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Structure 1. Models of the particulate nature of matter - Ideal gases

Grade 11IBChemistry

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

🔑Concepts

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The Kinetic Molecular Theory (KMT) describes an ideal gas as a collection of particles in constant, random motion with negligible volume and no intermolecular forces.

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Collisions between gas particles and the walls of the container are perfectly elastic, meaning there is no net loss of kinetic energy.

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The average kinetic energy of gas particles is directly proportional to the absolute temperature in Kelvin (TT).

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Pressure (PP) is defined as the force exerted by gas particles per unit area of the container walls.

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An ideal gas strictly obeys the gas laws (PV=nRTPV = nRT) under all conditions of temperature and pressure.

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Real gases deviate from ideal behavior at high pressures and low temperatures where intermolecular forces and the volume of the particles themselves become significant.

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Standard Temperature and Pressure (STP) for IB Chemistry is defined as 273 K273 \text{ K} (0∘C0^\circ \text{C}) and 100 kPa100 \text{ kPa} (105 Pa10^5 \text{ Pa} or 1 bar1 \text{ bar}).

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Molar volume (VmV_m) of an ideal gas at STP is 22.7 dm3 mol−122.7 \text{ dm}^3 \text{ mol}^{-1}.

📐Formulae

PV=nRTPV = nRT

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

n=mMn = \frac{m}{M}

PV=mMRTPV = \frac{m}{M}RT

ρ=PMRT\rho = \frac{PM}{RT}

Ptotal=P1+P2+⋯+PnP_{total} = P_1 + P_2 + \dots + P_n

💡Examples

Problem 1:

Calculate the volume occupied by 0.500 mol0.500 \text{ mol} of an ideal gas at a pressure of 150 kPa150 \text{ kPa} and a temperature of 25.0∘C25.0^\circ \text{C}. Take R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}.

Solution:

  1. Convert temperature to Kelvin: T=25.0+273.15=298.15 KT = 25.0 + 273.15 = 298.15 \text{ K}
  2. Convert pressure to Pascals (if using SI units) or keep in kPa if using dm3dm^3: P=150 kPaP = 150 \text{ kPa}
  3. Use the Ideal Gas Law: V=nRTPV = \frac{nRT}{P} V=0.500×8.31×298.15150V = \frac{0.500 \times 8.31 \times 298.15}{150} V=1238.81150≈8.26 dm3V = \frac{1238.81}{150} \approx 8.26 \text{ dm}^3

Explanation:

The units for PP and VV must be consistent with the gas constant RR. Using PP in kPakPa and RR as 8.31 J K−1 mol−18.31 \text{ J K}^{-1} \text{ mol}^{-1} results in volume in dm3dm^3.

Problem 2:

A sample of gas occupies 2.00 dm32.00 \text{ dm}^3 at 300 K300 \text{ K} and 100 kPa100 \text{ kPa}. What will be its volume if the pressure is increased to 200 kPa200 \text{ kPa} and the temperature is increased to 600 K600 \text{ K}?

Solution:

Using the combined gas law: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} Rearranging for V2V_2: V2=P1V1T2T1P2V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} V2=100×2.00×600300×200V_2 = \frac{100 \times 2.00 \times 600}{300 \times 200} V2=12000060000=2.00 dm3V_2 = \frac{120000}{60000} = 2.00 \text{ dm}^3

Explanation:

Since the pressure doubled and the absolute temperature also doubled, their effects on the volume cancelled each other out, leaving the volume unchanged.