DEFINITION: The total number of coordinates required to describe the position of a particle in space is called the degree of freedom. Or the ways in which the atoms in a molecule can move (translation, rotation, vibration) independently.  For example, if a particle or an atom is present only in the x-direction, then this particle has only one degree of freedom. Similarly, if a particle is present in a plane (i.e., in direction), then this particle has two degrees of freedom (two coordinates x, y will be required to describe its position). If a particle is present or moves in space, then 3 coordinates are required to describe the position of a particle. Thus, for any atom, molecule, or any body, the total degrees of freedom are 3. And if this substance or body has N atoms, then the total degree of freedom is 3N. Thus, 3N degrees of freedom = translational + rotational + vibrational motions. Or the total degrees of freedom (3N) = dof (trans) + dof (vib) + dof (rot). It is important in the calculation of internal energy (E = (f/2)nRT, entropy (ΔS = nCv ln(T₂/T₁), and heat capacity (Cv = (f/2)R, (Cp = (f/2) + 1)R     

Translational degrees of freedom

The translational motion is due to the change in the centre of gravity (the point in the substance in which mass is constant) of a molecule. If there is a change in the centre of gravity of a molecule, that molecule will show translational motion. For all types of molecules (monoatomic, diatomic, or polyatomic), there are 3 degrees of translational motion. Thus,

3N-3 = dof (vib) + dof (rot).

Rotational degrees of freedom

The rotational degree of freedom describes the ways in which a molecule rotates around the centre of gravity so that its orientation changes.

• For monoatomic molecules, there is no change in orientation during rotation, so the dof (rot) = 0.
• For diatomic or polyatomic linear molecules, there are only 2 coordinates, i.e., y and z axes, in which orientation is changed, and on the x-axis, there is no change in the centre of gravity. Thus, it is concluded that in diatomic or polyatomic linear molecules, there are only two rotational degrees of freedom.
• For polyatomic non-linear molecules, there are 3 rotational degrees of freedom (i.e., x,y,z).

Vibrational degrees of freedom

The number of unique ways in which the atoms within the molecules may move relative to one another is called vibrational degree of freedom.

• For monoatomic gaseous molecules, as there is no other atom related to the first, the vibrational degree of freedom = 0
• For diatomic molecules or linear polyatomic molecules, the vibrational degree of freedom = 3N-5
• For polyatomic nonlinear molecules, the vibrational degrees of freedom = 3N-6

For example, CO₂ (linear, N=3,  9 DOF total) 3 transl + 2 rotational + (3×3−5) = 4 modes 

H₂O (nonlinear, N=3, 9 DOF total): 3 transl + 3 rotational + (3×3−6)=3 vibrational modes 

Energy associated with each degree of freedom

According to the law of equipartition of energy, the energy associated with each degree of freedom is (½)K_BT (or (½) RT per mole). Thus, for all types of degrees of freedom, the total energy can be calculated as:

Translational degrees of freedom. In translational motions, three degrees of freedom exist. Therefore, the total energy is 3 x (½)K_BT = 1.5 K_BT.

 Rotational degrees of freedom. In rotational motions, three degrees of freedom exist for polyatomic non-linear. Therefore, the total energy is 3 x (½)K_BT = 1.5 K_BT. Whereas, two degrees of freedom exist for diatomic or polyatomic linear Therefore, the total energy is 2 x (½)K_BT = 1K_BT.

Vibrational degrees of freedom. In vibrational degrees of freedom, 1k_BT per mode exists (½k_BT kinetic + ½k_BT potential), but only at high temperature. At room temperature, most molecules only exhibit translational and rotational DOF, while vibrational DOF  are frozen out” at room temperature and become important only at ~10³–10⁴ K.

The values of  K_B  is 1.38×10⁻²³J/K, Room Temperature is (~300 K), R is 8.314J/mol⋅K. In the calculation of internal energy, only translational and rotational modes contribute at all temperatures. At high temperature, vibration modes also contribute; the electronic transition takes place at too high  temperature. Therefore, we can not consider the electronic transition in the total internal energy of any system. The energy required to a system for translational motion is ΔE_tran = 10^-37 J/molecule, the value of K_BT = 1.38 x 10^-23J/K x 298K  = 1.38 x 10^-21J. (K_B is the Boltzmann constant; this constant bridges temperature and energy and converts temperature into energy units. If we directly compare an energy to T, the units would not even match). Thus, ΔE_tran <<<  K_BT, and translational motion does not depend on the temperature. The ΔE_rot = 10^-23 J/molecule, which is also less than the energy of room temperature, i.e., 1.38 x 10^-21J.  Thus, this motion also does not depend upon the temperature and takes place at all temperatures. For the vibration motions ( or the energy gap between the vibrational levels) = 10^-21J. This energy is equal to the energy of room temperature. That is why vibrational motion does not contribute to internal energy. However, on increasing the T, the value of K_BT increases from 1.38 x 10^-21J and ΔE vibrations become ΔE_vib <<<  K_BT and vibrational levels now contribute to the total energy.   

In thermodynamics, γ is known as the heat “capacity ratio”, “adiabatic index”, or “specific heat ratio”, defined as: γ =Cp/Cv. On putting the values of  Cp and Cv, γ = 1 + (2/f) or conversely, f= 2/​γ−1. For example, for a monatomic gas (f = 3): γ=1+(2/3)=5/3≈1.67. Similarly, for a diatomic gas at room temperature (f ≈ 5): γ=1+(2/5=1.4.

The γ reflects how many independent modes of motion, like translation, rotation, and vibrations, contribute to a molecule’s energy storage. Because each degree of freedom adds energy (½ k_BT or 1/2RT). When classical equipartition applies, γ tells us which modes are “active” (contributing) versus “frozen out” (inactive due to quantum effects). Thus, when γ = 5/3 =1.67. Only translational modes are active; when γ = 7/5 =1.4, it includes translational + rotational modes. And when γ = 8/6 =1.33, translation + rotation, vibrations are still frozen. As temperature increases, vibrational degrees begin to activate, f values rise, and γ falls, indicating access to additional molecular energy pathways.

 Monatomic Gases (e.g., He, Ar)

• Only 3 translational DOF → f=3
• Per molecule: Ɛ=3 x (½)k_BT = 3/2 k_BT
• Molar internal energy: E= (f/2)RT = 3x (½)RT = (3/2) RT = 1.5RT
• C_V= ΔE / ΔT or (f/2) R = 3/2R = 1.5 R
•  Cp is Cv + R = (f/2) R + R =( f/2 +1)R = (3/2+1)R = (5/2)R = 2.5R
• γ = 1+( 2/f) = 5/3 = 1.67.

Diatomic Gases (e.g., N₂, O₂)

• Translational 3 + Rotational 2
• Vibrational = 3N-5 = 3X2-5 = 1 total = 6
• Per molecule: Ɛ =  3/2 k_BT + 2/2 k_BT + k_BT  = 7/2 k_BT = 3.5 k_BT
• Molar energy: E = 3.5RT
• Cv = (f/2) R or ΔE / ΔT  =  3.5R
•  Cp is Cv + R = (f/2) R + R =( f/2 +1)R = (7/2+1)R = 4.5R
• γ = 1+( 2/f) = =1.33  

Polyatomic Gases (e.g., CH₄)

• All modes active 3N= 3×5 = 15 dof: translational 3, rotational 3, vibrational (kinetic + potential) 3N-6 = 3×5 -6 = 15-6 = 9 modes of vibrations.
• Total energy per molecule: Ɛ =  3/2 k_BT + 3/2 k_BT + 9k_BT  = 12k_BT
• Molar energy E=12RT
•  C_V=15/2R = 7.5R or or ΔE / ΔT = 12R ( ΔE / ΔT is the general thermodynamic definition of C_V, whilef/2 R is a special case that follows from the equipartition for an ideal gas with constant f)   
• Cp = Cv + R = (f/2) R + R =( (15/2) +1)R = 8.5R.
• γ = 1+( 2/f) = 1.133.

EAT CAPACITY

Heat capacity is the amount of heat energy required to raise the temperature of a substance by a unit amount, i.e., the amount of heat required to raise the temperature of a substance by 1 ^oC or 1 K. It is a measure of the ability of a substance to absorb and store heat energy.  Mathematically, it is written as:    C = Q / ΔT

where Q is the amount of heat energy added to the substance, and ΔT is the resulting temperature change.

The unit of heat capacity is J/K or cal/°C. The value of heat capacity gives information about heat transfer, phase transitions, and any change in chemical reactions.

For gases, normally gases are studied under two conditions: i) isothermally, when T or ΔT = 0, and ii) adiabatically, when q = 0. So, the heat capacities in both conditions can be calculated as, Isothermally → c = q/mx0 = q/0 = ∞ (very large).

Adiabatic → c = 0/mxΔT = 0. Thus, for gases, the heat capacity is in any form i.e. 0→∞  

In thermodynamics and kinetic theory, the heat capacity of a gas depends on its degrees of freedom (denoted as f). Degrees of freedom represent the independent ways in which a molecule can store energy (translational, rotational, vibrational, etc.). There are two main categories of heat capacities: For 1 mole:

at constant volume, C_V = (f/2)R or ΔE/ΔT

at constant pressure, C_P = C_V + R =  (f/2)R+R = (f/2+1)R or ΔH/ΔT  

Cp – Cv = R

Specific heat capacity (c)

The amount of heat energy required to raise the temperature of a unit mass of a substance by a unit amount. Or the amount of heat energy required to raise the temperature of a unit mass of a substance by 1 ^oC or 1 K. Mathematically, it is written as,

c = q /mΔT, where ‘m’ is the mass of the substance. In terms of heat, it can be written as q=mcΔT, ‘c’ is the property or ability of a material to store the heat, while q is the amount of heat absorbed or liberated from the substance. Therefore, in terms of heat capacities, we always calculate the values of q. For example, the heat specific heat capacity of H₂O is 4.184 J/g°C and that of Fe is 4.184 J/g°C .

Molar heat capacity (C)

The amount of heat energy required to raise the temperature of a mole of a substance by a unit amount. Or the amount of heat energy required to raise the temperature of a mole of a substance by 1 ^oC or 1 K. Mathematically, it is written as, C= Q /nΔT, where ‘n’ is the number of moles of the substance. At constant volume and constant pressure, the molar heat capacities can be written as C_v= Q/nΔT and C_p= Q /nΔT, respectively.  The amount of heat absorbed at constant volume and constant pressure can be written as Q_p=nC_pΔT and Q_v=nC_vΔT.  For example, the molar heat capacities of H₂O are 75 J/mol.K, and that of Fe is 25 J/mol.K.

The specific heat capacities and molar heat capacities can be correlated as C_v =Mc_v and C_p =Mc_p. Where ‘C_v’ is the molar heat capacity at constant volume, c_v is the specific heat capacity at constant volume, ‘ C_p is the molar heat capacity at constant pressure, and c_p is the specific heat capacity at constant pressure, and ‘M’ is the molar mass of the compound. For any type of gas, the relation between molar heat capacities (C_p and C_v) can be written as C_p – C_v = R where R is the general gas constant.

Derivation of heat capacity

The heat capacity can be derived from the definition of internal energy (E) and the first law of thermodynamics. According to the first law of thermodynamics, energy cannot be created or destroyed, only converted from one form to another. Mathematically, it is written as, ΔE = Q + W (or ΔE = Q – W). at constant volume, W = 0 → ΔE = Q

The heat capacity at constant volume, C_v = (ΔE / ΔT)_v, where ΔE is the internal energy of a system. Similarly, the heat capacity at constant pressure, C_p = (ΔH / ΔT)_p, where ΔH is the enthalpy of a system at constant pressure. The values of heat capacities in terms of ‘Q’ can be derived as:

C=ΔQ/ΔT or ΔQ= CΔT  (common form)

ΔQ=ΔE+ΔW (according to the first law of thermodynamics)

At constant volume: ΔW=0, so, ΔQ=ΔE⇒C_V=(ΔQ/ΔT)_V=(ΔE/ΔT)_V​

At constant pressure: ΔQ=ΔE+PΔV⇒C_P=(ΔQ/ΔT)_P = (ΔE/ΔT)_P + P(ΔV/ΔT)

Heat capacity at constant volume(C_V)and constant pressure (C_P) depends on whether the process is carried out at constant volume or pressure.

The heat capacities can be calculated from any of the following formulas:

⇒ From the degree of freedom: C_v = (f/2)R, C_p = (f/2+1)R

⇒ From the internal energy: C_v = (ΔE/ΔT), C_p = (ΔH/ΔT)

⇒ From the heat of the system: C_v = Q/ΔT ,C_p = Q/ΔT (common form), C_v = Q/nΔT (molar form),  c_v = Q/mΔT (specific form).

IMPORTANT NUMERICALS RELATED TO HEAT CAPACITY

Q: How much heat is required to raise the temperature of 100 g of water from 25°C to 75°C?. (Given: Specific heat of water c = 4.18 J/g\K).

Soln: As in the question, heat is to be determined, and the value of heat capacity is provided. Thus, the value of heat is calculated by using the heat and heat capacity relationship, i.e., c = Q/mΔT ⇒ Q = mcΔT ⇒ 100×4.18×(75−25) =  100×4.18×50=20900J.     

Q: A metal object of mass 200 g absorbs 1000 J of heat and its temperature rises by 25°C. Find its specific heat.

Soln: As in the question, heat capacity is to be determined, and the value of heat is provided. Thus, the value of heat capacity is calculated by using the heat and heat capacity relationship i.e. c = Q/mΔT ⇒ c=1000/200×25 = 1000/5000=0.2 J/g\k

 Q: How much heat is required to raise the temperature of 2moles of oxygen gas from 300 K to 350 K at constant volume?.

Soln: As in the question, heat is to be determined, and the information about gas is provided, but no information about heat capacity is provided. (“Remember” whenever the name or molar mass of a gas is provided, then the value of Cv has to be calculated by the degree of freedom). Thus, the value of heat capacity is calculated by using the degree of freedom of the O₂ molecule. F =3N = 3X2=6  (As temperature is just above the room temperature i.e., not the high temperature, therefore the vibrational modes are considered to be frozen out, therefore only active degrees of freedom are 5⇒ f =5/2).  Cv = f/2R = q=nCVΔT ⇒ q=2×5/2×8.314×50=5×8.314×50=2078.5J.

Q: Calculate the molar heat capacity at constant pressure if CV=20.8 J/mol for a gas.

Soln: In the question, heat capacity at constant volume is provided, and heat capacity at constant pressure is to be determined. Thus, the following relation is used: C_P-C_V=R⇒ C_P=C_V+R ⇒ C_P=20.8+8.314=29.114 J/mol.K

Q: A gas sample exhibits a change in internal energy of Δu=1500  J/kg over a ΔT=25 K. The molar mass is M=20.0g/mol. Determine the molar heat capacity CV​ in J·mol⁻¹·K⁻¹.

Soln: In the question, the molar heat capacity at constant volume is to be determined, and internal energy is provided. Thus, by using the relation between internal energy and heat capacity, the molar heat capacity is to be calculated. c_v = (ΔE/ΔT),  c_v = 1500/25 = 60 J/Kg.K^-1. Now, this heat capacity is converted into molar heat capacity as: CV=Mcv ⇒ CV = 0.02kg/mol x 60J/Kg.K^-1. =0.02X60 Jmol⁻¹·K⁻¹.=1.2 Jmol⁻¹·K⁻¹

Q: In a rigid calorimeter, 0.300 mol of a substance is heated at constant volume. A heat input of Q=1800  J raises the temperature from 280 K to 330 K. Calculate the molar heat capacity CV​.

Soln: In the question, the molar heat capacity at constant volume is to be determined, and the values of heat is provided.Thus, by using the relation between heat absorbed and heat capacity, the molar heat capacity is to be calculated. CV = (ΔQ/nΔT),  CV = 1800/0.3×50 = 1800/15=120Jmol⁻¹·K⁻¹.