Question-1
Solution:
Diameter (size) of oxygen molecule = 3 Å
Radius of oxygen molecule, r = Å = 1.5 × 10^{-10}m
Therefore, volume of 1 mole of oxygen gas at S.T.P
= πr^{3} x N
= π(1.5 x 10^{-10})^{3} x 6.0225 x 10^{23 }^{ }= 8.514 × 10^{-6} m^{3} = 8.514 × 10^{-3} litre
Fraction of the molecular volume occupied = = 3.8 x 10^{-4}.
Question-2
Solution:
Standard temperature T = 273 K
Pressure P = 1 atm. = 1.013 x 10^{5} 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} m^{3}
1 litre = 1000 cm^{3}
= 10^{3} x 10^{-6} m^{3} = 10^{-3}
V =
V = 22.395 litre.
Question-3
(i) What does the dotted plot signify?
(ii) Which is true: T_{1} > T_{2} or T_{1} < T_{2}?
(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)?
Solution:
(i) The dotted plot corresponds to the ideal gas behavior.
(ii) T_{1} > T_{2}.
(iii) 0.26 J/K
(iv) No, 6.3 x 10^{-5} kg of H_{2} would yield the same value.
Question-4
Solution:
Case (i)
Volume V = 30 litres = 30 × 10^{-3} m^{3}
Pressure P = 15 × 1.013 x 10^{5} Nm^{-2}
Temperature T = 27 + 273 = 300 K
Gas constant R = 8.3 J mol^{-1} K^{-1}
We know that,
PV = nRT
n_{1} =
n_{1} = = 18.307 mole
Case (ii) Volume in this case will also be the same
V = 30 × 10^{-3} m^{3}
P = 11 × 1.013 × 10^{5} Nm^{-2}
T = 17 + 273 K
n_{2} =
= = 13.898 mole
n_{1} - n_{2} = 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.
Question-5
Solution:
Initial volume of the bubble V_{1} = 1 cm^{3} = 10^{-6}m^{3}
Initial pressure P_{1} = atmospheric pressure + pressure of 40 m water
= 1.013 x 10^{5} + 40 x 1000 x 9.8 (hρg)
= 101300 + 392000
= 4.933 x 10^{5} Pa
TemperatureT_{1} = 273 + 12 = 285 K
At the surface of the lake, P_{2} = 1 atm. = 1.013 x 10^{5} Pa
Temperature T_{2} = 273 + 35 = 308 K
=
V_{2} =
=
= 5.262 x 10^{-6} m^{3}.
Question-6
Solution:
Volume of the room V = 25.0 m^{3}
Temperature of the room T = 27°C
= 27 + 273 = 300 K
Pressure of the room P = 1 atm. = 1.013 x 10^{5} Nm^{-2}
Boltzmann's constant k =
Where N is the Avagadro's number
PV = nRT
Total no. of air molecules = n =
=
= 6.117 x 10^{26}.
Question-7
(1) Room temperature (27 °C)
(2) Temperature on the surface of the sun (6000 K).
Solution:
The energy per molecule E = kT
Boltzmann's constant = k = 1.38 x 10^{23} JK^{-1}
(i) At temperature T = 27 + 273 = 300 K
E_{1} = kT
= x 1.38 x 10^{-23} x 300
= 6.21 x 10^{-21} J
(ii) At temperature T = 6000 K
E_{2} = x 1.38 x 10^{-23} x 6000
= 1.242 x 10^{-19} J.
Question-8
Solution:
Applying Avagadro's law, we find that the vessels contain equal number of molecules of respective gases.
Also, v_{rms} =
Since M is different for the three gases therefore v_{rms} of three gases will be different. Since neon is the lightest of the three gases therefore v_{rms} for neon is largest.
Question-9
Solution:
RMS velocity of an atom v_{rms} =
Given vrms of argon at T_{1} = v_{rms} of He at T_{2}
=
T_{2} = -20 + 273 = 253 K
=
T_{1} =
= = 2523.7 K.
Question-10
N_{2} = 28.0).
Solution:
Pressure of the nitrogen inside the cylinder P = 2.0 atm = 2 × 1.013 × 10^{5} 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 λ =
=
= 1.103 × 10^{-7} m
rms velocity = v_{rms} =
=
= 507.83 ms^{-1}
Collision frequency v =
=
= 4.6 x 10^{9} Hz
Time taken for one collision =
=
= 3.93 × 10^{13} s
Time between two successive collisions =
=(1.103 × 10^{-7} ) / (507.83)
= 2.17 × 10^{-10} s.
Question-11
Solution:
From the figure 1,
P_{1} = Pressure of enclosed air = H = 76cm of mercury.
Volume of the enclosed air = 15 cm (l_{1})
By Boyle's law,
P_{1} × l_{1} = 76 × 15 -----------(1)
From figure 2,
P_{2} = 76 - h
l_{2} = 100 - h
.^{.}. P_{2} × l_{2} = (76 - h) × (100 - h) ---------------(2)
Equating (1) and (2) (Boyle's law) and simplifying we get
h^{2} - 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.
Question-12
Solution:
The mean kinetic energy of the particle at temperature T is given by
Here v^{2} is mean squared speed. Therefore, root mean squared speed of the particle is given by
v_{rms} =
where constant K = 1.38 ×10^{-23} J/mol/K
mass m = 10-4 kg
Temperature T = 27 + 273 = 300 K
v_{rms} =
= 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.
Question-13
(a) size of the Brownian particle
(b) density of the medium
(c) temperature of the medium
(d) viscosity of the medium
Solution:
(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.
Question-14
Solution:
The mean potential energy of the torsion pendulum
where α = couple required to twist the fibre of the pendulum by one radian
<θ^{2}> = mean square deflection of the mirror
As the pendulum is in thermal equilibrium with the surrounding air, the thermal energy
= KT
∴ = KT
where α = 1.8 ×10^{-17} J/rad^{2}
<θ^{2}> = 2.6 ×10^{-4} rad^{2}
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.
Question-15
Solution:
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 = MCl_{5}.
Question-16
Solution:
Diffusion rate of hydrogen R_{1} = 28.7 cm^{3}
Diffusion rate of another gas R_{2} = 7.2 cm^{3} s^{-1 15.9 = M}2^{ = 7.9 ~ 8}
Hence the gas is oxygen.
Question-17
(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^{-10}m) 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?
Solution:
(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^{-10}m), 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.
Question-18
Solution:
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.
Question-19
n_{2} = n_{1} exp
where n_{2}, n_{1} refer to number density at heights h_{2} and h_{1} respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_{2} = n_{1} exp
where ρ is the density of the suspended particle, and ρ that of surrounding medium. [N_{A} is Avogadro's number and R the universal gas constant]
Solution:
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 = ρ
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
-gρ dh = dn.kT (ρ = mn)
= ------(3)
where N_{A} = Avogadro's number
R = Universal constant.
Integrating equation (3), we get
n_{2} = n_{1} e
Effective mass of the particle =
where r is the radius of the particle and ρ is density of the fliud.
n_{2} = n_{1}
∴n_{2} = n_{1}
Question-20
Solution:
Temperature of gum-resin suspension in water T = 22 °C
= 273 + 22 = 295 K
Average radius of a grain of suspension r = 0.2 μm
Mass of the grain m = 6.2 ×10^{-17} kg
Average concentration in a layer n_{1} = 100.3 particles/unit area
Average concentration in another layer n_{2} = 43 particles/unit area
Height between the two layers, h_{2} - h_{1} = 11 μm
Avogadro's constant R = 8.31 J/mol K
n_{2} = n_{1}
=
N_{A} =
ρ = 100 kg/m^{3}
ρ =
ρ = 1851 kg/m^{3}
N_{A} =
=
∴ N_{A} = 6.2 ×10^{23}.
Question-21
Substance |
Atomic mass |
Density (10^{3} kg m^{-3}) |
Carbon (diamond) |
12.01 |
2.22 |
Gold |
197.0 |
19.32 |
Nitrogen (liquid) |
14.01 |
1.00 |
Lithium |
6.94 |
0.53 |
Fluorine (liquid) |
19.00 |
1.14 |
Solution:
Avogadro's number N_{A} = 6.02 ×10^{23}
For carbon:
Atomic mass A = 12.01 g
Density ρ = 2.22 ×10^{3} kg/m^{3}
∴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
Density ρ = 19.32 ×10^{3} kg/m^{3}
∴r =
= 1.593 ×10^{-10} m
For nitrogen:
Atomic mass A = 14.01 g
= 14.01 ×10^{-3} kg
Density ρ = 1.00 ×10^{3} kg/m^{3}
∴r =
= 2.331 ×10^{-10} m
For lithium:
Atomic mass A = 6.94 g
= 6.94 ×10^{-3} kg
Density ρ= 0.53 ×10^{3} kg/m^{3}
∴r =
= 1.732 ×10^{10} m
For fluorine:
Atomic mass A = 19.00 g
= 19 ×10^{-3} kg
Density = 1.14 × 10^{3} kg/m^{3}
∴ r =
= 2.362 ×10^{-10} m.
Question-22
Solution:
(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 =
where
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.