1 / 0

Graham’s Law

Graham’s Law. Diffusion & Effusion. Diffusion & Effusion. Diffusion is the gradual mixing of two gases. Effusion is the process by which a gas confined in a container passes through a tiny opening in the container. Both processes are due to the spontaneous, random motion of gas particles.

neveah
Download Presentation

Graham’s Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Graham’s Law

    Diffusion & Effusion
  2. Diffusion & Effusion Diffusion is the gradual mixing of two gases. Effusion is the process by which a gas confined in a container passes through a tiny opening in the container. Both processes are due to the spontaneous, random motion of gas particles.
  3. Graham’s Law Law states that the rates of effusion and diffusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. This law means that: Heavier gases move slower. Lighter gases move faster. Equation: =
  4. Applying Graham’s Law Sample Problem #1: Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Solution: = = 3.98 Interpret Solution: Hydrogen effuses approximately four times faster than oxygen.
  5. Applying Graham’s Law Sample Problem #2: A sample of hydrogen effuses through a hole in a container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas. Solution: = 9 9 = 12.79 = ? g/mol =163 g/mol Interpret Solution: The unknown gas has a molar mass of approximately 163 g/mol.
  6. Graham’s Law Video Demo Graham's Law Demonstration
  7. Graham’s Law Video Data
More Related