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*NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory*

**Class 11 Physics Chapter 13 Kinetic Theory****:**

Section Name | Topic Name |

13 | Kinetic Theory |

13.1 | Introduction |

13.2 | Molecular nature of matter |

13.3 | Behaviour of gases |

13.4 | Kinetic theory of an ideal gas |

13.5 | Law of equipartition of energy |

13.6 | Specific heat capacity |

13.7 | Mean free path |

**Introduction**

What is Kinetic Theory?

- Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules.
- In solids the molecules are very tightly packed as inter molecular space is not present In liquids inter molecular spaces are more as compared to solids and in gases the molecules are very loosely packed as intermolecular spaces are very large.
- The random movement of molecules in a gas is explained by kinetic theory of gases.
- We will also see that why kinetic theory is accepted as a success theory.
- Kinetic theory explains the following:-
- Molecular interpretation of pressure and temperature can be explained.
- It is consistent with gas lawsand Avogadro’s hypothesis.
- Correctly explains specific heat capacities of many gases.

**Molecular nature of matter**

**John Dalton**

- Atomic hypothesis was given by many scientists. According to which everything in this universe is made up of atoms.
- Atoms are little particles that move around in a perpetual order attracting each other when they are little distance apart.
- But if they are forced very close to each other then they rebel.
- For example: – Consider a block of gold. It consists of molecules which are constantly moving.

- Dalton’s atomic theory is also referred as the molecular theory of matter.This theory proves that matter is made up of molecules which in turn are made up of atoms.
- According to Gay Lussac’s law when gases combine chemically to yield another gas,their volumes are in ratios of small integers.
- Avogadro’s law states that the equal volumes of all gases at equal temperature and pressure have the same number of molecules.
- Conclusion: – All these laws proved the molecular nature of gases.
- Dalton’s molecular theory forms the basis of Kinetic theory.

__Why was Dalton’s theory a success?__

- Matter is made up of molecules, which in turn are made up of atoms.
- Atomic structure can be viewed by an electron microscope.

__Solids, Liquids, Gases in terms of molecular structure__

Basis of Difference | Solids | Liquids | Gases |

Inter Atomic Distance(distance between molecules). | Molecules are very tightly packed. Inter atomic distance is minimum. | Molecules are not so tightly packed. Inter atomic distance is more as compared to solids. | Molecules are loosely packed .Free to move. Inter atomic distance is maximum. |

Mean Free Path is the average distance a molecule can travel without colliding. | No mean free path. | Less mean free path. | There is mean free path followed by the molecules. |

**Behaviour of Gases**

- Gases at low pressures and high temperatures much above that at which they liquify (or solidify) approximately satisfy a relation between their pressure, temperature and volume:
- PV=KT (i)
- This is the universal relation which is satisfied by all gases.
- where P, V, T are pressure,volume and temperature resp. and
- K is the constant for a given volume of gas. It varies with volume of gas.

- K=Nk
_{B}where- N=number of molecules and
- k
_{B}= Boltzmann Constant and its value never change.

- From equation (i) PV= Nk
_{B} - Therefore PV/NT = constant=(k
_{B})(Same for all gases). - Consider there are 2 gases :- (P
_{1},V_{1},T_{1}) and (P_{2}, V_{2},T_{2}) where P, V and T are pressure, volume and temperature resp. - Therefore
**P**_{1},V_{1}/(N_{1}T_{1}) = P_{2}V_{2}/(N_{2}T_{2}) - Conclusion: – This relation is satisfied by all gases at low pressure and high temperature.

= (11.143×10^{5}x30x10^{-3})/ (8.314×290)

= 13.86

But, n_{2} = m_{2}/M Where, m_{2} is the mass of oxygen

remaining in the cylinder m_{2} = n_{2}M = 13.86 × 32

= 453.1 g

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

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

= m_{1} – m_{2}

= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

**Problem:- **An air bubble of volume 1.0 cm^{3} 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:- **Volume of the air bubble, V_{1} = 1.0 cm^{3} = 1.0 × 10^{–6} m^{3}

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T_{1} = 12°C = 285 K

Temperature at the surface of the lake, T_{2} = 35°C = 308 K

The pressure on the surface of the lake:

P_{2} = 1 atm = 1 ×1.013 × 10^{5} Pa

The pressure at the depth of 40 m:

P_{1} = 1 atm + dρg Where, ρ is the density ofwater = 10^{3} kg/m^{3} g is the acceleration due

to gravity = 9.8 m/s^{2}

Therefore, P_{1} = 1.013 × 10^{5} + 40 × 10^{3} × 9.8 = 493300 Pa

We have: P_{1} V_{1}/T_{1} = P_{2}V_{2}/T_{2}

Where, V_{2} is the volume of the air bubble when it reaches the surface

V_{2} = P_{1} V_{1}T_{2}/T_{1}P_{2}

= (493300) (1.0×10^{-6})308/(285×1.013×10^{5})

= 5.263 × 10^{–6} m^{3} or 5.263 cm^{3}

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm^{3}.

**Justification of the Avogadro’s hypothesis from equation of gas**

- Avogadro’s hypothesis states that equal volumes of all gases at equal temperature and pressure have the same number of molecules.
- Consider the equation PV/NT = constant and if P,V and T are same for 2 gases then N(number of molecules) is also same.
- According to Avogadro’s hypothesis number of molecules per unit volume is same for all gasesat a fixed P and T.
- Avogadro number is denoted by N
_{A}. Where_{A}denotes Avogadro number. **N**. It is universal value._{A}= 6.02×10^{23}- Experimentally it has been found that the mass of 22.4 litres of any gas is equal to molecular weight in grams at standard temperature and pressure.

**Perfect Gas Equation**

- Perfect gas equation is given by
**PV=μRT,**- WhereP,V are pressure, volume, T =absolute temperature,μ = number of moles and R =universal gas constant.
- R= k
_{B}N_{A}where k_{B}= Boltzmann constant and N_{A}= Avogadro’s Number

- This equation tells about the behaviour of gas at a particular situation.
- If a gas satisfies this equation then the gas is known as Perfect gas or an ideal gas.

__Different Forms of Perfect Gas Equation__

- PV=μRT (i)

- Where μ (no. of moles) = N/N
_{A}where N=no of molecules and N_{A}= Avogadro number(no of molecules in 1 mole of gas).Orμ = M/M_{o}where M=mass of sample of gas and M_{o}= molar mass. - PV = (N/N
_{A})RT(putting μ=N/N_{A}in equation(i)) - By simplifying PV = Nk
_{B}T - PV=Nk
_{B}T => P = (N/V) k_{B}T => P=nk_{B}T where n(number density) =N/V where N=number of molecules and V=volume. - Therefore we get
**PV=nk**_{B}T

- Substitute μ = M/M
_{o}in equation(i)

- PV=(M/M
_{o}) RT => P=(M/V)1/M_{o}RT where- ρ(mass density of the gas) = M/V

- Therefore
**P=ρRT/M**_{o}

**Ideal Gas**

- A gas that satisfies the perfectgas equation exactly at all pressures and temperatures.
- Ideal gas is atheoretical concept.
- No real gas is truly ideal.A gas which is ideal is known as real gas.
- Real gases approach the ideal gas behaviour for low pressures and high temperatures.

** Problem:-** ( Kinetic Theory Notes )

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour

and other constituents) in a room of capacity 25.0 m^{3} at a temperature of 27 °C and 1 atm

pressure.

__Answer:-__

Volume of the room, V = 25.0 m^{3}

Temperature of the room, T = 27°C = 300 K

Pressure in the room, P = 1 atm = 1 × 1.013 × 10^{5} Pa

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T)

can be written as:

PV = k_{B}NT

Where,

K_{B} is Boltzmann constant = 1.38 × 10^{–23} m^{2} kg s^{–2} K^{–1}

N is the number of air molecules in the room

N=PV/ K_{B}T

=1.013×10^{5}x25/1.38×10^{-23}x300

= 6.11 × 10^{26} molecules

Therefore, the total number of air molecules in the given room is 6.11 × 10^{26}

__Real gases deviation from ideal gas__

- Real gases approach the ideal gas behaviour for low pressures and high temperatures.
- Ideal gas equation PV=μRT, for 1 mole ,μ=1,PV=RT
- =>PV/RT=constant
- Graph should be a straight line(parallel to x-axis) for ideal gas.
- This means it has constant value at all temperature and all pressure.

But in case of real gases graph approach ideal gas behaviour at high temperature and low pressure.

- At high temperature and low pressure molecules are far apart. When temperature is increased the molecules will move randomly far from each other.
- As a result molecular interaction decreases the gas behaves as an ideal gas.
- The ideal behaviour comes into picture when the molecular present inside the gas don’t interact with each other.

Charles’s law:-Consider If P(Pressure) is constant, then

**Deducing Dalton’s Law of partial pressures**

- Dalton’s law of partial pressure states that the total pressure of a mixture of ideal gases is the sum of partial pressures.
- Consider if there are several ideal gases mixed together in a vessel,then the total pressure of that vessel is equal to sum of partial pressure.
- Partial pressure is the pressure exerted by a particular gas if only that gas is present in the vessel.
- For example: –
- Consider if in a vessel there is a mixture of 3 gases, A,B and C.So the partial pressure of A is equal to pressure exerted only by Aand considering B and C are not present.
- Similarly partial pressure of B is equal to the pressure exerted only by B and considering A and C are not there.
- Similarly for C.

- According to Dalton’s law the total pressure of mixture is sum of partial pressure of A, partial pressure of B and partial pressure of C.

__To show how perfect gas equation concludes Dalton’s law of partial pressure:-__

- Suppose there is a mixture of ideal gases which means these gases do not interact with each other.
- By perfect gas equation PV=μRT
- Where V=volume of vessel,P=Pressure and T=temperature andμ (no. of moles).

- As there are mixture of gases therefore μ=μ
_{1}+μ_{2}+ —- so on. - PV=( μ
_{1}+μ_{2})RT - P=( μ
_{1}+μ_{2})RT/V =>μ_{1}RT/V + μ_{2}RT/V + —- P=P
_{1}+P_{2}+—-- Where P
_{1}=partial pressure of gas 1 and P_{2}=partial pressure of gas 2.

- Where P
- Therefore
**P= P**total pressure due to the mixture of gases is equal to the sum of the partial pressure of the gas._{1}+P_{2}+—

- P=P

- By perfect gas equation PV=μRT

**Problem**: -Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of the oxygen to be 3Å.

**Answer:- **At STP, Temperature = 273K and Pressure=1atm, diameter=3Å =3×10^{-8} cm.

Actual volume by 1mole O_{2} gas at STP =22.4l = 22400cm^{3}.

Molecular volume of O_{2} = (4/3) πr^{3}xN_{A}

= (4/3) 3.14x (1.5×10^{-8})^{3}x6.02×10^{23} =8.51cm^{3}

Therefore, Ratio =8.51/22400 = 3.8×10^{-4}

**Problem****:- **The density of water is 1000kg m^{–3}. The density of water vapour at 100 °Cand 1 atm pressure is 0.6 kg m^{–3}. Thevolume of a molecule multiplied by the totalnumber gives, what is called, molecularvolume. Estimate the ratio (or fraction) ofthe molecular volume to the total volumeoccupied by the water vapour under theabove conditions of temperature andpressure.

**Answer**:- For a given mass of water molecules,the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 ×10 ^{-4} )times larger. If densities of bulk water and watermolecules are same, then the fraction ofmolecular volume to the total volume in liquidstate is 1. As volume in vapour state hasincreased, the fractional volume is less by thesame amount, i.e. 6×10^{-4}.

**Problem:** – Estimate the volume of awater molecule if the density of water is 1000kg m^{–3}. The density of water vapour at 100 °Cand 1 atm pressure is 0.6 kg m^{–3}?

**Answer**: – In the liquid (or solid) phase, themolecules of water are quite closely packed. Thedensity of water molecule may therefore, beregarded as roughly equal to the density of bulkwater = 1000 kg m^{–3}. To estimate the volume ofa water molecule, we need to know the mass ofa single water molecule. We know that 1 moleof water has a mass approximately equal to(2 + 16)g = 18 g = 0.018 kg.

Since 1 mole contains about 6 × 10^{23}molecules (Avogadro’s number), the mass ofa molecule of water is (0.018)/(6 × 10^{23}) kg =3 × 10^{–26} kg. Therefore, a rough estimate of thevolume of a water molecule is as follows:

Volume of a water molecule

= (3 × 10^{–26} kg)/ (1000 kg m^{–3})

= 3 × 10^{–29} m^{3}

= (4/3) π (Radius)^{3}

Hence, Radius ≈ 2 ×10^{-10} m = 2 Å.

**Problem:- **What is the averagedistance between atoms (interatomicdistance) in water?

**Answer:** A given mass of water in vapour statehas 1.67×10^{3} times the volume of the same massof water in liquid state. This is alsothe increase in the amount of volume availablefor each molecule of water. When volumeincreases by 10^{3} times the radius increases byV1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So theaverage distance is

= 2 × 20 = 40 Å.

**Problem:- ** A vessel contains two nonreactivegases: neon (monatomic) andoxygen (diatomic). The ratio of their partialpressures is 3:2. Estimate the ratio of (i)number of molecules and (ii) mass densityof neon and oxygen in the vessel. Atomicmass of Ne = 20.2 u, molecular mass of O_{2}= 32.0 u.

**Answer**: – Partial pressure of a gas in a mixture isthe pressure it would have for the same volumeand temperature if it alone occupied the vessel.

(The total pressure of a mixture of non-reactivegases is the sum of partial pressures due to itsconstituent gases.) Each gas (assumed ideal)obeys the gas law.

Since V and T are common tothe two gases, we have

P_{1}V = μ_{1} RT and P_{2}V =μ_{2}RT, i.e. (P_{1}/P_{2}) = (μ_{1} / μ_{2}).

Here 1 and 2 referto neon and oxygen respectively. Since (P_{1}/P_{2}) =(3/2) (given), (μ_{1}/ μ_{2}) = 3/2

(i) By definition μ_{1} = (N_{1}/N_{A} ) and μ_{2} = (N_{2}/N_{A})where N_{1} and N_{2} are the number of moleculesof 1 and 2, and NA is the Avogadro’s number.

Therefore, (N_{1}/N_{2}) = (μ_{1} / μ_{2}) = (3/2).

(ii) We can also write μ_{1} = (m_{1}/M_{1}) and μ_{2} =(m_{2}/M_{2}) where m_{1} and m_{2} are the masses of 1 and 2; and M_{1} and M_{2} are their molecularmasses. (Both m_{1} and M_{1}; as well as m_{2} and M_{2} should be expressed in the same units).If ρ_{1} and ρ_{2} are the mass densities of 1 and 2 respectively, we have,

= (½) (m_{1}/V)/m_{2}/V = (m_{1}/m_{2}) (1/2) (M_{1}/M_{2})

= (3/2) (20.2/32.0)

=0.947

**Problem:- **An air bubble of volume 1.0cm^{3} rises from the bottom of a lake 40m deep at a temperature of 12^{0}C. To what volume does it grow when it reaches the surface, which is at temperatures of 35^{0}C?

**Answer:- **Let the depth (d) of the lake = 40m. Temperature (T_{1}) at the bottom = 12^{0}C = 285K and (V_{1}) Volume of air bubble= 1.0cm^{3} =1×10^{-6}m^{3}.Temperature at the surface (T_{2}) =35^{0}C = 308K, Pressure at the surface (P_{2}) = 1 atm= 1.013×10^{5} Pa.

Pressure at the bottom P_{1} = P_{2}+ρdg

=1+10^{3}x9.8×40

=493300 Pa

Also (P_{1}V_{1})/T_{1}= (P_{2}V_{2})/T_{2}

V_{2} = (P_{1}V_{1}T_{2})/P_{2}T_{1}

After calculating,

V_{2}=5.263 cm^{3}

**Kinetic Theory of an Ideal Gas**

** Basis of Kinetic Theory: **–

- Molecules of gas are in incessant random motion,colliding against one another

andwith the walls of the container.

- All collisions are elastic.
- Total Kinetic energy is conserved.
- Total momentumis conserved.

- In case of an elastic collision total Kinetic energy and momentum before collision is equal to the total Kinetic energy and momentum after collision.

__What does Kinetic Theory tells?__

- At ordinary temperature and pressure the molecular size is very small as compared to inter molecular distance between them.
- In case of gas, molecules are very far from each other and the size of molecules is small as compared to the distance between them.
- As a result, interaction between them is negligible. As there is no interaction between the molecules, there will be no force between the molecules.
- As a result molecules are moving freely as per newton’s first law of motion.
- The molecules should move along straight line but when they come closer they experience the intermolecular forces and as a result their velocities change.
- This phenomenon is known as collision.These collisions are elastic.

__Justifying the assumptions:-__

- We have assumed the container containing the gas is a cube. The shape of the container is immaterial.
- For a vessel of any arbitrary shape, we can choose a small infinitesimal (planar) area and can prove the above derivation.
- We will see A and Δt are not there in the final result.
- By Pascal’s law pressure in one portion of gas in equilibrium is the same as anywhere else.
- All collisions are neglected.
- The number of molecules hitting the wall in time Δt was found to be ½ n AvxΔt, with random collisionsand asteady state of gas.
- Thus, if a molecule with velocity (v
_{x}, v_{y},v_{z})acquires a different velocity due to collision withsome molecules, there will always be some othermolecule with a different initial velocity whichafter a collision acquires the velocity (v_{x}, v_{y}, v_{z}). - Molecular collision, when they are not too frequent and the time spent in collision is very small compared to the time between collisions, will not have any affect in the above calculation.

__Problem:-__

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 millionKelvin (the typical core temperature in the case of a star).

__Answer:-__

At room temperature, T = 27°C = 300 K

Average thermal energy= (3/2) kT

Where k is Boltzmann constant = 1.38 × 10^{–23} m^{2} kg s^{–2} K^{–1}

(3/2) kT= (3/2) x1.38×10^{-38}x300= 6.21 × 10^{–21}J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is

=6.21× 10^{–21} J.

On the surface of the sun, T = 6000 K

Average thermal energy= (3/2) kT

= (3/2) x1.38×10^{-38}x6000

= 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.

At temperature, T = 10^{7} K

Average thermal energy= (3/2) kT

= (3/2) x1.38×10^{-23}x10^{7}

= 2.07 × 10^{–16} J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 ×10^{–16}

(ii) Now ½ m vrms2 = average kinetic energy permolecule = (3/2) k_{B}T where m is the massof a molecule of the gas. Therefore,

v^{2}_{rmsAr}/ = v^{2}_{rmscl} =M_{cl}/M_{AR} =70.9/39.9 =1.77

where M = the molecular mass of the gas.(For argon, a molecule is just an atom of argon.)

Taking square root of both sides, v_{rmsAr} = v_{rmscl} = 1.33

** Problem:- **Uranium has two isotopesof masses 235 and 238 units. If both arepresent in Uranium hexafluoride gas whichwould have the larger average speed? Ifatomic mass of fluorine is 19 units,estimate the percentage difference inspeeds at any temperature.

** Answer:- ** At a fixed temperature the averageenergy = ½ m <v2> is constant. So smaller the mass of the molecule, faster will be the speed.The ratio of speeds is inversely proportional tothe square root of the ratio of the masses. Themasses are 349 and 352 units. Sov

_{349}/ v

_{352}= (352/ 349)

^{1/2}= 1.0044.

Hence difference= 0.44 %.[^{235}U is the isotope needed for nuclear fission.

To separate it from the more abundant isotope^{238}U, the mixture is surrounded by a porouscylinder.

**Law of Equipartition of energy:Degrees of Freedom**

- Degrees of Freedom can be defined as independent displacements or rotations that specify the orientation of a body or system.
- A molecule free to move in space needs three coordinates to specify its location.
- If it is constrained to move in a plane it needs to.
- If constrained to move along a line, it needs just one coordinate to locate it.
- For example:-Consider a room and if we tie a thick rope from one wall to another.
- Take a ball which is moving straight on the rope.
- The ball has only 1 degree of freedom. It can move only in one particular dimension.
- Consider if the ball is on the floor which is two-dimensional, then the ball can move along 2 directions.
- The ball has 2 degree of freedoms.
- Consider if we throw the ball in space which is 3 dimensional. Then the ball can move in 3 dimensions.

- Therefore degree of freedom tells us in how many ways a body can move or rotate or vibrate.

**Translational degree of freedom:-**

- Translation means motion of the body as a whole from one point to another.

- For example:
- Consider the oxygen molecule;it has 2 oxygen atoms which are bonded together.
- The 2 oxygen atoms along with the bond are considered as whole body.
- When the body as a whole is moving from one point to another is known as translational.
- Consider a molecule which is free to move in space and so it will need 3 coordinates(x, y, z) to specify its location.
- Therefore it has 3 degrees of freedom.
- Similarly a molecule which is free to move in a plane which is 2 dimensional and so it needs 2 coordinates to specify its location.
- Therefore it has 2 degrees of freedom.
- Similarly a molecule which is free to move in line it needs 1 coordinate to specify its location.
- Therefore it has 1 degree of freedom.

- Molecules of monoatomic gas have only translational degrees of freedom.This means gases which have only one atom.
- For example:-Helium atom it consists of only one He atom.It will have translationaldegrees of freedom.
- Each translational degree of freedom contributes a term that contains square of some variable of motion.
- The variable of motion means the velocity (v
_{x},v_{y},v_{z}). - The term (
**1/2) mv**will contribute to energy.This is Kinetic energy which is involved with the motion of the molecule from one point to another._{x}^{2}

- The variable of motion means the velocity (v

In thermal equilibrium, the average of each such term is (**1/2) k _{B}T**.

**Rotational Degree of freedom**

- Independent rotations that specify the orientation of a body or system.
- There is rotation of one part of the body with respect to the other part.

- Rotational degree of freedom happens only in diatomic gas.
- Diatomic molecules have rotational degrees of freedom in addition to translational degrees of freedom.
- It is possible in diatomic molecules as 2 atoms are connected together by a bond.So the rotation of one atomw.r.t to other atom.
- In diatomic there is translational in addition to that they have rotational degree of freedom also.
- For example: – Two oxygen atoms joined together by a bond. There are two perpendicular axes.
- There are 2 rotations possible along the two axes.
- They have 3 translational degrees of freedom and also 2 rotational degree of rotation.

- Therefore Rotational degree of freedom contributes a term to the energy that contains square of a rotational variable of motion.
- Rotational variable of motion comes from angular momentum ω.
- Linear velocity is v
_{x},v_{y},v_{z}. Whereas angular velocity is w_{x},w_{y},w_{z}. - E
_{R}(rotational) = (1/2)(I_{1}ω_{1})+(1/2)I_{2}ω_{2}. These are 3 rotationaldegrees of freedom along the 2 perpendicular axes.

- Linear velocity is v
- The total energy contribution due to the degrees of freedom for oxygen molecule.
- There will be 3 translational degree of freedom (1/2)m
_{x}v_{x}^{2},(1/2)m_{y}v_{y}^{2},(1/2)m_{z}v_{z}^{2}) - 2 rotational degree of freedom (1/2)I
_{1}^{2}ω_{1}^{2},(1/2)I_{2}^{2}ω_{2}^{2}

- There will be 3 translational degree of freedom (1/2)m

**Vibrational degree of freedom**

- Some molecules have a mode of vibration,i.e. its atoms oscillate along the inter-atomic axis like a one-dimensional oscillator.
- This vibration is observed in some molecules.
- For example:- CO atoms oscillate along the interatomic axis like a

one-dimensional oscillator.

- Consider two 2 atoms they vibrate along the inter-atomic axis.
- The vibrational energy terms contain square of vibrational variables of motion.
- Total vibrational energy term E
_{v}= (1/2) m (dy/dt)^{2}+ (1/2) ky^{2}where

(1/2) m(dy/dt)^{2}=Kinetic energy and (1/2)ky^{2} =Potential energy and k=force constant one-dimensional oscillator.

- The vibrational degree of freedom contributes 2 terms.

__Law of Equipartition of energy__

According to this law, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to (1/2)k_{B}T.

- Each translational degree of freedom contributes (1/2) k
_{B} - Each rotational degree of freedom contributes (1/2) k
_{B} - Each vibrational degree of freedom contributes 2x (1/2)k
_{B}

**Specific Heat Capacity for monoatomic gases**

- Monoatomic gases will only have translational degree of freedom.
- Maximum they can have is three translational degrees of freedom.
- Each degree of freedom will contribute (1/2) k
_{B} - Therefore 3 degrees of freedom will contribute (3/2) k
_{B} - By using law of equipartition of energy, the total internal energy of 1 mole of gas U=(3/2) k
_{B}TxN_{A}=(3/2) RT - Specific heat capacity at constant volume C
_{V}= dU/dT=(**3/2) R**(i) - For an ideal gas
**C**By using equation(i)_{P}-C_{V}=R,**C**_{P}=(5/2)R - Ratio of specific heats
**γ=C**_{P}/C_{V}=(5/3)

**Specific Heat of Diatomic gases (rigid)**

- A rigid diatomic gas means they will have translational as well as rotational degree of freedom but not vibrational.
- They are rigid oscillator.
- A rigid diatomic molecule will have 3 translational degrees of freedom and 2 rotational degrees of freedom. Total 5 degrees of freedom.
- By law of equipartition of energy, each degree of freedom will contribute (1/2) k
_{B} - Therefore 5 degree of freedom will contribute (5/2) k
_{B} - Therefore the total internal energy of 1 mole of gas, U=(5/2) k
_{B}TxN_{A}=(5/2)RT - Specific heat capacity at constant volume
**(C**=dU/dT=(_{V})**5/2)R** - Specific heat capacity at constant pressure of a rigid diatomic is given as
**C**_{P}=(7/2)R - Ratio of specific heats
**γ=C**=(7/5)_{P}/C_{V}

**Specific Heat of Diatomic gases (non-rigid)**

- A no-rigid diatomic gas has translational, rotational as well as vibrational degrees of freedom.
- There will be 3 translational degrees of freedom and 2 rotational degrees of freedom and 1 vibrational degree of freedom.
- Total contribution by translational= (1/2) k
_{B}T, rotational=2x (1/2)k_{B}T and vibrational =k_{B} - Total Internal energy for 1 mole =(5/2)k
_{B}T+k_{B}T = (7/2)k_{B}T= (**7/2)RT**. - C
_{V}=dU/dT = (**7/2) R**. - C
_{P}=C_{V}+R= (**9/2) R**. - γ= C
_{P}/C_{V}=(**9/7)**

**Specific Heat Capacity for polyatomic gases**

- Polyatomic gases will have 3 translational degree of freedom, 3 rotational degrees of freedom and ‘f’ number of vibrational modes.
- Total internal energy of 1 mole of gas =(3x(1/2)k
_{B}T + 3x(1/2)k_{B}T+ fk_{B}T)x N_{A}

((3/2) + (3/2) +f)RT = **(3+f) RT**.

- C
_{V}=dU/dT =**(3+f)R** - C
_{P}=C_{V}+R=**(4+f)R** - γ= C
_{P}/C_{V}=**(4+f)/(3+f)**

**Specific Heat Capacity for solids**

- Consider there are N atoms in a solid. Each atom can oscillate about its mean position.
- Therefore vibrational degree of freedom = k
_{B}T - In one-dimensional average energy=k
_{B}T, in three-dimensional average energy =3K_{B}T - Therefore total internal energy (U) of 1 mole of solid = 3K
_{B}TxN_{A}= 3RT - At constant pressure, ΔQ = ΔU + PΔV change in volume is very less in solids .Therefore ΔV = 0.
- =>ΔQ = ΔU
- C
_{V}=(dU/dT)_{v} - C
_{P}=(dQ/dT)_{V}as ΔQ = ΔU, Therefore C_{V}=dU/dT=3R - Therefore
**C**_{P}=C_{V}=3R

**Specific Heat Capacity of water**

- Consider water as solid,so it will have ‘N’ number of atoms.
- Therefore for each atom average energy =3k
_{B}T - No of molecules in H
_{2}O= 3 atoms. - Total internal energy U=3k
_{B}Tx3xN_{A}=9RT. - C
_{V}=C_{P}=9R.

__Conclusion on Specific heat__

- According to classical mechanics, the specific heat which is calculated based on degree of freedom should be independent of temperature.
- However T->0,degree of freedom becomes inefficient.
- This shows classical mechanics is not enough; as a result quantum mechanics came into play.
- According to quantum mechanics minimum non-zero energy is required for degree of freedom to come into play.
- Specific heats of all substances approach zero as T->0.

**Mean free path**

- Mean free path is the average distance between the two successive collisions.
- Inside the gas there are several molecules which are randomly moving and colliding with each other.
- The distance which a particular gas molecule travels without colliding is known as mean free path.

__Expression for mean free path__

- Consider each molecule of gas is a sphere of diameter (d).The average speed of each molecule is<v>.
- Suppose the molecule suffers collision with any other molecule within the distance (d). Any molecule which comes within the distance range of its diameter this molecule will have collision with that molecule.
- The volume within which a molecule suffer collision =<v>Δtπd
^{2}. - Let number of molecules per unit volume =n
- Therefore the total number of collisions in time Δt =<v>Δtπd
^{2}xn - Rate of collision =<v>Δtπd
^{2}xn/Δt=<v>πd^{2}n - Suppose time between collision τ =1/<v>πd
^{2}n - Average distance between collision = τ<v> = 1/πd
^{2} - 1/πd
^{2}n this value was modified and a factor was introduced. - Mean free path(l) =
**1/√2 π d**^{2}n

**Conclusion**: – Mean free path depends inversely on:

- Number density (number of molecules per unit volume)
- Size of the molecule.

Temperature inside the cylinder, T = 17°C =290 K

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

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

Molecular mass of nitrogen, M = 28.0 g = 28 × 10^{–3} kg

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

v_{rms}=√3RT/M

Where,

R is the universal gas constant = 8.314 J mole^{–1} K^{–1}

Therefore v_{rms}=√ (3×8.314×290)/ (28×10^{-3})

=508.26m/s

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

l= KT/√2xd^{2}xP

Where, k is the Boltzmann constant = 1.38 × 10^{–23} kgm^{2} s^{–2}K^{–1}

Therefore l= 1.38×10^{-23}x290/√2x 3.14x (2×10^{-10})^{2}x2.026×10^{5}

= 1.11 × 10^{–7} m

Collision frequency=v_{rms}/l

=508.26/1.11×10^{-7}

= 4.58 × 10^{9} s^{–1}

Collision time is given as:

T=d/v_{rms}

= 2×10^{-10}/508.26 = 3.93 × 10^{–13} s

Time taken between successive collisions:

T’=l/v_{rms} = 1.11×10^{-7}m/508.26m/s= 2.18 × 10^{–10} s

T’/T= 2.18 × 10^{–10} s/3.93×10^{-13} = 500

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

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