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The Particulate Nature of Matter - Gas Laws

Grade 12IBPhysics

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

πŸ”‘Concepts

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An Ideal Gas is a theoretical model where particles are considered point masses with negligible volume and no intermolecular forces except during perfectly elastic collisions.

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The Kinetic Molecular Theory assumes: 1. Large number of particles in random motion. 2. Identical particles. 3. Negligible volume of particles compared to the gas volume. 4. Perfectly elastic collisions. 5. No forces between particles except during collisions.

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The Absolute Temperature TT of an ideal gas is a measure of the average random kinetic energy of its particles: Ek∝TE_k \propto T.

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Boyle's Law: For a fixed mass of gas at constant temperature, P∝1VP \propto \frac{1}{V}, meaning PV=constantPV = \text{constant}.

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Charles's Law: For a fixed mass of gas at constant pressure, V∝TV \propto T, meaning VT=constant\frac{V}{T} = \text{constant}.

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Gay-Lussac's (Pressure) Law: For a fixed mass of gas at constant volume, P∝TP \propto T, meaning PT=constant\frac{P}{T} = \text{constant}.

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The Internal Energy UU of an ideal gas consists only of the random kinetic energy of its molecules, as potential energy is assumed to be zero due to the lack of intermolecular forces.

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Standard units: Pressure PP in Pascals (PaPa), Volume VV in m3m^3, and Temperature TT in Kelvin (KK). Note that T(K)=t(∘C)+273T(K) = t(^{\circ}C) + 273.

πŸ“Formulae

PV=nRTPV = nRT

PV=NkBTPV = Nk_B T

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Eˉk=32kBT=32RNAT\bar{E}_k = \frac{3}{2} k_B T = \frac{3}{2} \frac{R}{N_A} T

n=NNA=mMn = \frac{N}{N_A} = \frac{m}{M}

U=32nRT=32PVU = \frac{3}{2} nRT = \frac{3}{2} PV

πŸ’‘Examples

Problem 1:

A gas cylinder with a fixed volume contains an ideal gas at a pressure of 3.0Γ—105Β Pa3.0 \times 10^5 \text{ Pa} at a temperature of 27∘C27^{\circ}C. If the cylinder is heated to 127∘C127^{\circ}C, what will be the new pressure of the gas?

Solution:

  1. Convert temperatures to Kelvin: T1=27+273=300Β KT_1 = 27 + 273 = 300 \text{ K}, T2=127+273=400Β KT_2 = 127 + 273 = 400 \text{ K}.
  2. Use the Pressure Law since volume is fixed: P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}.
  3. Rearrange for P2P_2: P2=P1Γ—T2T1P_2 = P_1 \times \frac{T_2}{T_1}.
  4. Substitute values: P2=(3.0Γ—105)Γ—400300=4.0Γ—105Β PaP_2 = (3.0 \times 10^5) \times \frac{400}{300} = 4.0 \times 10^5 \text{ Pa}.

Explanation:

Since the volume is constant, the pressure is directly proportional to the absolute temperature (Pressure Law). Increasing the temperature increases the average kinetic energy of the molecules, leading to more frequent and more forceful collisions with the walls.

Problem 2:

Calculate the average random kinetic energy of a molecule in an ideal gas at a temperature of 300Β K300 \text{ K}. (Given kB=1.38Γ—10βˆ’23Β JΒ Kβˆ’1k_B = 1.38 \times 10^{-23} \text{ J K}^{-1})

Solution:

  1. Use the kinetic energy formula: Eˉk=32kBT\bar{E}_k = \frac{3}{2} k_B T.
  2. Substitute the constants: EΛ‰k=32(1.38Γ—10βˆ’23)(300)\bar{E}_k = \frac{3}{2} (1.38 \times 10^{-23}) (300).
  3. Result: EΛ‰k=6.21Γ—10βˆ’21Β J\bar{E}_k = 6.21 \times 10^{-21} \text{ J}.

Explanation:

The average kinetic energy of a single molecule depends only on the absolute temperature of the gas, not on its pressure, volume, or the type of gas molecule.

Gas Laws - Revision Notes & Key Formulas | IB Grade 12 Physics