Unit 5: Solids, Liquids and Gases: Chapter 19: Solids, Liquids and Gases

Ideal Gas Molecules

  • Random Motion and Pressure: Molecules in a gas have random motion and exert a force, hence pressure, on the walls of a container.

  • Absolute Zero:

    • There's an absolute zero of temperature: 273-273°C.

  • Kelvin Scale:

    • Describes the Kelvin scale of temperature.

    • Able to convert between Kelvin and Celsius scales.

  • Temperature and Speed: An increase in temperature results in an increase in the average speed of gas molecules.

  • Kelvin Temperature and Kinetic Energy: The Kelvin temperature of a gas is proportional to the average kinetic energy of its molecules.

  • Qualitative Relationships (Fixed Gas Amount):

    • Pressure and volume at constant temperature.

    • Pressure and Kelvin temperature at constant volume.

  • Pressure and Kelvin Temperature Relationship (Fixed Mass, Constant Volume):

    • P<em>1T</em>1=P<em>2T</em>2\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}

  • Pressure and Volume Relationship (Fixed Mass, Constant Temperature):

    • P<em>1V</em>1=P<em>2V</em>2P<em>1V</em>1 = P<em>2V</em>2

Solids, Liquids, and Gases

  • Matter in Motion: All matter is made of continuously moving particles; arrangement and movement determine material properties.

  • Gas Laws: Laws describe the relationship between pressure, temperature, and volume.

  • Brownian Motion: Observation made by botanist Robert Brown, leading to the kinetic theory of matter.

The Gas Laws

  • Particle Movement: Gases consist of molecules moving in a random, haphazard way.

  • Pressure Exertion: Molecules hitting container walls exert a force; combined collisions result in pressure.

Boyle's Law

  • Robert Boyle's Discovery: Air is "squashy," and springs back when compressed; demonstrated using a bicycle pump or syringe.

  • Experiment: Boyle studied how gas volume depends on exerted pressure, keeping temperature constant.

  • Pressure Measurement: Pressure is force per unit area, measured in N/m² or pascals (Pa).

  • Gas Compression: Gases can be compressed because molecules are spread out; squeezing into a smaller container increases pressure.

  • Particle Theory Explanation: At constant temperature, average particle speed is the same; squeezing into smaller volume causes more frequent collisions, increasing average force and pressure.

  • Observation: Doubling the pressure halves the volume.

  • Boyle's Law Equation:P<em>1V</em>1=P<em>2V</em>2P<em>1V</em>1 = P<em>2V</em>2

    • For a fixed gas mass, changing pressure or volume results in the same product of p×Vp \times V if temperature remains constant.

  • Example Calculation:

    • Atmospheric pressure is 100100 kPa. Air in a sealed container has a volume of 2m32 m^3 at atmospheric pressure. Reducing the volume to 0.2m30.2 m^3 would result in a pressure of 10001000 kPa.

Absolute Zero

  • Temperature Impact: Temperature affects gas pressure. Experiment involves measuring pressure of a fixed gas volume at varying temperatures.

  • Pressure-Temperature Relationship: Gas pressure increases with temperature and decreases with cooling.

  • Absolute Zero Concept: Cooling gas leads to decreased pressure; a temperature exists below which cooling is impossible: absolute zero.

  • Absolute Zero Value: Approximately 273-273°C.

Kelvin Temperature Scale

  • Starting Point: Starts from absolute zero.

  • Proportionality: Kelvin temperature is proportional to the average kinetic energy of molecules.

  • Conversion Formulas:

    • Temperature in K = Temperature in °C + 273

    • Temperature in °C = Temperature in K - 273

  • Example Conversions:

    • Water freezes at 00°C = 273273 K.

    • Room temperature (20°C) = 293 K.

    • 400400 K = 127127°C.

  • Pressure vs. Kelvin Temperature Graph: Straight line through the origin, indicating proportionality.

  • Temperature Doubling: Heating gas from 200200 K (-73°C) to 400400 K (127°C) doubles its pressure.

  • Example Calculation

    • An empty tin is heated using a Bunsen burner until the temperature of the air inside is 5050 °C. Find the pressure of the air inside the tin. The temperature of the room is 2020°C.

    • p2 = 100kPa×323K293K=110kPa100 kPa \times \frac{323 K}{293 K} = 110 kPa

  • Explanation for relationship:

    • Heating a gas increases particle speed, resulting in harder and more frequent collisions with container walls, increasing average pressure.

  • Cooling Effect:

    • Cooling decreases kinetic energy; at absolute zero, particles have no thermal/movement energy, so they cannot exert pressure.

  • Temperature Symbol:

    • TT for Kelvin, remember to convert Celsius to Kelvin when using gas law equations involving temperature changes.

  • Fixed Mass, Constant Volume Relationship:

    • P<em>1T</em>1=P<em>2T</em>2\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}

NB T must be measured in Kelvin.

Formulae:

The formulas included in the note are:

  • P<em>1T</em>1=P<em>2T</em>2\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}

  • P<em>1V</em>1=P<em>2V</em>2P<em>1V</em>1 = P<em>2V</em>2

  • Temperature in K = Temperature in °C + 273

  • Temperature in °C = Temperature in K - 273