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Boyle’s Law

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. Data from Boyle’s Law Exp’t. P (kPa) V(mL) 1/V (mL -1 ) P*V (kPa*mL)

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Boyle’s Law

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  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) 100 100 0.01 10 000 125 80 0.0125 10 000 167 60 0.017 10 000 250 40 0.025 10 000 500 20 0.050 10 000

  3. We can plot two graphs of these data: P1V1 = P2V2 or PV = k ( a constant) (T,n unchanged) Check this . . .

  4. 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)

  5. 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.

  6. Graphing Charles’ Data . . . V α T (P, n unchanged) or V = (constant)*T or V = constant or T V1=V2 (P, n const) T1 T2

  7. Charles’ Law states that for a sample of gas at constant P, V1=V2 (P, n const) T1 T2

  8. 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 = -272oC This is the Absolute zero of temperature. (Accepted value 0 K = - 273oC)

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

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

  11. What is the conversion between Celsius and Kelvin? K = oC + 273

  12. ALWAYS do calculations using absolute, or KELVIN TEMPERATURE K = oC + 273

  13. 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.

  14. data obtained in class . . .

  15. P α T (V, n unchanged) or P = (constant)*T or P = constant or T P1 = P2 (V, n unchanged) T1 T2 Graphing these data gives:

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

  17. 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 = -283oC not bad for class data! Absolute zero (-273oC or 0 K )can be determined from either V vs TorP vs T.

  18. 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?

  19. So far: gas laweq’nunchanged Boyle’s P1V1 = P2V2 n, T Charles V1=V2 n, P T1 T2 Gay-Lussac’s P1 = P2 n, V T1 T2 Put these together to get the Combined Gas Law: P1V1 = P2V2 n unchanged T1 T2

  20. Combined Gas Law P1V1 = P2V2 for constant n T1 T2 Useful for calculations involving changes in T, P, V for a fixed amount of gas.

  21. example When a 12.0 L sample of H2S(g), originally at 101kPa and 25oC is subjected to a pressure of 205 kPa at 78oC, what will be the “new” volume of the gas? Use combined gas law: P1 = 101 kPa V1 = 12.0L T1 = (25 + 273 = 298K) P2 = 205 kPa T2 = (78 + 273 = 351K)

  22. P1V1 = P2V2 T1T2 rearranging gives V2 = P1V1T2 P2T1 = (101kPa)(12.0L)(351K) (205 kPa)(298K) = 6.96 L is the expected volume.

  23. 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

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