krit.club logo

States of Matter: Gases and Liquids - Kinetic Molecular Theory of Gases

Grade 11ICSEChemistry

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

🔑Concepts

•

The Kinetic Molecular Theory (KMT) explains the macroscopic properties of gases like pressure and temperature based on their molecular composition and motion.

•

Gases consist of large numbers of tiny particles (atoms or molecules) that are in constant, rapid, and random motion.

•

The actual volume of the gas molecules is negligible compared to the total volume of the container, making gases highly compressible.

•

There are no intermolecular forces of attraction or repulsion between gas particles in an ideal gas.

•

Collisions between gas molecules and between molecules and the container walls are perfectly elastic; there is no net loss of kinetic energy during collisions.

•

The pressure of a gas is caused by the collisions of the gas molecules with the walls of the container.

•

The average kinetic energy of gas molecules is directly proportional to the absolute temperature (TT) of the gas, expressed as K.E.∝TK.E. \propto T.

•

Real gases deviate from ideal behavior at high pressures and low temperatures because molecular volume and intermolecular forces become significant.

📐Formulae

PV=nRTPV = nRT

PV=13mnu2PV = \frac{1}{3} m n u^{2}

urms=3RTMu_{rms} = \sqrt{\frac{3RT}{M}}

uavg=8RTπMu_{avg} = \sqrt{\frac{8RT}{\pi M}}

ump=2RTMu_{mp} = \sqrt{\frac{2RT}{M}}

Average K.E. (per mole)=32RT\text{Average } K.E. \text{ (per mole)} = \frac{3}{2} RT

Average K.E. (per molecule)=32kBT\text{Average } K.E. \text{ (per molecule)} = \frac{3}{2} k_B T

💡Examples

Problem 1:

Calculate the Root Mean Square (RMS) speed of O2O_2 molecules at 27∘C27^{\circ}C. (Given: R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}, Molar mass of O2=32 g mol−1O_2 = 32 \text{ g mol}^{-1})

Solution:

  1. Convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300 \text{ K}.
  2. Convert molar mass to kg: M=32×10−3 kg mol−1M = 32 \times 10^{-3} \text{ kg mol}^{-1}.
  3. Use the formula urms=3RTMu_{rms} = \sqrt{\frac{3RT}{M}}.
  4. urms=3×8.314×30032×10−3≈483.6 m/su_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{32 \times 10^{-3}}} \approx 483.6 \text{ m/s}.

Explanation:

The RMS speed represents the square root of the average of the squares of the speeds of all molecules, providing a measure of the average speed adjusted for kinetic energy calculations.

Problem 2:

Find the total kinetic energy of 2 moles2 \text{ moles} of an ideal gas at 25∘C25^{\circ}C.

Solution:

  1. Temperature T=25+273=298 KT = 25 + 273 = 298 \text{ K}.
  2. n=2 molesn = 2 \text{ moles}.
  3. Total K.E.=n×32RT=2×32×8.314×298K.E. = n \times \frac{3}{2} RT = 2 \times \frac{3}{2} \times 8.314 \times 298.
  4. K.E.=3×8.314×298=7432.7 JK.E. = 3 \times 8.314 \times 298 = 7432.7 \text{ J} or 7.43 kJ7.43 \text{ kJ}.

Explanation:

According to KMT, the kinetic energy of a gas depends only on the number of moles and the absolute temperature, regardless of the chemical identity of the gas.

Kinetic Molecular Theory of Gases Revision - Class 11 Chemistry ICSE