAr

= (253 / 4) x 39.9

We get,

= 2523.675

= 2.52 x 103 K

Hence, the temperature of the argon atom is 2.52 x 103 K

Question 10 : Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 170 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 10 :

Mean free path = 1.11 x 10-7 m

Collision frequency = 4.58 x 109 s-1

Successive collision time  500 x(Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm =2.026 x 105 Pa

Temperature inside the cylinder, T = 170 C= 290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 x 1010 m

Diameter, d = 2 x 1 x 1010 =2 x 1010 m

Molecular mass of nitrogen, M = 28.0 g = 28 x 10-3 kg

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

Vrms=√3RT / M

Where,

R is the universal gas constant = 8.314 J mol-1 K-1

Hence,

Vrms=3 x 8.314 x 290 / 28 x 10-3

On calculation, we get,

= 508.26 m/s

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

l = KT / √2 x π x d2 xP

Where,

k is the Boltzmann constant = 1.38 x 10-23 kgm2 s-2 K-1

Hence,

l = (1.38 x 10-23 x290) / (√2 x 3.14 x (2 x 10-10)2 x 2.026 x 105

We get,

= 1.11 x 10-7 m

Collision frequency = Vrms / l

= 508.26 / 1.11 x 10-7

On calculation, we get,

= 4.58 x 109 s-1

Collision time is given as:

T = d / Vrms

= 2 x 10-10 /508.26

On further calculation, we get

= 3.93 x 10-13 s

Time taken between successive collisions:

T’ = l / Vrms =1.11 x 10-7 / 508.26

We get,

= 2.18 x 10-10

Hence,

T’ / T = 2.18 x 10-10 /3.93 x 10-13

On calculation, we get,

= 500

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


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