QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 11.2NCERT Class 11 · Physics · Chapter 11

First Law of Thermodynamics & Specific Heat Capacities

The First Law of Thermodynamics is the macroscopic generalization of the law of conservation of energy applied to thermodynamic systems. It establishes that heat and mechanical work are mutually convertible forms of energy, and energy can neither be created nor destroyed.

1. Mathematical Formulation of the First Law

When a quantity of heat ΔQ is supplied to a thermodynamic system, it is utilized in two distinct ways:

  1. To increase the internal energy of the system by ΔU.
  2. To perform external work ΔW on the surroundings.
ΔQ = ΔU + ΔW   ⇒   ΔU = ΔQ − ΔW

In differential form for an infinitesimal quasi-static process:

dQ = dU + dW = dU + P dV
Key Deduction: Although dQ and dW individually depend on the path taken between states 1 and 2, their difference dU = dQ − dW is entirely independent of path and depends exclusively on initial state 1 and final state 2.

2. Molar Specific Heat Capacities of Gases

For solids and liquids, thermal expansion upon heating is negligible, so specific heat capacity has a unique value. For gases, however, the amount of heat required to raise the temperature of 1 mole by 1 K depends heavily on how the volume and pressure are allowed to change during heating:

C = (1 / n) · (dQ / dT)

Depending on the process, C can range from −∞ to +∞. Two principal molar specific heats are defined:

  • Molar Heat Capacity at Constant Volume (Cv): The amount of heat required to raise the temperature of 1 mole of gas by 1 K while keeping volume constant. Since dV = 0, dW = 0, we have:
    Cv = (1 / n) (dU / dT)   ⇒   dU = n Cv dT
  • Molar Heat Capacity at Constant Pressure (Cp): The amount of heat required to raise the temperature of 1 mole of gas by 1 K at constant pressure. Here, heat supplied increases internal energy AND does expansion work against constant pressure. Thus, Cp > Cv always.

3. Derivation of Mayer's Relation (Cp − Cv = R)

Consider 1 mole (n = 1) of an ideal gas governed by the equation of state P V = R T:

  1. At constant volume: dQv = dU = Cv dT.
  2. At constant pressure: dQp = Cp dT. From the first law:
    dQp = dU + P dV = Cv dT + P dV
  3. Differentiating the ideal gas equation P V = R T at constant pressure P:
    P dV = R dT
  4. Substituting P dV into the first law equation:
    Cp dT = Cv dT + R dT
  5. Dividing throughout by dT yields Mayer's equation:
Cp − Cv = R
Important Note on Specific Heats per Gram: If specific heat capacities are expressed per unit mass (cp and cv in J/(kg·K)), then:
cp − cv = R / M = r
where M is the molar mass of the gas and r is the specific gas constant.

4. Degrees of Freedom (f) & Law of Equipartition of Energy

The degrees of freedom (f) of a molecule is the total number of independent coordinates or quadratic terms required to specify its position and configuration in space. According to the Law of Equipartition of Energy, the average kinetic energy associated with each degree of freedom per molecule in thermal equilibrium at temperature T is (1/2) kB T, or (1/2) R T per mole.

Total internal energy of 1 mole of gas: U = (f / 2) R T.

Differentiating with respect to T:

Cv = (dU / dT) = (f / 2) R

Cp = Cv + R = ((f + 2) / 2) R

γ = Cp / Cv = 1 + 2 / f

5. Comparison Table for Monoatomic, Diatomic & Polyatomic Gases

Atomicity of GasExample GasesDegrees of Freedom (f)CvCpγ = Cp/Cv
Monoatomic He, Ne, Ar, Kr 3 (3 translational) (3/2) R (5/2) R 5/3 ≈ 1.67
Diatomic (Rigid) H2, O2, N2, CO (room temp) 5 (3 trans + 2 rot) (5/2) R (7/2) R 7/5 = 1.40
Diatomic (Vibrating) Cl2, Br2 (high temp) 7 (3 trans + 2 rot + 2 vib) (7/2) R (9/2) R 9/7 ≈ 1.29
Polyatomic (Non-linear) H2O, CH4, NH3 6 (3 trans + 3 rot) 3 R 4 R 4/3 ≈ 1.33
Polyatomic (Linear) CO2, C2H2 5 (3 trans + 2 rot) + vib (5/2) R + vib (7/2) R + vib Depends on T

6. Mixture of Non-Reacting Ideal Gases

When n1 moles of a gas with molar heat capacity Cv1 are mixed with n2 moles of a gas with molar heat capacity Cv2:

Cv,mix = (n1 Cv1 + n2 Cv2) / (n1 + n2)

Cp,mix = Cv,mix + R = (n1 Cp1 + n2 Cp2) / (n1 + n2)

γmix = Cp,mix / Cv,mix = (n1 Cp1 + n2 Cp2) / (n1 Cv1 + n2 Cv2)
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