2. Charles’ Law Timberlake, Chemistry 7 th Edition, page 259 V 1 V 2 = T 1 T 2 (Pressure is held constant)

3. Charles' Law If n and P are constant, then V = (nR/P) = kT This means, for example, that Temperature goes up as Pressure goes up. Jacques Charles (1746 - 1823) Isolated boron and studied gases. Balloonist. A hot air balloon is a good example of Charles's law. VT V and T are directly related. T 1 T 2 V 1 V 2 = (Pressure is held constant)

4. Raising the temperature of a gas increases the pressure if the volume is held constant. The molecules hit the walls harder. The only way to increase the temperature at constant pressure is to increase the volume. Temperature

5. If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K one of 2 things happen 300 K

6. Either the volume will increase to 2 liters at 1 atm. 300 K 600 K

7. 300 K 600 K the pressure will increase to 2 atm.

8. Charles’ Law Timberlake, Chemistry 7 th Edition, page 259 (Pressure is held constant) T 1 T 2 V 1 V 2 =

9. V vs. T (Charles’ law) At constant pressure and amount of gas, volume increases as temperature increases (and vice versa). Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. T 1 T 2 V 1 V 2 = (Pressure is held constant)

10. Charles’ Law

11. The volume and absolute temperature (K) of a gas are directly related –at constant mass & pressure V T Charles’ Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Volume (mL) Temperature (K) V / T (mL / K) 40.0 44.0 47.7 51.3 273.2 298.2 323.2 348.2 0.146 0.148 0.147

12. The volume and absolute temperature (K) of a gas are directly related –at constant mass & pressure V T Charles’ Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

13. Charles’ Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

14. Volume vs. Kelvin Temperature of a Gas at Constant Pressure 0 100 200 300 400 500 Temperature (K) -273 -200 100 0 100 200 Temperature ( o C) Trial Temperature (T) Volume (V) o C K mL 1 10.0 283 100 2 50.0 323 114 3 100.0 373 132 4 200.0 473 167 180 160 140 120 100 80 60 40 20 0 origin (0,0 point) Trial Ratio: V / T 10.35 mL / K 20.35 mL / K 30.35 mL / K 40.35 mL / K Volume (mL)

15. Volume vs. Kelvin Temperature of a Gas at Constant Pressure 0 100 200 300 400 500 Temperature (K) -273 -200 100 0 100 200 Temperature ( o C) Trial Temperature (T) Volume (V) o C K mL 1 10.0 283 100 2 50.0 323 114 3 100.0 373 132 4 200.0 473 167 180 160 140 120 100 80 60 40 20 0 origin (0,0 point) Trial Ratio: V / T 10.35 mL / K 20.35 mL / K 30.35 mL / K 40.35 mL / K Volume (mL) 180 160 140 120 100 80 60 40 20 0

16. Volume vs. Kelvin Temperature of a Gas at Constant Pressure 0 100 200 300 400 500 Temperature (K) -273 -200 100 0 100 200 Temperature ( o C) Trial Temperature (T) Volume (V) o C K mL 1 10.0 283 100 2 50.0 323 114 3 100.0 373 132 4 200.0 473 167 origin (0,0 point) Trial Ratio: V / T 1 0.35 mL / K 2 0.35 mL / K 3 0.35 mL / K 4 0.35 mL / K Volume (mL) 20 40 60 80 100 120 140 160 20 40 60 80 100 120 140 160 absolute zero

17. Plot of V vs. T (Different Gases) Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 408 He CH 4 H2OH2O H2H2 N2ON2O 6 5 4 3 2 1 -200-1000 100200300 T ( o C) -273 o C V (L) Low temperature Small volume High temperature Large volume

18. Plot of V vs. T (Kelvin) Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 408 He CH 4 H2OH2O H2H2 N2ON2O 6 5 4 3 2 1 73173273 373473573 T (K) 0 V (L)

19. Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 428 Charles' Law

20. Temperature and Volume of a Gas Charles’ Law At constant pressure, by what fraction of its volume will a quantity of gas change if the temperature changes from 0 o C to 50 o C? T 1 = 0 o C + 273 = 273 K T 2 = 50 o C + 273 = 323 K V 1 = 1 V 2 = X 1 273 K = X 323 K X = 323 / 273 or 1.18 x larger V 1 = V 2 T 1 T 2

21. VT Calculation (Charles’ Law) At constant pressure, the volume of a gas is increased from 150 dm 3 to 300 dm 3 by heating it. If the original temperature of the gas was 20 o C, what will its final temperature be ( o C)? T 1 = 20 o C + 273 = 293 K T 2 = X K V 1 = 150 dm 3 V 2 = 300 dm 3 150 dm 3 293 K = 300 dm 3 T 2 T 2 = 586 K o C = 586 K - 273 T 2 = 313 o C

22. Temperature and the Pressure of a Gas High in mountains, Richard checked the pressure of his car tires and observed that they has 202.5 kPa of pressure. That morning, the temperature was -19 o C. Richard then drove all day, traveling through the desert in the afternoon. The temperature of the tires increased to 75 o C because of the hot roads. What was the new tire pressure? Assume the volume remained constant. What is the percent increase in pressure? P 1 = 202.5 kPa P 2 = X kPa T 1 = -19 o C + 273 = 254 K T 2 = 75 o C + 273 = 348 K 202.5 kPa 254 K = P 2 348 K P 2 = 277 kPa % increase = 277 kPa - 202.5 kPa x 100 % 202.5 kPa or 37% increase

23. Gas Laws with One Term Constant Keys Gas Laws with One Term Constant http://www.unit5.org/chemistry/GasLaws.html