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Behavior of Perfect Gas and Kinetic Theory - Kinetic Theory of Gases (Pressure, Temperature and RMS Speed)

Grade 11ICSEPhysics

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

🔑Concepts

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Postulates of Kinetic Theory: Gases consist of large numbers of identical, tiny, spherical particles (atoms or molecules). They are in constant, random motion, and collisions between them and the walls of the container are perfectly elastic.

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Pressure of an Ideal Gas: Pressure is exerted due to the change in momentum of molecules colliding with the container walls. It is expressed as P=13ρvrms2P = \frac{1}{3} \rho v_{rms}^2.

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RMS Speed (vrmsv_{rms}): The square root of the mean of the squares of the speeds of the individual gas molecules. It represents the speed of a molecule possessing the average kinetic energy.

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Kinetic Interpretation of Temperature: The absolute temperature (TT) of a gas is directly proportional to the average translational kinetic energy per molecule of the gas. At 0 K0\ K, molecular motion ceases.

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Boltzmann Constant (kBk_B): The gas constant per molecule, defined as kB=RNAk_B = \frac{R}{N_A}, where RR is the universal gas constant and NAN_A is Avogadro's number (kB≈1.38×10−23 J/Kk_B \approx 1.38 \times 10^{-23}\ J/K).

📐Formulae

P=13MVvrms2=13ρvrms2P = \frac{1}{3} \frac{M}{V} v_{rms}^2 = \frac{1}{3} \rho v_{rms}^2

vrms=3PVM=3RTM=3kBTmv_{rms} = \sqrt{\frac{3PV}{M}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_B T}{m}}

Average Kinetic Energy per molecule=32kBT\text{Average Kinetic Energy per molecule} = \frac{3}{2} k_B T

Total Kinetic Energy of n moles=32nRT\text{Total Kinetic Energy of } n \text{ moles} = \frac{3}{2} nRT

Relation between vrms and density: vrms=3Pρ\text{Relation between } v_{rms} \text{ and density: } v_{rms} = \sqrt{\frac{3P}{\rho}}

💡Examples

Problem 1:

Calculate the root mean square speed of H2H_2 molecules at 27∘C27^{\circ}C. (Given: R=8.314 J mol−1K−1R = 8.314\ J\ mol^{-1}K^{-1} and Molar mass of H2=2×10−3 kg/molH_2 = 2 \times 10^{-3}\ kg/mol)

Solution:

Given T=27+273=300 KT = 27 + 273 = 300\ K, M=2×10−3 kg/molM = 2 \times 10^{-3}\ kg/mol. Using vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}: vrms=3×8.314×3002×10−3v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{2 \times 10^{-3}}} vrms=7482.62×10−3=3741300≈1934.25 m/sv_{rms} = \sqrt{\frac{7482.6}{2 \times 10^{-3}}} = \sqrt{3741300} \approx 1934.25\ m/s.

Explanation:

The temperature must always be converted to Kelvin. The molar mass must be in kg/molkg/mol to ensure the velocity is in m/sm/s.

Problem 2:

At what temperature will the RMS speed of oxygen molecules (O2O_2) be double their RMS speed at 27∘C27^{\circ}C?

Solution:

We know vrms∝Tv_{rms} \propto \sqrt{T}. Let v1v_1 be speed at T1=300 KT_1 = 300\ K and v2=2v1v_2 = 2v_1 at T2T_2. v2v1=T2T1  ⟹  2=T2300\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \implies 2 = \sqrt{\frac{T_2}{300}} Squaring both sides: 4=T2300  ⟹  T2=1200 K4 = \frac{T_2}{300} \implies T_2 = 1200\ K. In Celsius: T2=1200−273=927∘CT_2 = 1200 - 273 = 927^{\circ}C.

Explanation:

Since vrmsv_{rms} is proportional to the square root of the absolute temperature, doubling the speed requires the absolute temperature to increase by a factor of four.

Kinetic Theory of Gases (Pressure, Temperature and RMS Speed) Revision - Class 11 Physics ICSE