Temperature and Volume Relationship of Gases (Charles’s Law) Temperature and Pressure Relationship of Gases (Gay Lussac’s Law) A.10/A.11 In text

Using the data you gathered in investigation A.9 fill in the table: Based on A.9 and A.1 investigations: what is the relationship between Temperature and Volume in gases? Temp (oC) Volume (in mm) 90o 8.5mm 80o 7.9mm 70o 7.3mm

A.1 Investigation 2: Balloons in Water of Differing Temperatures What happened to the volume of the balloon in hot water: What happened to the volume of the balloon in ice water:

Summarize the relationship between pressure and volume for a gas at a constant temperature. What is the quantitative way to summarize this relationship? (the mathematical formula?)

When pressure and amount of gas molecules is held constant: As temperature goes up volume goes up As temperature goes down, volume goes down DIRECT RELATIONSHIP

**Unit for temperature MUST be in KELVINS! Charles’ Law:

What about the relationship between temperature and pressure at a constant volume? (Gay-Lussac’s Law) Temperature must still be in units of Kelvin! Direct relationship

Problem #1 Temperature will affect airships in Tyria – a fantasy world. The small airships have an initial volume of 4,200L. To replicate atmospheric conditions in Tyria scientists place it in a chamber and reduce the temperature from 20 to -60oC. What will the final volume of the airship be?

Steps to Solve : First predict how the change will affect the volume (will it go up or down)? Second: Convert any temperatures to Kelvin (oC + 273) Third: define each variable (should be 1 unknown) V1: T1: V2: T2: Fourth: “Plug” the values into the equation

Problem #1:

4200L = V2 -> (4200L) * (213K) = V2 293K 213K (293K) V2 = 3100L First predict how the change will affect the volume (will it go up or down)? If the temperature goes down the volume should be smaller than 4,200L. Second: Convert any temperatures to Kelvin (oC + 273) T1: 20 + 273 = 293 K, T2: -60 + 273 = 213 K V1: 4200L T1: 293K V2: ? T2: 213K Fourth: “Plug” the values into the equation. (Cross multiply) 4200L = V2 -> (4200L) * (213K) = V2 293K 213K (293K) V2 = 3100L

When problem has been checked off: 1. Complete A. 11 1-4 AND A When problem has been checked off: 1. Complete A.11 1-4 AND A.11 Extra Practice 1-6 2. Concept Check 1-3 All are due next class meeting