1st PUC Physics Question Bank Chapter 13 Kinetic Theory

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Karnataka 1st PUC Physics Question Bank Chapter 13 Kinetic Theory

Molecular Nature of Matter

Question 1.
Give the major differences between solids, liquids and gases.
Answer:
In solids, atoms are tightly packed and rigidly fixed in their position. They cannot move around.
In liquids, atoms are not tightly packed and not rigidly fixed in their position. They can move around.
In solids and liquids, the interparticle distances are almost same.
In gases, the interatomic distances are large compared to solids and liquids.

Behaviour of Gases

Question 1.
State Avogadro’s hypothesis.
Answer:
The hypothesis states that “the number of molecules per unit volume is same for all gases at a fixed temperature and pressure.

Question 2.
What is the value of Avogadro’s number?
Answer:
NA = 6.02 × 10²³

Question 3.
How many molecules are present in 22.4 litre at S.T.P?
Answer:
6.02 × 10²³

Question 4.
What is the volume occupied by 6.02 × 10²³ molecules at S.T.P.?
Answer:
22.4 litre

Question 5.
Define a mole.
Answer:
The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P. is called a mole.

Question 6.
Write the perfect gas equation. What does the symbols stands for?
Answer:
PV = μRT
P = pressure of gas, V = Volume of gas
μ = number of moles of gas, T = absolute temperature in Kelvin
μ = \(\frac{M}{M_{0}}=\frac{N}{N_{A}}\) where M is the mass of the gas containing N molecules. M₀ is the molar mass containing N₀ molecules.
R = N_Ak_B = Universal gas constant.
k_B = Boltzmann constant.
PV = \(\frac{N}{N_{A}}\) x N_Ak_BRT
PV = k_BNT
or P = K_BnT
when n is the number density.

Question 7.
What is the value of R in S.I. units?
Answer:
R = 8.314 J mol^-1 K^-1

Question 8.
Define number density.
Answer:
It is defined as number of molecules per unit volume.

Question 9.
What is. the value of Boltzmann constant in S.I. units?
Answer:
k_B = 1.38 × 10^-23 J K^-1

Question 10.
Obtain the expression to calculate the pressure exerted by the gas from perfect gas equation if density is provided.
Answer:

where ρ is the density of gas.

Question 11.
What is ideal gas?
Answer:
A gas that satisfies perfect gas equation PV = μRT at all pressures and temperatures is called ideal gas.

Question 12.
Under what conditions of temperature and pressure real gases behave ideal one? Explain.
Answer:
At high temperatures and low pressures, real gases behave like ideal gases. This is because at high T and low P, the molecules are far apart and molecular interactions are negligible.

Question 13.
State Boyle’s law.
Answer:
It states that the pressure of the given mass of gas is inversely proportional to its volume provided temperature is constant.
constant
P ∝\(\frac{1}{V}\)
P = \(\frac{\text { constant }}{V}\)
PV = constant
P₁V₁ = P₂V₂

Question 14.
State Charle’s law.
Answer:
It states that the volume of a given mass of gas is directly proportional to its temperature provided pressure is kept constant.
V ∝T
V = constant × T
\(\frac{V}{T}\) = constant
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

Question 15.
State and explain Dalton’s law of partial pressure.
Answer:
It states that “the total pressure of a mixture of ideal gases is the sum of partial pressures.”
Explanation: Consider a mixture of non-interacting ideal gases: μ₁ moles of gas 1, μ₂ moles of gas 2 etc in a vessel of volume V at temperature T and pressure P.
Total number of moles of gas = (μ₁ + μ₂ + μ₃ + ) and PV = (μ₁ + μ₂ + μ₃ + …… )RT
i.e., \(P=\frac{\mu_{1} R T}{V}+\frac{\mu_{2} R T}{V}+\frac{\mu_{3} R T}{V}+\)
P = P₁ + P₂ + P₃ + …… P_n
where P₁ + P₂ + P₃ + ……, are partial pressures of 1, 2, 3 respectively.

Kinetic Theory of an Ideal Gas

Question 1.
Write the postulates of Kinetic theory of gases.
Answer:

• The gas molecules are of different size and different shape and are in ‘n’ number
• The gas consists of very large number of molecules and some of these molecules are identical, perfectly elastic, spherical in shape, and their sizes are negligible.
• These molecules travel in random motion and they do not have a preferred direction of motion
• Some gas molecules travel in a straight line and are free in motion most of the times.
• The time interval of collision between any two molecules is very small.
• The collision between the molecules and the container is perfectly elastic, i.e., kinetic energy and linear momentum are conserved.

Question 2.
Derive an expression for pressure of an ideal gas.
Answer:
Consider an ideal gas enclosed in a cube of length ‘l’. Consider a gas molucule of mass ‘m’ moving with a velocity v_x, v_y, v_z colliding with the wall of area ‘A’.
Initial momentum of the gas molecule (momentum before collision) = mv_x
Final momentum of the gas molecule (momentum after collision) = -mv_x
Change in momentum = Final momentum – Initial momentum = -mv_x – mv_x
= -2 mv_x

Momentum imparted on wall = 2mv_x
Time taken for the gas molecule to move from one end to other end = \(\frac{\text { Distance travelled }}{\text { Velocity }}\)
Δt = \(\frac{l}{v_{x}}\)
Time taken by the gas molecule to move from other end to the original position = Δt_y2 = \(\frac{l}{v_{x}}\)
Therefore total time taken ⇒ Δt = Δt₁ + Δt₂ = \(\frac{l}{v_{x}}+\frac{l}{v_{x}}=\frac{2 l}{v_{x}}\)
∴l = \(\frac{\Delta t v_{x}}{2}\)
Number of gas molecules present in the given volume = Area x length = \(A\left[\frac{\Delta t v_{x}}{2}\right]\)

Question 3.
Show that average kinetic energy of gas molecule is directly proportional to temperature of the gas.
Answer:
Consider a gas molecule to be enclosed in a container with pressure ‘p’.
WKT,
P = \(\frac{1}{3} n m \overline{v^{2}}\)
multiply V on both sides
PV = \(\frac{1}{3} n m \overline{v^{2}} \cdot V\)
multiply and divide by 2 on RHS.
PV = \(\frac{2}{3} n m \overline{v^{2}} \times V \times \frac{1}{2}\)
∴\(P V=\frac{2}{3}(n V) \cdot \frac{1}{2} m \overline{v^{2}}\)
Since nV = Total number of gas molecules present in the given volume
nV = N
∴ \(P V=\frac{2}{3} N\left(\frac{1}{2} m \bar{v}^{2}\right)\) ………(1)
WKT,
PV = Nk_BT ……..(2)
From (1) and (2),

\(\frac{2}{3}(E)=k_{B} T\) where E = average kinetic energy
⇒E ∝ T

Question 4.
State the law of equipartition of energy.
Answer:
“Any system in thermal equilibrium, the total energy of the system is distributed equally among all the degrees of freedom and the energy associated with each degree of freedom is given \(\frac{1}{2} k_{B} T\).
where k_B ⇒ Boltzmann constant
T ⇒ Temperature.

Question 5.
Briefly discuss the law of equipartition of energy.
Answer:
The energy of molecule corresponding to translational mode is given by
\(E_{T}=\frac{1}{2} m \overline{v_{x}^{2}}+\frac{1}{2} m \overline{v_{y}^{2}}+\frac{1}{2} m \overline{v_{z}^{2}}=\frac{3}{2} k T\)
so that along x, y and z average translational may be taken independently as
\(\left\langle\frac{1}{2} m \bar{v}_{x}^{\overline{2}}\right\rangle=\left\langle\frac{1}{2} m \overline{v_{y}^{2}}\right\rangle=\left\langle\frac{1}{2} m \overline{v_{z}^{2}}\right\rangle=\frac{1}{2} k T\)
Energy of molecule in rotational mode of a molecule is given by
\(E_{T}=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}\)
For vibratory motion of a molecule.
\(E_{v}=\frac{1}{2} m\left(\frac{d y}{d t}\right)^{2}+\frac{1}{2} k_{f} y^{2}\)
where kf is force constant of the atomic oscillator and y the vibrational coordinate.
Since for each mode of absorption of energy, \(\frac{1}{2} k_{B} T\) is associated, the net vibrational energy per degree of freedom
= \(2 \times \frac{1}{2} k_{B} T=k_{B} T\)
we can easily note that \frac{1}{2} k_{f} y^{2} corresponds to potential energy and \(\frac{1}{2} m\left(\frac{d y}{d t}\right)^{2}\) corresponds to kinetic energy.
For polyatomic gases, one mole of gas has energy.
\(U=\left[\frac{3}{2} k_{B} T+\frac{3}{2} k_{B} T+k_{B} T\right] N_{A}\)
where N_A is Avogadro’s number.

Question 6.
Represent graphically that real gases approach ideal gas behaviour at low pressures and high temperatures.
Answer:

Question 7.
Relative Boltzmann constant and Universal gas constant.
Answer:
If ‘N_A‘ represents Avogadro’s number, ‘k_B’ Boltzmann’s constant and ‘R’ universal gas constant, then
k_BN_A = R
\(k_{B}=\frac{R}{N_{A}}\)

Question 8.
What is meant by partial pressure?
Answer:
The pressure exerted by a gas at the same conditions of volume and temperature if no other gases were present is called partial pressure.

Question 9.
Define and explain degree of freedom.
Ans: It is defined as the number of co-ordinates required to specify the configuration of a system of molecules.
In general, degrees of freedom is calculated by the formula f = 3N – K
where N → Number of gas molecules
K → Number of independent relations between the constituent atoms.

Question 10.
Find out the number of degrees of freedom for mono atomic gas, diatomic gas and tri-atomic gas.
Answer:
For Mono atomic gas
f = 3N – K
[N = 1, K = 0]
f = 3(1) = 0
f = 3 ⇒ 3 degrees of freedom
For diatomic gas
f = 3N – K
[N = 2, K = 1]
f = 3(2) – 1
f = 6 – 1
f = 5 ⇒ 5 degrees of freedom
For triatomic gas
Non Linear
\(\left[\begin{array}{l}
{K=3} \\
{N=3}
\end{array}\right]\)
f = 3N – K
f = 3(3) – 3
f = 9 – 3
f = 6
∴6 degrees of freedom

Linear
\(\left[\begin{array}{l}
{K=2} \\
{N=3}
\end{array}\right]\)
f = 3N – K
f = 3(3) – 2
f = 9 – 2
f = 7
∴7 degrees of freedom

Question 11.
Give the expression for total internal energy of mono atomic gas.
Answer:
\(U=\frac{3}{2} R T\)
or
\(U=\frac{3}{2} k_{B} T N_{A}\)

Question 12.
How many degrees of freedom are there in the translatory motion of atoms?
Answer:
Three degree of freedom are possible which is equal to straight line motion along three coordinates (x, y, z) axes.

Question 13.
How many degrees of freedom are there in a molecule of mono atomic gas?
Answer:
A molecule of mono atomic gas has only three translational degrees of freedom.

Question 14.
Write the expression of molar heat at constant volume in mono atomic gas.
Answer:
\(C_{v}=\frac{d U}{d T}=\frac{3}{2} R T\)

Question 15.
Write the expression of molecule heat at constant pressure in mono atomic gas.
Answer:
\(C_{p}=\frac{5}{2} R\)

Question 16.
What is the ratio of specific heats at constant P and V for mono atomic gas?
Answer:
\(\gamma=\frac{C_{P}}{C_{V}}=\frac{5}{3}\)

Question 33.
Write the relation between CP and Cv for an ideal gas.
Answer:
C_p – C_v = R

Question 17.
How many degrees of freedom are there in diatomic gases?
Answer:
5 degrees of freedom are possible in diatomic gases. Among them 3 translational and 2 rotational.

Question 18.
Write the expression of the total internal energy of a mole of rigid diatomic gas.
Answer:
\(U=\frac{5}{2} k_{B} T \times N_{A}\)
\(U=\frac{5}{2} R T\)

Question 19.
Write the expressions of Cv and CPand ratio of specific heats (y) for rigid and non rigid diatomic gases.
Answer:
C_p (rigid diatomic) = \(\frac{7}{2} R\)
C_v (rigid diatomic) = \(\frac{5}{2} R\)
γ(rigid diatomic) = \(\frac{7}{5}\)

Question 20.
What is y for a gas?
Answer:
γ is the rate of specific heat of a gas at constant pressure to the specific hat at constant volume.
\(\gamma=\frac{C_{p}}{C_{v}}\)

Question 21.
What is the importance of y of a gas?
Answer:
The knowledge of γ helps in finding the atomicity of a gas as well as the molecular structure of the gas.

Question 22.
Write the expressions of total internal energy, C_P, C_V and γ for non rigid diatomic gas.
Answer:

Question 23.
How many translational and rotational degrees of freedoms are possible in a poly atomic gas?
Answer:
Translational degrees of freedom = 3
Vibrational degree of freedom = 3
Many number of vibrational modes (f) are possible in poly atomic gas.

Question 24.
Write the expressions of total internal energy, C_p, C_v and γ for a poly atomic gas.
Answer:

Question 25.
Show that specific heat for a solid is equal to 3R (or) show that for a solid, C = 3R.
Answer:
According to the law of equipartition of the energy, the energy distributed is KBT. The energy can be distributed in 3 different dimension with NA molecules. Therefore, total energy of a solid.
U=3N_AK_BT
U = 3RT
ΔU = 3RΔT
\(\frac{\Delta U}{\Delta T}=3 R\) ……..(1)
When heat is supplied to a solid the change in volume is totally negligible. Therefore, heat supplied is utilised only to increase the internal energy.
Therefore, ΔQ = ΔU.
But accordingly to specific heat,
\(C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{\Delta T}\)
\(C=\frac{\Delta U}{\Delta T}\)
From (1) and (2),
C = 3R

Question 26.
What is the total internal energy and specific heat capacity of water?
Answer:
U = 3 + 3k_BT × N_A = 9 RT
\(C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{\Delta t}=9 R\)

Question 27.
Write a note on mean free path.
Answer:
According to kinetic theory of gases, the molecules of a gas are in a state of random motion and collide frequently with each other.
The path travelled by a molecule between two successive collisions with other molecules is called the fee path.
The average distance travelled by a molecule between two successive collisions is called the mean free path. It can be shown that mean free path (l) is given by
\(l=\frac{1}{\sqrt{2} n \pi d^{2}}\)
Where n is the number of molecules per unit volume and d is the diameter of gas spheres.

Numerical Problems

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 A°.
Answer:
Consider 1 mole of oxygen gas arbitrary STP
1 mol = 6.023 × 10²³ molecules
= 22400 cm³ of gas
Diameter of oxygen molecule = 3 A°

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:
P = 1 atm = 1.013 × 10⁵ Nm^-2, n = 1 mol, Fr = 8.314 J mol^-1 k^-1, Y = 0°C = 273.15 k
The ideal gas state equation is given by
PV = nRT
∴\(V=\frac{n R T}{P}=\frac{1 \times 8 \cdot 314 \times 273 \cdot 15}{1 \cdot 013 \times 10^{5}}\)
V = 0.0224 m³ = 0.0224 × 10⁶cm³ = 22400 cm³ =22.4 litres
The volume occupied by 1 mol of gas at 1 atm pressure and 0°C is 22.4 litres.

Question 3.
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 O₂ = 32 u).
Answer:
Initial Volume, V₁ = 30 litres = 30 × 10^-3 m³
Initial Pressure P₁ = 15 atm =15 × 101.3 KPa
Initial temperature, T = 27°C = 300 K
Let ‘μ’ be the initial number of moles of oxygen.
From ideal gas law,
P₁V₁ = μ₁R₁T₁

Final volume, V₂ = 30 litres = 30 × 10^-3
Final pressure, P₂ = 11 atm =11.143 × 10^-5
Final temperature, T₂ = 17°C = 290 K
∴Final number of moles, n₂ is given by.

∴Moles of oxygen taken out, Δμ = μ₁ – μ₂ = 18.276 – 13.865 = 4.411 moles
∴Mass of oxygen taken out = Δμ × M
where M is the molecular mass of oxygen.
∴mass of oxygen taken out = 4.411 × 32 = 141.152 g
∴141.152 g of oxygen was taken out of cylinder.

Question 4.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep arbitrary 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:
Initial volume of air bubble, V₁ = 1.0 cm³ = 10^-6 m³
Initial temperature, T₁ = 12°C = 285 K
Initial Pressure, P₁ = P_s + ρgh
where P₁ = depth of air bubble = 40 m
P_s = Pressure at lake surface = 1 atm
ρ = density of water = 10² kg/cm³
g = acceleration due to gravity = 9.8 ms²
P₁ =1.013 × 10 + 103 × 9.8 × 40 =493300 Pa = 4.933 × 10⁵Pa
Let final volume of air bubble = V₂
Final Pressure = P_s = 1.013 × 105 Pa
Final Temperature, T₂ = 35°C = 308 K
By combined gas equation

∴The volume of air bubble grows to 5.263 cm when it reaches the lake surface.

Question 5.
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:
From ideal gas law PV = µRT
P = atm = 1.013 × 10⁵ Pa.
V = Volume = 25 m³
T = Temperature = 27 °C = 300 k
µ = number of moles of gas
R = Gas constant = 8.314 J mol^-1 k^-1
∴\(\mu=\frac{P V}{R T}=\frac{1 \cdot 013 \times 10^{5} \times 25}{8 \cdot 314 \times 300}\)
= 0.01015 × 10 = 1015 mol
Since 1 mol = 6.023 × 10²³ molecules
N= 1015 mol = 6.023 × 10²³ × 1015
N = 6.113 × 10²⁶ molecules.
The total number of molecules in the given room is 6.113 × 10²⁶ molecules.

Question 6.
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:
(i) Average thermal energy of an atom \(E=\frac{3}{2} K T\)
where, K = Boltzmann constant = 1.38 × 10^-23 m² kgs^-2 K^-1
T = absolute temperature = 27°C = 300 K
∴E = \(\frac{3}{2}\) × 1.38 x 10^-23 × 300 =1.242 x 10^-21J

(ii) The temperature on the surface of the sun = T = 6000 K
∴E = \(\frac{3}{2}\) × (1.38 × 10^-23) × 6000 = 1.242 x 10^-19J

(iii) On core of a star, T= 10 x 10⁶ = 10⁷ K
∴E = \(\frac{3}{2}\) × (1.38 x 10^-23) × 10⁷ = 2.07 × 10^-16J.

Question 7.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains (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 the largest?
Answer:
Since all vessels are of same capacity and are at same pressure and temperature,
P = P₁ = P₂ = P₃ ……(1)
V = V₁ = V₂ = V₃ …..(2)
T = T₁ = T₂ = T₃ …..(3)
where P_i, V_i and T_i are pressure, volume and temperature of i vessel respectively.
From ideal gas law
P_iV_i = µ_iRT_i
where i = 1, 2, 3
From equation (1), (2) and (3).
PV = µ_iRT
or µ = \(\frac{P V}{R T}\) = constant for all the three vessels.
RT
∴All the three vessels have same number of moles of respective gases.
From Avagadro’s law, equal mols of gases contain equal number of molecules.
The root mean square (rms) speed of a molecule is given by
\(V_{r m s}=\sqrt{\frac{3 k_{B}}{M}}\)
where k₈ = Boltzmann constant, T = temperature M = molecular mass of the gas.
As \(V_{r m s} \alpha \frac{1}{M}\), the rms speed of gas molecules in the three vessels are different. Since neon has the least molecular mass of the three gases, it has highest rms speed.

Question 8.
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:
Temperature of helium gas, T₁ =- 20°C = 253 K
Gram molecular mass of Helium, m₁ = 4.0 G
Gram molecular mass of Argon, m₂ 39.9 gm2
Let temperature of Argon = T₂, The rms velocity of a gas atom is given by
\(V_{r m s}=\sqrt{\frac{3 R T}{M}}\)
where T = Temperature of the gas, R = Universal gas constant, M = Molar mass of the gas.

∴At 2523.675 K, the rms speed of an atom of argon will be equal to that of helium at – 20°C.

Question 9.
Estimate the mean free path and collision frequency of a nitrogen molecule in a atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Answer:
The mean free path is given by
\(l=\frac{k_{B} T}{\sqrt{2} \pi d^{2} \cdot P}\)
where k B= Boltzmann constant = 1.38 × 10^-23 kg² m2 s^-2K^-1
d = diameter of the atom / molecule = 2 × 1 A° = 2 x 10⁵ m
P = Pressure of the gas = 2 atm = 2.026 × 105 Pa

∴mean free path =1.112 × 10^-7 m
RMS speed of nitrogen molecules is given by


∴the time taken between successive collisions is about 500 times the time spent on collision.

Question 10.
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?
Answer:
Total length of bore = 1 m = 100 cm.
Length of air column = 15 cm
Length of mercury thread = 76 cm.
∴length of open space = 100 – (15 + 76) = 9 cm
Initial pressure P₁ = 76 cm of mercury
Initial volume, V₁ = 15 A cm³
where, A = sectional area of the bore in cm²
Let h cm³ of mercury flow out of tube when held vertically
∴ New length of air column = 15 + 9 + h = 24 + h cm .
New length of mercury = 76 -h cm
∴ Final pressure, P₂ = 76 – (76 – h) = h cm of mercury
Find volume, V₂ = (24 + h) × A cm³
Since temperature is constant and no air escapes the bore, P₁V₁ = P₂V₂
∴76 × 15A = h(24 + h)A
∴76 × 15 = h(24 + h)
h = [/latex]\frac{-24 \pm \sqrt{576+4560}}{2}[/latex]
∴\(h=\frac{-24 \pm 71 \cdot 67}{2}\)
h = 23-83, -47.83
Since height can only be positive, the amount of mercury that flows out of bore is 23.83 cm.
Therefore, mercury left in the bore = 52.2 cm and length of air column = 47.8 cm.

Question 11.
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. [Hint: Use Graham’s law of diffusion: R₁/R₂ = where R₁, R₂ are diffusion rates of gases 1 and 2, and Mi and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer:
According to graham’s law of diffusion,
\(\)\frac{R_{1}}{R_{2}}=\left(\frac{M_{1}}{M_{2}}\right)^{\frac{1}{2}}[/ latex]
R₁ = rate of diffusion of hydrogen = 28.7 cm³ s^-1
R₂ = Rate diffusion of unknown gas = 7.2 cm³ s^-1
M₁ = Molecular mass of hydrogen = 2.02
M₂ = Molecular mass of unknown gas.

Since molecular mass of oxygen gas is 32, the given unknown gas is oxygen.

Question 12.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, use the known value of Avogadro’s number. You should, however, 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 A°].
Answer:
Assuming all molecules to be spherical, and solids and liquids are tightly packed,
volume, V=\frac{4}{3} \pi r^{3}
where r = radius of a molecules
Also, V=\frac{\text { mass }}{\text { density }}
For 1 mole of a substance,
V=\frac{4}{3} \pi r^{3}, N_{A}=\frac{M}{\rho}
where NA = Avogadro number = 6.023 × 1023, M = gram atomic mass and ρ = density of the substitute
r=\left(\frac{3 M}{4 \pi \rho N_{A}}\right)^{\frac{1}{3}}

(1) Carbon (diamond):
M = 12.01 g = 12.01 × 10^-3 kg, f = 2.2 × 103 kg m^-3

(2) Gold:
M = 197 g = 197 × 10^-3kg, f = 19.32 × 10³kg ^-3

(3) Nitrogen (liquid):
M = 14.01 g = 14.01 × 10^-3 kg, f = 10³ kg m^-3 ( 3 × 6.94 × 10^-3)

(4) Fluorine (liquid)
M=19 × 10^-3kg, f =1.14 × 10³ kgm^-3.

Substance	Radius (A°)
Carbon (diamond)	1.29
Gold	1.59
Nitrogen (liquid)	1.77
Lithium	1.73
Fluorine (liquid)	1.88