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Behavior of Perfect Gas and Kinetic Theory - Specific Heat Capacities of Gases

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Molar Specific Heat at Constant Volume (CvC_v): It is defined as the amount of heat required to raise the temperature of one mole of an ideal gas by 1 K1 \text{ K} while keeping its volume constant. Formula: Cv=(dUdT)C_v = \left( \frac{dU}{dT} \right).

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Molar Specific Heat at Constant Pressure (CpC_p): It is the heat required to raise the temperature of one mole of an ideal gas by 1 K1 \text{ K} while keeping its pressure constant. CpC_p is always greater than CvC_v because heat supplied at constant pressure is used to increase internal energy AND perform work against external pressure.

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Mayer's Relation: For an ideal gas, the relationship between the two specific heats is given by Cp−Cv=RC_p - C_v = R, where RR is the Universal Gas Constant (8.314 J mol−1K−18.314 \text{ J mol}^{-1} \text{K}^{-1}).

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Degrees of Freedom (ff): The number of independent ways in which a molecule can possess energy. For monoatomic gases, f=3f=3; for diatomic gases (at room temperature), f=5f=5; for polyatomic non-linear gases, f=6f=6.

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Law of Equipartition of Energy: This law states that for any dynamic system in thermal equilibrium, the total energy is divided equally among all the degrees of freedom, and the energy associated with each molecule per degree of freedom is 12kBT\frac{1}{2} k_B T.

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Adiabatic Index (γ\gamma): The ratio of specific heats γ=CpCv\gamma = \frac{C_p}{C_v} determines the atomicity of the gas and its behavior during adiabatic processes.

📐Formulae

Cp−Cv=RC_p - C_v = R

Cv=f2RC_v = \frac{f}{2}R

Cp=(f2+1)RC_p = \left( \frac{f}{2} + 1 \right)R

γ=CpCv=1+2f\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}

ΔU=nCvΔT\Delta U = n C_v \Delta T

Cv,mix=n1Cv1+n2Cv2n1+n2C_{v, mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}

💡Examples

Problem 1:

Calculate the values of CvC_v, CpC_p, and γ\gamma for Oxygen (O2O_2) gas at room temperature.

Solution:

Oxygen is a diatomic gas, so its degrees of freedom f=5f = 5. Using the formulae:

  1. Cv=f2R=52RC_v = \frac{f}{2}R = \frac{5}{2}R
  2. Cp=Cv+R=52R+R=72RC_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R
  3. γ=CpCv=7/25/2=1.4\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = 1.4

Explanation:

At room temperature, diatomic molecules like O2O_2 have 3 translational and 2 rotational degrees of freedom, totaling f=5f=5. The ratio γ=1.4\gamma = 1.4 is characteristic of diatomic gases.

Problem 2:

If the ratio of specific heats γ\gamma for a gas is 1.671.67, identify the atomicity of the gas and calculate its CvC_v in terms of RR.

Solution:

Given γ=1.67\gamma = 1.67, which is approximately 53\frac{5}{3}. We know γ=1+2f\gamma = 1 + \frac{2}{f}. 53=1+2f  ⟹  23=2f  ⟹  f=3\frac{5}{3} = 1 + \frac{2}{f} \implies \frac{2}{3} = \frac{2}{f} \implies f = 3. Since f=3f=3, the gas is monoatomic. Cv=f2R=32RC_v = \frac{f}{2}R = \frac{3}{2}R.

Explanation:

A value of γ≈1.67\gamma \approx 1.67 indicates a monoatomic gas (like Helium or Argon) where only translational motion contributes to the internal energy.

Specific Heat Capacities of Gases - Revision Notes & Key Formulas | ICSE Class 11 Physics