1. Peer Instruction: A Way for In-Class Interactions by Adham Ramadan Department of Chemistry

2. What is Peer Instruction? Learning from peers: student-student interactions leading to learning. Examples include: – In-class group work for case studies (Chem 104) – In-class group activities for problem analysis and numerical solutions (Chem 105) – Peer evaluation (Chem 104, 206, 304, 414) – Concept tests (Chem 106)

3. Why Peer Instruction? Students who have just acquired an understanding of a concept are better aware of the challenges in understanding it, and are therefore better equipped to explain it. Explaining a concept improves understanding it. Support of communication skills, interpersonal skills, critical thinking skills, team work, etc…

4. Concept tests: a structured peer instruction method In-class interactions Concept Tests Concept question Time for individual thinking & answer Record answer Student – student discussion Answer again Record answer

5. Personal experience with CTs Chem 106: a General Chemistry course covering fundamentals of Physical Chemistry, for science majors. Sci 150L: a General Science laboratory covering various basic concepts in Chemistry and Physics, for non-science/engineering majors.

6. Are Concept Tests effective? Actual Effectiveness I No. of correct answers: Increase in no. of correct answers after student-student discussions Level of confidence: Improvement in level of confidence BEFORE AFTER PS NQS PS G NQS PS G NQS PS G NQS PS G NQS PS G NQS PSG NQS PS G NQS PS NQS PS G NQS G PS NQS PS NQS PS Perceived Effectiveness II CTs identified by students as helping them learn in class, more so than other learning tools.

7. Commonly-cited challenges This takes too much class time. This takes too much time for preparation and marking. Students would not be confident enough to do it. It cannot be used for non-science courses. Analysis of results.

8. Peer Instruction promoting active learning Reading assignments Reading quizzes Concept tests

9. Concept Test: Example 1 A certain amount “n” of air is used to inflate a balloon located in a chamber where the pressure is 1 atmospheric pressure. This results in a volume of the balloon = 1 m 3. If another balloon located in another chamber, where the pressure is 2 atmospheric pressure, is to be inflated to the same volume of 1 m 3 :  “n” needed for balloon 2 will be the same as that used for balloon 1  “n” needed for balloon 2 will be the double that used for balloon 1  “n” needed for balloon 2 will be half that used for balloon 1  Balloon 2 cannot be inflated to 1 m 3 1 atm air 1 m 3 2 atm air 1 m 3

10. Concept Test: Example 2 Two bodies are moving from left to right. Snapshots are taken of their positions at 2 second intervals, and are numbered sequentially. Which of the following statements is correct: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9  They are both moving with the same speed  The red body is moving with a constant speed  After 10 seconds the two bodies have covered the same distance  The white body is moving with a constant speed