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Kinetic Theory - Kinetic Interpretation of Temperature

Grade 11CBSEPhysics

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

🔑Concepts

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The Kinetic Theory of Gases relates macroscopic properties like pressure and temperature to microscopic properties like molecular speed and mass.

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Pressure exerted by an ideal gas is given by P=13ρv2‾P = \frac{1}{3} \rho \overline{v^2}, where ρ\rho is the density and v2‾\overline{v^2} is the mean square speed of molecules.

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The average translational kinetic energy of a molecule is directly proportional to the absolute temperature TT of the gas, expressed as E∝TE \propto T.

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Temperature is a macroscopic manifestation of the average molecular kinetic energy. At absolute zero (T=0 KT = 0 \text{ K}), the translational kinetic energy of molecules becomes zero.

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The Boltzmann constant kBk_B serves as the bridge between the macroscopic temperature and microscopic energy, where kB=RNA≈1.38×10−23 J/Kk_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23} \text{ J/K}.

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The root mean square (RMS) speed vrmsv_{rms} is the square root of the average of the squares of the speeds of the molecules. It depends on the temperature and the molar mass of the gas: vrms∝TMv_{rms} \propto \sqrt{\frac{T}{M}}.

📐Formulae

P=13Nmv2‾VP = \frac{1}{3} \frac{Nm\overline{v^2}}{V}

K.E.ˉ=32kBT\bar{K.E.} = \frac{3}{2} k_B T

Etotal=32nRTE_{total} = \frac{3}{2} nRT

vrms=3kBTm=3RTMv_{rms} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}

kB=RNAk_B = \frac{R}{N_A}

💡Examples

Problem 1:

Calculate the average kinetic energy of a molecule of an ideal gas at 27∘C27^\circ C. (Given kB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23} \text{ J/K})

Solution:

  1. Convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300 \text{ K}.
  2. Use the formula for average kinetic energy per molecule: E=32kBTE = \frac{3}{2} k_B T.
  3. Substitute the values: E=32×(1.38×10−23)×300E = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300.
  4. E=1.5×1.38×10−23×300=6.21×10−21 JE = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J}.

Explanation:

The kinetic interpretation of temperature states that the average kinetic energy depends only on the absolute temperature, not on the nature of the gas.

Problem 2:

At what temperature will the root mean square speed of Oxygen molecules (O2O_2) be twice its value at 0∘C0^\circ C?

Solution:

  1. Let the initial temperature T1=273 KT_1 = 273 \text{ K} and final temperature be T2T_2.
  2. The RMS speed is given by vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}, which implies vrms∝Tv_{rms} \propto \sqrt{T}.
  3. We are given v2=2v1v_2 = 2v_1. Therefore, v2v1=T2T1\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}.
  4. 2=T2273  ⟹  4=T22732 = \sqrt{\frac{T_2}{273}} \implies 4 = \frac{T_2}{273}.
  5. T2=4×273=1092 KT_2 = 4 \times 273 = 1092 \text{ K}.
  6. In Celsius: t2=1092−273=819∘Ct_2 = 1092 - 273 = 819^\circ C.

Explanation:

To double the vrmsv_{rms} of a gas, the absolute temperature must be increased by a factor of four (222^2) because speed is proportional to the square root of temperature.

Kinetic Interpretation of Temperature Revision - Class 11 Physics CBSE