Kinetic Theory

Given:

Diameter of an oxygen molecule (D) = 3 Å

∴ the radius of an oxygen molecule (r) = 1.5 Å (∵ r = D/2)

Fraction = ?

Explanation:We know that the volume occupied by 1 mole of oxygen gas at STP =22.4 litres. The molecular volume occupied 1 mole of oxygen molecules can be found by multiplying the volume of a single oxygen molecule (considering it to be a sphere) with the

Avogadro's constant. Then dividing the latter by the former the required result is obtained.

Volume of 1 mole of oxygen gas at S.T.P(V_molecular)

V_molecular = πr³ × N_A ---(1)

The actual volume occupied by oxygen gas at STP (V_actual)

V_actual = 22.4 litres = 0.0224 m³

V_actual = 0.0224 m³ ---(2)

(Where N_A = 6.02214086 × 10²³ mol^-1 is Avogadro's constant.)

Substituting for ‘r’ and ‘N_A’ in equation (1) we get

V_molecular = πr³ × N_A = π(1.5 × 10^-10)³ × ( 6.02214086 × 10²³)

= 0.000008513≈ 8.51 × 10^-6 m³

(∵ 1 Å = 1 × 10^-10 m)

∴ V_molecular = 8.52 × 10^-6 m³ ---(3)

Fraction = = = 0.00038007 ≈ 3.8 × 10^-4

Fraction = 3.8 × 10^-4

The required ratio/fraction is3.8 × 10^-4.

Given:

P = 1 atm = 101 kPa

T = 273.15 K

n = 1 mole

V = ?

Explanation:We know that the volume occupied by 1 mole of oxygen gas at STP =22.4 litres. By using the Ideal Gas Law (PV = nRT) we can obtain the required result.

Wkt…

PV = nRT ---( Ideal Gas Law )

∴ V =

⇒ V = ≈ 0.0224 m³ = 22.4 litres

⇒ V = 22.4 litres

The required volume is22.4 litres.

(a) Seeing that the dotted line doesn’t vary with pressure and by using theIdeal Gas Law (PV = nRT ) we can say that it implies the ideal behaviour of Gas.

Wkt…

PV = nRT

⇒ nR = PV/T = constant

The dotted plot impliesPV/T = constant.

(b) T₂is heavily deviates from ideal behaviour than T₁.

⇒ T₁ > T₂ is true.

T₁>T₂is true.

(c) Using PV = nRT

⇒ PV/T = nR

Given: Mass of the gas = 1 × 10^-3 kg = 1 g

Molecular mass of Oxygen = 32g/mole

So, number of moles = given weight/molecular weight

⇒ number of moles = 1/32

So, nR = 1/32 × 8.314 J / mol /K = 0.26 J/K

Hence, value of PV/T = 0.26 J/K.

(d) 1g of Hydrogen doesn't represent the same number of mole. molecular mass of H₂ = 2g/mol

Hence, number of moles of Hydrogen required is 1/32

So, mass of H₂ required = no of moles of H₂ × molecular mass of H₂

⇒ mass of H₂ required = 1/32mole × 2 g/mol = 1/16 g = 0.0625 g

⇒Hence themass of H₂ required = 0.0625 g

Explanation: We have to consider 2 cases here and find the masses of O₂in each case and then by calculating the difference of those masses gives thethe mass of oxygen taken out of the cylinder.

Given:

Initially, Case1

V₁ = 30 litres = 0.03 m³

P₁ = 15 atm = 15 × 1.01 × 10⁵ Pa

T₁ = 27+273 = 300 K

If n₁ is the moles O₂ in the cylinder, then by using

P₁V₁ = n₁RT₁ ---( Ideal Gas Law )

⇒ n₁ = =

⇒ n₁ = 18.276 moles

Molecular weight of O₂ (M) = 32 g

Initial mass of cylinder (m₁) = n₁M

⇒ m₁ = n₁M

⇒ m₁ = (18.276 moles) × ( 32 g)

⇒ m₁ = 584.84 g

Now,

Let n₂ be the moles of O₂ left in the cylinder

V₂ = 30 litres = 0.03 m³

P₂ = 11 atm = 11 × 1.01 × 10⁵ Pa

T₂ = 17+273 = 290 K

Thus n₂ = =

⇒ n₁ = 13.86 moles

Therefore, m₂ be the final mass of O₂ in the cylinder

⇒ m₂ = n₂M

⇒ m₂ = (13.86 moles) × ( 32 g)

⇒ m₂ = 453.1 g

∴ the mass of oxygen taken out of the cylinder

Δm = m₁-m₂

⇒ Δm = 584.84 g - 453.1 g

⇒ Δm = 131.74 g

The mass of oxygen taken out of the cylinder is 131.74 g.

This problem can by solved using the fact that ‘the number of moles/molecules of the air bubble remains unchanged as the bubbles rises from bottom to the surface’.

Given:

Volume of the air bubble (V₁) = 1.0 cm³ = 1.0 × 10 ^–6 m³

Bubble rises to height (h) = 40 m

Temperature at a depth of 40 m (T₁) = 12°C = 285 K

Temperature at the surface (T₂) = 35°C = 308 K

The pressure on the surface (P₂) = 1 atm = 1 × 1.01 × 10⁵ Pa

The pressure at the depth of 40 m (P₁) = 1 atm + ρgh

(Where, ρ is the density of water = 10³ kg/m³

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

∴ P₁ = (1.013 × 10 ⁵ pa)+ (40 m × 10 ³ kg/m³ × 9.8 m/s² )

⇒ P₁ = 493300 Pa

We have = ( = number of moles/molecules of the air bubble)

(Where, V₂ is the volume of the air bubble when it reaches the surface )

∴ V₂ = =

⇒ V₂ = 5.263 × 10 ^–6 m³

⇒ V₂ = 5.263 cm³

When the air bubble reaches the surface, its volume is 5.263 cm³

We have to consider the Ideal Gas Law with theBoltzmann constant to solve this problem.

Given:

Volume of the room (V) = 25 m³

Temperature of the room (T) = 27°C = 300 K

Pressure in the room (P) = 1 atm = 1 × 1.013 × 10⁵ Pa

The ideal gas equation also be written as

PV = k_BNT

(Where, K_B is Boltzmann constant = 1.38 × 10^–23 m² kg s^–2 K^–1 and N is the number of air molecules in the room)

∴ N = =

⇒ N = 6.11 × 10 ²⁶ molecules

The total number of air molecules in the given room is 6.11 × 10²⁶

This problem can be solved using the relation between kinetic energy and temperature i.e. averagethermal energy = (3/2)k_BTper atom.

(i) Average thermal energy of a helium atom at room temperature (27 °C)

At room temperature (T) = 27°C = 300 K

Average thermal energy (E_Th) = (3/2)k_BT

(Where k_B is Boltzmann constant = 1.38 × 10^–23 m² kg s^–2 K^–1 )

∴ E_Th = (3/2)k_BT

⇒ E_Th = (3/2) 1.38 × 10^–23 × 300

⇒ E_Th = 6.21 × 10^–21J

The average thermal energy of a helium atom at room temperature is 6.21 × 10^–21 J.

(ii) The surface of the sun has T = 6000 K

∴ E_Th = (3/2)k_BT

⇒ E_Th = (3/2) × 1.38 × 10^-23 × 6000

⇒ E_Th = 1.241 × 10 ^-19 J

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10^–19 J.

(iii) The temperature of 10 million

kelvin Average thermal energy = (3/2)k_BT

⇒ E_Th = (3/2) × 1.38 × 10^-23 × 10⁷

⇒ E_Th = 2.07 × 10^-16 J

Hence, the average thermal energy of a helium atom(the typical core temperature in the case of a star)is 2.07 × 10^–16 J.

All the three vessels have the same capacity, which implies they have the same volume. Thus, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

Which is equal to Avogadro’s number, N_A = 6.023 × 10²³.

The root mean square speed (V_rms) of a gas of mass m, and temperature T, is given by the relation:

mV_rms² = k_BT

⇒ V_rms =

where, k_B is Boltzmann constant For the given gases, k and T are constants. Hence V_rms depends only on the mass of the atoms, i.e., V_rms∝

The root mean square speed of the molecules in the vessels is not the same. The mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

All the three vessels have the same capacity, which implies they all have the same volume. Thus, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

Given:

Temperature of the helium atom(T_h)

T_h = –20°C = 253 K

Atomic mass of argon(M_a) = 39.9 u

Atomic mass of helium(M_h) = 4u

Let, V_a be the rms speed of argon and let V_h be the rms speed of helium.

Since, MV² = k_BT (let k_B be simply k)

∴ V =

The rms speed of argon is given by-

V_a = ---(1)

The rms speed of helium is given by-

V_h = ---(2)

∴ since V_a = V_h

⇒ =

⇒ =

⇒ T_a = M_a

⇒ T_a = × 39.9 u

⇒ T_a = 2.52 × 10³ K

Therefore, the temperature of the argon atom is 2.52 × 10³ K.

First we have to find the rms speed of the nitrogen molecule, then by finding the collision time and the time the between two successive collisions, and comparing the two by dividing the latter by former we get the required result.

Given:

Mean free path = 1.11 × 10^–7 m

Collision frequency = 4.58 × 10⁹ s^–1

Pressure inside the cylinder containing nitrogen (P)

P = 2 atm = 2.026 × 10⁵ Pa

Temperature inside the cylinder (T) = 17°C = 290 K

Radius of a nitrogen molecule (r) = 1 Å = 1 × 10¹⁰ m

Diameter (d) = 2r

d = 2 × 10¹⁰ m

Molecular mass of nitrogen (M) = 28 g

M = 28 × 10^–3 kg

The root mean square speed of nitrogen is given by

V_rms =

⇒ V_rms =

⇒ V_rms = 508.26 m/s

Let the mean free path be L, given as

L =

⇒ L =

⇒ L = 1.11 × 10^-7 m

Collision frequency (f) = V_rms /L

⇒ f = 508.26 / (1.11 × 10^-7 )

⇒ f = 4.58 × 10⁹ s^-1

The Collision time is given as (T)

T = d / V_rms

⇒ T = (2 × 10 ^-10 ) / (508.26)

⇒ T = 3.93 × 10^-13 s

Time taken between successive collisions be T_c

T_c = l / V_rms

⇒ T_c = (1.11 × 10^-7 ) / (508.26)

⇒ T_c = 2.18 × 10^-10 s

∴ T_c / T = (2.18 × 10^-10) / (3.93 × 10^-13)

⇒ T_c / T = 500

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

If the temperature is constant, then the Boyle’s law state that,

P₁ V₁ = P₂ V₂

Where,

‘P₁’ is the initial pressure.

‘V₁’ is the initial pressure.

‘P₂’ is the final pressure.

‘V₂’ is the final pressure.

Given,

The length of the tube = 1 m = 100 cm

The length of the mercury thread = 76 cm

The length of the air column = 15 cm

Therefore,

The length of 9 cm is left in the open end when the tube is kept in a horizontal position. We suppose ‘a’ cm² as the cross section area of the tube.

⇒ When the tube is held horizontally,

Initial pressure, P₁ = 76 cm (of mercury)

Initial Volume, V₁ = 15 a cm³

⇒ When the tube is held vertically,

The length of the air column = 15 cm + 9 cm = 24 cm

(The two air column on the both sides of the mercury thread adds up)

The length of the mercury thread = 76 cm

(at the open end of the tube)

Since, the pressure of the mercury and the air adds up to be greater than the atmospheric pressure. Some mercury flows out of the open end until the total pressure inside reaches the atmospheric pressure.

Suppose, The length of the mercury thread that is reduced = ‘h’ cm

The length of the air column now = ( 24 + h ) cm

The length of the mercury thread now = ( 76 – h ) cm

The new final pressure, P₂ = 76 cm – (76 – h ) cm

⇒ P₂ = h cm ( of mercury)

The new final Volume, V2 = (24 + h) a cm³

By Boyle’s law, we have

P₁V₁ = P₂V₂

⇒ 76 × 15 a = h × (24 + h) a

⇒ 1140 = 24h + h²

⇒ h² + 24h – 1140 = 0

Solving the quadratic equation we get,

⇒ cm

⇒ cm

⇒

⇒ h = 23.83 cm or h = -47.83

Since, the height cannot be negative, h = 23.83 cm

Thus,if the tube is held vertically with the open end at the bottom the mercury will flow out of the tube to equalize the pressure with the atmospheric pressure and the thread of mercury will reduce by h = 23.83 cm.

The Graham’s Law of Diffusion states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. We have,

Where,

M is the molecular mass of the first gas,

M’ is the molecular mass of the second gas

R is the diffusion/ effusion rate of the first gas,

R’ is the diffusion/ effusion rate of the second gas

Given,

Molecular mass of hydrogen, M = 2.020 g

Rate of diffusion of hydrogen, R = 28.7 cm³ /s

Rate of diffusion of unknown gas, R’ = 7.2 cm³ /s

Thus, From the Graham’s law of diffusion, we get:

⇒

⇒

⇒ M’ = 32.09 g

32 g is the molecular mass of Oxygen .

The unknown gas is identified to be Oxygen.

Let us suppose:

The density of the suspended particle is ρ

The density of the medium is ρ’

Mass of the medium which is displaced by the suspended particle is m

Mass of the suspended particle is m’

Volume of the suspended particle is V

Since, the equation for sedimentation equilibrium of a suspension in a liquid column is asked we will use the Archimedes’ principle.

The Archimedes’ Principle for a suspended particle in a liquid column, the effective weight W’ of the suspended particle in the liquid is given as:

W’ = Weight of the displaced medium – Weight of the suspended particle

⇒ W’ = mg – m’g

⇒ W’ = mg – Vρ’g =

⇒ ……..(i)

And we have,

Gas Constant, R = k_BN

⇒ k_B = ………………..(ii)

Given,

The law of atmosphere states that:

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] ……………(iii)

where,

n₁ is the number density at the height ‘h₁’

n₂ is the number density at the height ‘h₂’

‘mg’ is the apparent weight of the particle suspended in the gas column.

Replacing the values from equation (i) and (ii) in equation (iii) we get:

⇒ (h₂ – h₁)

⇒(h₂ – h₁)

This is the equation for sedimentation equilibrium of a suspension in a liquid column.

Let us suppose that the atoms are ‘tightly packed’ in a solid or liquid phase and

The Density of the substance = D

Atomic mass of the substance = M

Avogadro’s Number, N = 6.022 × 10²³

Radius of an atom of the substance = r

Volume of an atom =

Therefore, the volume of the N molecules = ……(1)

Also, the volume of one mole of the substance = …….(2)

Equating the equation (1) and (2) we get :

=

⇒

Thus,

(i) Carbon

Given,

Atomic mass of carbon, M = 12.01× 10^-3 kg

The Density of the carbon, D = 2.22× 10³ kg / m³

⇒

⇒ r = 1.29× 10^-10 m = 1.29 Å

The radius of the carbon atom is 1.29Å.

(ii) Gold

Given,

Atomic mass of Gold atom, M = 197.0× 10^-3 kg

The Density of the Gold atom, D = 19.32× 10³ kg / m³

⇒

⇒ r = 1.59× 10^-10 m = 1.59 Å

The radius of the Gold atom is 1.59Å.

(iii) Liquid Nitrogen

Given,

Atomic mass of Liquid nitrogen, M = 14.01× 10^-3 kg

The Density of the Liquid nitrogen,D = 1.0× 10³ kg / m³

⇒

⇒ r = 1.77× 10^-10 m = 1.77 Å

The radius of the Liquid nitrogen atom is 1.77 Å.

(iv) Lithium

Given,

Atomic mass of Lithium, M = 6.94× 10^-3 kg

The Density of the Lithium, D = 0.53× 10³ kg / m³

⇒

⇒ r = 1.73× 10^-10 m = 1.73 Å

The radius of the Lithium atom is 1.73 Å.

(v) Liquid Fluorine

Given,

Atomic mass of Liquid flourine, M = 19.0× 10^-3 kg

The Density of the Liquid flourine,D = 1.14× 10³ kg / m³

⇒

⇒ r = 1.88× 10^-10 m = 1.88 Å

The radius of the Liquid flourine atom is 1.88 Å.

The Rough estimates of the size of the atoms thus are: