1. Boyle’s Law Imagine: Hold your finger over the hole at the end of a syringe. Now depress the plunger. How does the pressure of the gas in the syringe change as the volume of gas decreases? The P increases.

2. Data from Boyle’s Law Exp’t. P(kPa)V(mL)1/V (mL -1 )P*V (kPa*mL) 1001000.0110 000 125 800.012510 000 167 600.01710 000 250 400.02510 000 500 200.05010 000

3. We can plot two graphs of these data: P 1 V 1 = P 2 V 2 or PV = k ( a constant) (T,n unchanged) Check this...

5. We can also make a reciprocal plot: P α 1/V or P = k * (1/V) or P = k/V or PV = k (T, n unchanged)

6. Charles’ Law Jacques Charles collected the following data for a sample of gas maintained at constant Pressure: He obtained the same data when different gases were used. V (L) T ( o C) 0.75-192 1.0-165 1.5-111 2.0-57 2.5-3.6 3.050.2 3.5104

8. Graphing Charles’ Data... V α T (P, n unchanged) or V = (constant)*T or V = constant or T V 1 = V 2 (P, n const) T 1 T 2

9. Charles’ Law states that for a sample of gas at constant P, V 1 = V 2 (P, n const) T 1 T 2

10. Look again at Charles’ Law Data: Equation of line is: V = 0.0093T + 2.5328 What will be the temperature when the gas occupies zero volume? T = - 2.5328/0.0093 = -272 o C This is the Absolute zero of temperature. (Accepted value 0 K = - 273 o C)

11. Temperature Scales Charles’ Law states V α T (@ constant n, P) If T  ↑ 2X, therefore V ↑  2X. But what is 2X (-10 o C) ? -20 o C ??? Colder??? Doesn’t make sense! What is 2X 0 o C? 0 o C??? Doesn’t make sense!

12. So what’s up with the Celsius Temperature scale? It’s relative to water: 0 o C is fp of water; 100 o C is bp of water. But—the Kelvin Scale is

13. What is the conversion between Celsius and Kelvin? K = o C + 273

14. ALWAYS do calculations using absolute, or KELVIN TEMPERATURE K = o C + 273

15. Gay-Lussac’s Law Henri Gay-Lussac studied the relationship between P and T for a fixed volume of gas. Refer to class demo using above apparatus.

16. data obtained in class... Temp ( o C) Pressure (psi) 7916.9 2414.7 013.6 -1964.0 -799.8

17. Graphing these data gives: P α T (V, n unchanged) or P = (constant)*T or P = constant or T P 1 = P 2 (V, n unchanged) T 1 T 2

18. Looking at the graph again... Equation of line is P = 0.0474*T + 13.42 Predict the T at which P gas = 0?

19. P = 0.0474*T + 13.42, set P = 0 and solve for T 0 = 0.0474*T + 13.42 T = -13.42/0.0474 T = -283 o C not bad for class data! Absolute zero (-273 o C or 0 K )can be determined from either V vs T or P vs T.

20. check this... “... circulate to all engineers showing the folly of poor design and/or incorrect operating procedures. Apparently the rail car had been steamed out and was still hot inside when it started to rain. The tank had a vent designed to release pressure, not for a vacuum. “ What do you think happened to the rail car?

22. So far: gas laweq’nunchanged Boyle’sP 1 V 1 = P 2 V 2 n, T CharlesV 1 = V 2 n, P T 1 T 2 Gay-Lussac’sP 1 = P 2 n, V T 1 T 2 Put these together to get the Combined Gas Law: P 1 V 1 = P 2 V 2 n unchanged T 1 T 2

23. Combined Gas Law P 1 V 1 = P 2 V 2 for constant n T 1 T 2 Useful for calculations involving changes in T, P, V for a fixed amount of gas.

24. example When a 12.0 L sample of H 2 S(g), originally at 101kPa and 25 o C is subjected to a pressure of 205 kPa at 78 o C, what will be the “new” volume of the gas? Use combined gas law: P 1 = 101 kPa V 1 = 12.0L T 1 = (25 + 273 = 298K) P 2 = 205 kPa T 2 = (78 + 273 = 351K)

25. P 1 V 1 = P 2 V 2 T 1 T 2 rearranging gives V 2 = P 1 V 1 T 2 P 2 T 1 = (101 kPa)(12.0 L)(351 K) (205 kPa)(298 K) = 6.96 L is the expected volume.

26. HW Text p 511 to 542 P 514 PP – do a few P 515 RQ #1 – 14 do the ones that challenge you P 518 LC #13 – 18 P 522 PP – do a few P 525 PP – do a few P 542 PP 1 – 10 do a few