Understanding States of Matter, Energy Changes, and Phase Transitions

Slide Note
Embed
Share

Explore the properties of the three states of matter - solid, liquid, and gas - learn about energy changes during phase transitions when substances are heated or cooled, and understand how temperature, pressure, and volume play a role in describing these physical phenomena.


Uploaded on Oct 02, 2024 | 0 Views


Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

E N D

Presentation Transcript


  1. States of Matter & Gases

  2. States of Matter The three states of matter include: Solid: Liquid: Gas State Volume Shape Solid definite definite Liquid definite indefinite Gas indefinite indefinite

  3. Energy and phase change When Heated: Solid liquid gas When Cooled: Gas liquid solids

  4. Energy & phase change

  5. Heating Curve & phase changes Substances can exist in three states. They are solid, liquid, and gas. They (the substances) change when the kinetic energy of a substance changes. When a substance changes from a solid to a liquid, it is called melting. When a substance changes from a liquid to a gas, it is called vaporization. Gas particles have the most kinetic energy of the three states, but they also have the lowest potential energy. This is opposite of a solid which has high potential energy but low kinetic energy. There is one special phase change which is not common. The change is from a solid to a gas. This is called sublimation.

  6. Heating Curve of Water: 100 C -- Temperature 0 C -- Energy added

  7. Heating Curve of Water Each plateau represents a physical change . The energy that is added is used to change the state of water. Each slope represents the increase in temperature. The energy that is added is used to heat the of water.

  8. Temperature (T): must be described using the unit Kelvin (K). K = C + 273 C = K 273 You try: 1. 25 C = _______ K 2. 450K = _________ C 298 K 177 C

  9. Pressure (P): measures the force/area in units of atm., mm Hg or kPa 1 atm = 760 mm Hg = 101.3 kPa Now you try! 1. Convert 0.95 atm to mm Hg 722 mmHg 2. Convert 0.95 atm to kPa 96.2 mmHg 3. Convert 100.5 kPa to mmHg 754 mmHg

  10. Volume (V): is defined using milliliters (mL) or liters (L) 1000 mL = 1L Now you try! 1. 150 mL = _________ L 2. 34.5 L = __________ mL 34,500 mL .150L

  11. Atmospheric pressure: is the pressure exerted by the weight of air in the atmosphere. Traditionally it is measure using a barometer. https://www.youtube.com/watch?v=jmQ8FWnM0fA

  12. : Objective- to prove that atmospheric pressure exists. Fill in the remaining part of this experiment together in class: Background to experiment: Observations: Interpretation of findings: https://www.youtube.com/watch?v=JsoE4F2Pb20

  13. Boyles Law Describes the inverse ( )relationship of pressure: volume P1V1= P2V2

  14. A gas has an initial pressure of 1.0 atm and a volume of 1.5L. What is the new volume when the pressure is increased to 1.2 atm? P1 = 1.0atm V1 = 1.5L P2= 1.2 atm V2= ? Answer: V2= 1.25L

  15. Charles Law Describes the direct relationship between temperature and volume ( V1 = V2 T1 T2 ). Remember: Temperature is defined by Kelvin (K = C + 273)

  16. A gas has an initial temperature of 275K and a volume of 0.45L. What is the new volume when the temperature is increased to 300K? V1 = 0.45L T1 = 275K T2 = 300K V2 = ? Answer : 0.49L

  17. The combined gas law The combined gas law pulls together the inverse relationship of P:V and the direct relationship of V:T The combined gas law:

  18. Combined gas law: P1= 1.5 atm V1 = 0.75L T1 = 276K P2 = ? V2 = 1.25L T2 = 295K V2 = P1V1 T2 T1 P2 Answer = 0.96 atm

  19. The Ideal Gas Law: Includes the amount of a gas in moles (n) PV = nRT P= pressure (atm) V= volume (L) n = moles R = gas constant: 0.08206 L atm/ mole K T = temperature (K)

  20. Practice problem: Calculate the pressure of 0.500 moles nitrogen gas in a 10.0L container, held at 298K. P= ? V = 10.0L n = 0.500 mole T = 298K R = 0.08206 L atm/mole K P = nRT V Answer = 1.22 atm

  21. Gas law applications: Mountaineering Scuba Diving Tire pressure Hot Air balloons

More Related Content