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Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the radius of an oxygen molecule to be roughly 3 Å.

Diameter (size) of oxygen molecule = 3 Å
Radius of oxygen molecule, r = Å = 1.5 × 10-10m
Therefore, volume of 1 mole of oxygen gas at S.T.P
πr3 x N
π(1.5 x 10-10)3 x 6.0225 x 1023
   = 8.514 × 10-6 m3 = 8.514 × 10-3 litre
Fraction of the molecular volume occupied =   = 3.8 x 10-4.


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.

Standard temperature T = 273 K
Pressure P = 1 atm. = 1.013 x 105 Pa
Universal gas constant R = 8.31 J mol-1 K-1
Number of moles n = 1
                        PV = nRT
          Therefore V =  
                            = 22.3955 x 10-3 m3
                   1 litre = 1000 cm3  
                            = 103 x 10-6 m3 = 10-3
                         V =
                         V = 22.395 litre.


Fig below shows plots of PV/T versus P for 1.00 x 10-3 kg of oxygen gas at two different temperatures.
(i) What does the dotted plot signify?
(ii) Which is true: T1 > T2 or T1 < T2?
(iii) What is the value of PV/T where the curves meet on the Y-axis?
(iv) If we obtained similar plots for 1.00 x 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)?


(i) The dotted plot corresponds to the ideal gas behavior.

(ii) T1 > T2.

(iii) 0.26 J/K

(iv) No, 6.3 x 10-5 kg of H2 would yield the same value.


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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27° C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 17° C. Estimate the mass of oxygen taken out of the cylinder.

Case (i)
Volume V = 30 litres = 30
× 10-3 m3
Pressure P = 15
× 1.013 x 105 Nm-2
Temperature T = 27 + 273 = 300 K
Gas constant R = 8.3 J mol-1 K-1
We know that,
 PV = nRT
 n1 = 18.307 mole

Case (ii) Volume in this case will also be the same
V = 30
× 10-3 m3
P = 11
× 1.013 × 105 Nm-2
T = 17 + 273 K
    =  = 13.898 mole
n1 - n2 = 18.307 - 13.898 = 4.419 mole
Mass of one mole of oxygen = 32 g = 32
× 10-3 kg
Mass of oxygen taken out =  = 0.1414 kg.


An air bubble of volume 1cubic cm rises from the bottom of a lake 40m 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.

Initial volume of the bubble V1 = 1 cm3 = 10-6m3
Initial pressure P1 = atmospheric pressure + pressure of 40 m water
                          = 1.013 x 105 + 40 x 1000 x 9.8 (h
                          = 101300 + 392000
                          = 4.933 x 105 Pa

TemperatureT1    = 273 + 12 = 285 K
At the surface of the lake, P2 = 1 atm. = 1.013 x 105 Pa
Temperature T2  = 273 + 35 = 308 K


= 5.262 x 10-6 m3.


Estimate the total number of air molecules in a room of capacity 25 cubic metre at a temperature of 27° C and 1 atm pressure.

Volume of the room V = 25.0 m3
Temperature of the room T = 27°C
                                        = 27 + 273 = 300 K
Pressure of the room P = 1 atm. = 1.013 x 105 Nm-2
Boltzmann's constant k = 
Where N is the Avagadro's number
                       PV = nRT
Total no. of air molecules = n = 
                                     = 6.117 x 1026.


Estimate the average thermal energy of a helium atom at

(1) Room temperature (27 °C) 

(2) Temperature on the surface of the sun (6000 K).

The energy per molecule E = kT
Boltzmann's constant = k = 1.38 x 1023 JK-1

(i) At temperature T = 27 + 273 = 300 K
                         E1 = kT
                             = x 1.38 x 10-23 x 300
                             = 6.21 x 10-21 J

(ii) At temperature T = 6000 K
                      E2 = x 1.38 x 10-23 x 6000
                          = 1.242 x 10-19 J.


Three vessels of equal capacity have gasses at the same temperature and pressure. The first vessel contains Neon (monoatomic), the second contains chlorine (diatomic) and the third contains uranium hexa-fluoride (poly-atomic). Do the vessels contain equal number of respective molecules? Is the RMS velocity same in all the three cases? If not in which case is vrms the largest.

Applying Avagadro's law, we find that the vessels contain equal number of molecules of respective gases.

Also, vrms
Since M is different for the three gases therefore vrms of three gases will be different. Since neon is the lightest of the three gases therefore vrms for neon is largest.


At what temperature is the RMS Velocity of an atom in and Argon gas cylinder equal to the RMS Velocity of a helium gas atom at -20 °C? ( Atomic mass of Argon = 39.9, of Helium = 4).

RMS velocity of an atom vrms
Given vrms of argon at T1 = vrms of He at T2
T2 = -20 + 273 = 253 K
T1 =  
= 2523.7 K.


Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atoms and temperature 17 ° C. Take the radius of 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).

Pressure of the nitrogen inside the cylinder P = 2.0 atm = 2 × 1.013 × 105 Pa
Temperature of the nitrogen inside the cylinder T = 17°C = 17 + 273 = 290 K
Diameter of the nitrogen molecule = 2
× 1 × 10-10 m (1Å= 10-10m)
We know that the 'Universal gas constant', R = 8.3 J mol-1 K-1
Boltzman's constant of nitrogen k = 1.37
× 10 -23 JK-1
Molecular mass of nitrogen M = 28 gm = 0.028 kg
Mean free path
                        = 1.103
× 10-7 m
     rms velocity = vrms
                        = 507.83 ms-1
Collision frequency v =  
                               = 4.6 x 109 Hz
Time taken for one collision =  
                              = 3.93
× 1013 s
Time between two successive collisions =  
× 10-7 ) / (507.83)
                                                     = 2.17
× 10-10 s.


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.


From the figure 1,
P1 = Pressure of enclosed air = H = 76cm of mercury.
Volume of the enclosed air = 15 cm (l1)
By Boyle's law,
× l1 = 76 × 15                  -----------(1)


From figure 2,
P2 = 76 - h
l2 = 100 - h
...  P2
× l2 = (76 - h) × (100 - h)      ---------------(2)
Equating (1) and (2) (Boyle's law) and simplifying we get
h2 - 176h + 6460 = 0 which yields that h = 123.8 cm or 52.2 cm
Of these h = 52.2 cm is alone admissible. (as 123.8 cm is greater than the length of the tube)
Therefore the amount of mercury flown out = 76 - 52.2 = 23.8 cm.


Estimate the root mean square speed of the suspended particles in Brownian motion if the particle mass is 10-6 kg and the temperature of the liquid is 27 °C. Would you expect the answer to change if the liquid were replaced by another liquid of different density and viscosity keeping the temperature fixed?

The mean kinetic energy of the particle at temperature T is given by
Here v2 is mean squared speed. Therefore, root mean squared speed of the particle is given by

                         vrms =
where constant K = 1.38
×10-23 J/mol/K
          mass m = 10-4 kg
          Temperature T = 27 + 273 = 300 K

vrms =
       = 1.114
×10-7 m/s
The root mean squared speed of the suspended particle depends on mass of the particle and temperature of the liquid and does not depend upon density and viscosity when the liquid is replaced by another liquid.


Explain qualitatively how the extent of Brownian motion is affected by the

(a) size of the Brownian particle

(b) density of the medium

(c) temperature of the medium

(d) viscosity of the medium

(a) Brownian motion decreases with increase in the size of the particle.

(b) Brownian motion increases with decrease in the density of the medium.

(c) Brownian motion increases with the increase in temperature of the medium.

(d) Brownian motion decreases with increase in the value of viscosity of the medium.


In an experiment on Brownian motion using a torsion pendulum (a small mirror of area 1 mm2mounted on a thin torsion fibre with torsion constant = 1.8 ×10-1 7 J/rad2) the mean value of angular displacement θwas found to be nearly zero, and the fluctuation in θ, that is the mean square of θ was found to be 2.6 ×104 rad2. Estimate the value of Boltzmann constant from this data and compare it with the correct value. The temperature is 300 K.

The mean potential energy of the torsion pendulum
α = couple required to twist the fibre of the pendulum by one radian  
θ2> = mean square deflection of the mirror

As the pendulum is in thermal equilibrium with the surrounding air, the thermal energy
= KT

      = KT
α = 1.8 ×10-17 J/rad2
θ2> = 2.6 ×10-4 rad2
          T = 300 K

×1.8 ×10-17 ×2.6 ×10-4 = ×K ×300
K = = 1.584
×10-23 J/K

Hence the value of the Boltzmann's constant = 1.584
×10-23 J/K

Standard value of the Boltzmann's constant = 1.58
×10-23 J/K.


The specific heat at constant volume of a certain metal is approximately 0.1 cal/g °C. Write the chemical formula of its chlorine if it contains 0.345 fraction of the metal.

Approximate atomic mass of the metal =
Equivalent weight of the metal =         (atomic mass of chlorine = 35.5 u)
Velocity of metal = = 5
Therefore the formula of the chloride = MCl5.


From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s-1. Identify the gas.

Diffusion rate of hydrogen R1 = 28.7 cm3
Diffusion rate of another gas R2 = 7.2 cm3 s-1
                        15.9 =
2 = 7.9 ~ 8
Hence the gas is oxygen.


For Brownian motion of particles of suspension in a liquid, 

(a) What should be the typical size of suspended particles? Why should not the size of the particles be too small (say of atomic dimensions 10-10m) or too large (say of the order of 1m)?

(b) Bombardment of the suspended particles by molecules of the liquid are random. We should then expect equal number of molecules hitting a suspended particle from all directions. Why is not the net impact zero?

(c) Can the assembly of suspended particles be considered a 'gas' of heavy molecules? If so, what is the temperature of this 'gas', if the temperature of the liquid is T?

(a) For Brownian motion, the typical size of the suspended particle should be about 10-16 m to 10-5 m. The size of the particle is too small or too large, no Brownian motion takes place. When particle is too small (i.e.,10-10m), then chances of bombardment of the particle by the molecules of the liquid becomes negligible. On the other hand, when particles are too large ( l m), the suspended particle is bombarded equally from all sides and in the absence of a net unbalanced force, Brownian motion does not occur.

(b) Brownian motion is observed due to fluctuation in the number of molecules actually striking a suspended particle from the average number of molecules striking the particle.

(c) Yes, the assembly of suspended particles can be treated as a gas. Since, the suspended particles are in thermal equilibrium with the liquid at temperature T, the temperature of the suspended particles will also be T.


Brownian motion is generally regarded as one of the most compelling evidences (though necessary indirect, since we can never hope to 'see' the atoms directly) of the existence of atoms. Do you appreciate this claim? Why is it a better evidence than that, obtained from the laws of chemical combinations or from the heats of gases and solids at ordinary temperatures?

Brownian motion is generally regarded as one of the most compelling evidence of the existence of atoms. It is for the reason that it can be directly observed with the help of a microscope. The laws of chemical combination or specific heats of solids and gases also provide evidences for the existence of the atoms but the evidences are not visual ones as in the case of Brownian motion.


A gas in equilibrium has uniform density 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
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
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]

Assume that Brownian motion particles form a perfect gas which is in equilibrium under the action of the force due to gravity and assume that the gas, composed of Brownian particles is present in a vertical isothermal column.


Consider the layer of this gas. This layer is bounded by the surface at heights h and h + dh on which the pressures are p and p + dp respectively. 
Suppose h increases upwards, then consider the equilibrium of a unit area of the layer.
Let the density of the gas in the layer =
Mass of the particle = m
Number of particles per unit volume = n
The net force acting vertically down on the layer
                     = p + dp +
ρ gdh - p
                     = dp +
ρg dh
As the layer is in equilibrium, the net force should be zero.
    dp +
ρg dh = 0
                dp =
gρ dh       -----(1)

As the gas us assumed to be perfect
                 p = nkT            -----(2)
where k is Boltzmann's constant

Differentiating equation (2), we get
               dp = dn.kT

Substituting for dp from equation (1), we get
ρ dh = dn.kT        (ρ = mn)
                    =   ------(3)
where NA = Avogadro's number
          R = Universal constant.

Integrating equation (3), we get

     n2 = n1 e
Effective mass of the particle =
where r is the radius of the particle and
ρ is density of the fliud.

                n2 = n1 

n2 = n1


Given below are data on a gum-resin suspension in water at 22 °C. Average radius of a grain of suspension = 0.2 μm, 1 μm = 10-6 m, average mass of a grain = 6.2 ×1017 kg, average concentration in a layer = 43 particles per unit area at some reference level, average concentration in a layer = 100.3 particles per unit area at a level 11 μm lower. Estimate the value of Avogadro's number from this data and compare your answer with the correct value.

Temperature of gum-resin suspension in water T = 22 °C
                                                                       = 273 + 22 = 295 K
Average radius of a grain of suspension r = 0.2
Mass of the grain m = 6.2
×10-17 kg
Average concentration in a layer n1 = 100.3 particles/unit area
Average concentration in another layer n2 = 43 particles/unit area
Height between the two layers, h2 - h1 = 11
Avogadro's constant R = 8.31 J/mol K

n2 = n1
NA =

ρ = 100 kg/m3
ρ =
ρ = 1851 kg/m3
NA =
        NA = 6.2 ×1023.


Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms.


Atomic mass

Density (103 kg m-3)

Carbon (diamond)






Nitrogen (liquid)






Fluorine (liquid)



Avogadro's number NA = 6.02 ×1023 

For carbon:
Atomic mass A = 12.01 g
ρ = 2.22 ×103 kg/m3
The size of the atom of carbon r =
                                                 = 0.598
×10-10 m

For gold:
Atomic mass A = 197.0 g
                       = 197
×10-3 kg
ρ = 19.32 ×103 kg/m3
r =
     = 1.593
×10-10 m

For nitrogen:
Atomic mass A = 14.01 g
                      = 14.01
×10-3 kg
ρ = 1.00 ×103 kg/m3
r =
     = 2.331
×10-10 m

For lithium:
Atomic mass A = 6.94 g
                      = 6.94
×10-3 kg
ρ= 0.53 ×103 kg/m3
r =
     = 1.732
×1010 m

For fluorine:
Atomic mass A = 19.00 g
                      = 19
×10-3 kg
Density = 1.14
× 103 kg/m3
r =
     = 2.362
×10-10 m.


What is the simplest evidence in nature that you can think of to suggest that atoms are not point particles, but have finite (non-zero) size? (b) Does atomic size increase monotonically with increase in atomic mass. If not, can you guess why not? (c) What determines fundamentally the size of, say, a hydrogen atom? Why is not a hydrogen atom say 100 times bigger or smaller than its actual size 0.5 Å?

(a) Volume of a substance (liquid, solid or gas) can be made approximately equal to zero. This shows that atoms have finite size.

(b) In an atom, the electrostatic force between its nucleus and the electron is balanced by the centripetal force due to orbital motion of the electron around nucleus. The balancing condition is

r =
From Bohr's theory of structure of atom,
r =
m and e are mass and charge of an electron,
h is Planck's constant
Z is the atomic number
r is the radius of the circular orbit in which the electron is revolving around the nucleus.

The radius of the atom can be taken to be equal to r.
If A is the atomic mass of an atom and z is the number of protons inside the nucleus, the number of neutrons = A - Z

Thus Z does not increase monotonically with increase in A. Therefore, atomic radius does not increase monotonically with atomic mass.

(c) For an hydrogen atom r =
For an unexcited hydrogen atom,  n = 1
r =

The radius of the orbit of an electron around the nucleus of an hydrogen atom determines its size. m, h, e are constants, hence the size of an hydrogen atom us fundamentally determined by the values of the constants.

If the value of Planck's constant were 10 times its present value, the size of the hydrogen atom would have been 100 times more than its actual value.

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