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Kinetic Theory - Specific Heat Capacity

Grade 11CBSEPhysics

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

🔑Concepts

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Specific Heat Capacity is defined as the amount of heat required to raise the temperature of a unit mass of a substance by 1 K1 \text{ K} or 1∘C1^\circ \text{C}.

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Molar Specific Heat Capacity at constant volume (CvC_v) is the heat required to raise the temperature of 1 mole1 \text{ mole} of gas by 1 K1 \text{ K} while keeping volume constant.

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Molar Specific Heat Capacity at constant pressure (CpC_p) is the heat required to raise the temperature of 1 mole1 \text{ mole} of gas by 1 K1 \text{ K} while keeping pressure constant.

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Degrees of Freedom (ff) refer to the number of independent ways in which a molecule can possess energy (translational, rotational, and vibrational).

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The Law of Equipartition of Energy states that the total energy is equally distributed among all possible degrees of freedom, and the energy associated with each degree of freedom per molecule is 12kBT\frac{1}{2} k_B T.

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Monatomic gases (e.g., Helium) have f=3f = 3 (only translational).

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Diatomic gases (e.g., O2O_2, N2N_2) have f=5f = 5 at room temperature (3 translational + 2 rotational).

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Polyatomic non-linear gases have f=6f = 6 (3 translational + 3 rotational).

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Mayer's Relation states that for an ideal gas, the difference between CpC_p and CvC_v is equal to the Universal Gas Constant (RR).

📐Formulae

U=f2nRTU = \frac{f}{2} nRT

Cv=dUdT=f2RC_v = \frac{dU}{dT} = \frac{f}{2} R

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

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

Cv=Rγ−1C_v = \frac{R}{\gamma - 1}

💡Examples

Problem 1:

Calculate the value of CvC_v, CpC_p, and γ\gamma for a diatomic gas like Hydrogen at room temperature.

Solution:

For a diatomic gas at room temperature, the number of degrees of freedom is f=5f = 5. Using the formulae: Cv=52RC_v = \frac{5}{2} R Cp=Cv+R=52R+R=72RC_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R γ=CpCv=7/2R5/2R=75=1.4\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4

Explanation:

Since the gas is diatomic and at room temperature, we ignore vibrational modes and consider only 3 translational and 2 rotational degrees of freedom, giving f=5f=5.

Problem 2:

If the molar specific heat of a gas at constant volume is 20.8 J mol−1 K−120.8 \text{ J mol}^{-1} \text{ K}^{-1} and the gas constant R=8.3 J mol−1 K−1R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}, calculate the molar specific heat at constant pressure.

Solution:

Using Mayer's relation: Cp=Cv+RC_p = C_v + R Given values: 20.8+8.329.1\begin{array}{r} 20.8 \\ + 8.3 \\ \hline 29.1 \end{array} Therefore, Cp=29.1 J mol−1 K−1C_p = 29.1 \text{ J mol}^{-1} \text{ K}^{-1}.

Explanation:

Mayer's formula Cp−Cv=RC_p - C_v = R allows us to find one specific heat if the other and the gas constant are known.

Specific Heat Capacity Class 11 Notes & Examples