3. Gases have some interesting characteristics that have fascinated scientists for 300 years. Regardless of their chemical identity, gases tend to exhibit similar physical behaviors

4. The Nature of Gases Gas particles can be monatomic (Ne), diatomic (N 2 ), or polyatomic (CH 4 ) – but they all have these characteristics in common: 1) Gases have mass. 1) Gases have mass. 2) Gases are compressible. 3) Gases fill their containers. 4) Gases diffuse 5) Gases exert pressure. 6) Pressure is dependent on Temp.

5. KINETIC MOLECULAR THEORY (KMT) Deriving the Ideal Gas Law Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion(Brownian motion). P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.

7. Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres)

8. Common Units of Pressure Unit Atmospheric Pressure Scientific Field pascal (Pa); 1.01325 x 10 5 Pa SI unit; physics, kilopascal(kPa) 101.325 kPa chemistry atmosphere (atm) 1 atm* Chemistry millimeters of mercury 760 mmHg Chemistry, medicine, ( mm Hg ) biology torr 760 torr* Chemistry bar 1.01325 bar Meteorology, chemistry, physics

9. Converting Units of Pressure Problem: A chemist collects a sample of carbon dioxide from the decomposition of limestone (CaCO 3 ) in a closed end manometer, the height of the mercury is 341.6 mm Hg. Calculate the CO 2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors to find the pressure in the other units. Solution: P CO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr 1 mm Hg converting from mmHg to torr: converting from torr to atm: P CO2 ( atm) = 341.6 torr x = 0.4495 atm 1 atm 760 torr converting from atm to kPa: P CO2 (kPa) = 0.4495 atm x = 45.54 kPa 101.325 kPa 1 atm

10. Gas Pressure It can be defined as the force exerted by a gas per unit surface area of an object Due to: a) force of collisions, and b) number of collisions No particles present? Then there cannot be any collisions, and thus no pressure – called a vacuum

11. Since the Kinetic Molecular Theory states that the volume of the gas particles is zero, then the equation simplifies. As a result, the amount of available space for the gas particles to move around in is approximately equal to the size of the container. Thus, as stated before, the variable V is the volume of the container. 1 L = 1dm 3 = 1000 mL = 1000 cm 3 = 10 -3 m 3

12. The temperature variable is always symbolized as T (absolute temperature that is in K). Consequently, T values must be converted to the Kelvin Scale. To do so; Kelvin = Celsius + 273

13. The Gas Laws are a mathematical interpretation of the behavior of gases.

14. 1. Boyle- Mariotte’s Law If n and T are constant, then PV = k This means, for example, that P goes up as V goes down and k is dependent on n and T. Robert Boyle (1627- 1691). Son of Early of Cork, Ireland.

15. Effect of changing pressure on volume The pressure can be changed by adding or removing green weights from the top of the red piston. This is an example of Boyle's law.

17. Boyle’s Law Hyperbolic Relation Between Pressure and Volume (courtesy F. Remer) n constant

18. Boyle’s Law Hyperbolic Relation Between Pressure and Volume p V p – V Diagram n1n1 n2n2 n3n3 n 3 >n 2 >n 1 (courtesy F. Remer) T constant

20. PV P or V

24. Variation of gas volume with temperature at constant pressure. V  TV  T T (K) = t ( 0 C) + 273 n constant Charles’ Law Temperature must be in Kelvin!!! V/T = k

25. Charles’s Mathematical Law: since V/T = k Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C? V 1 V 2 T 1 T 2 = What if we had a change in conditions?

26. 1)determine which variables you have: T and V = Charles’s Law 2)determine which law is being represented:  T 1 = 127°C + 273 = 400K  V 1 = 3.0 L  T 2 = 227°C + 273 = 5ooK  V 2 = ?  T 1 = 127°C + 273 = 400K  V 1 = 3.0 L  T 2 = 227°C + 273 = 5ooK  V 2 = ?

27. 4) Plug in the variables: (500K)(3.0L) = V 2 (400K) V 2 = 3.8L 3.0L V 2 400K 500K = = 5) Cross multiply and chug

28. absolute zero = no molecular motion no molecular motion = zero force in the container

29. Gay Lussac’s Law Old man Lussac determined the relationship between temperature and pressure of a gas. He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law. During his experiments volume of the system and amount of gas were held constant. Old man Lussac determined the relationship between temperature and pressure of a gas. He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law. During his experiments volume of the system and amount of gas were held constant.

31. Temp Pressure How does Pressure and Temperature of gases relate graphically? P/T = k Volume, # of particles remain constant Volume, # of particles remain constant

32. Lussac’s Mathematical Law: What if we had a change in conditions? since P/T = k P 1 P 2 T 1 T 2 = Eg: A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?

33. T and P = Gay-Lussac’s Law  T 1 = 127°C + 273 = 400K  P 1 = 3.0 atm  T 2 = 227°C + 273 = 500K  P 2 = ?  T 1 = 127°C + 273 = 400K  P 1 = 3.0 atm  T 2 = 227°C + 273 = 500K  P 2 = ? 1)determine which variables you have: 2)determine which law is being represented:

34. 4) Plug in the variables: (500K)(3.0atm) = P 2 (400K) P 2 = 3.8atm 3.0atm P 2 400K 500K = = 5) Cross multiply

36. Gay-Lussac’s Law

37. Linear Relation Between Temperature and Pressure P T (K) 0100200300 P – T Diagram V1V1 V2V2 V3V3 V 1 <V 2 <V 3 (courtesy F. Remer)

39. The mass is changed by injecting molecules at the left. The density (mass/volume) remains a constant for constant pressure and temperature. Effect of changing mass on volume

40. LAW RELAT- IONSHIP LAW CON- STANT Boyle’s P VP VP VP V P 1 V 1 = P 2 V 2 T, n Charles’ V TV TV TV T V 1 /T 1 = V 2 /T 2 P, n Gay- Lussac’s P TP TP TP T P 1 /T 1 = P 2 /T 2 V, n Avogadro’s nVnVnVnV V 1 /n 1 = V 2 /n 2 P,T

41. The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written R = ideal gas constant PV = nRT

43. R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm /mol-K

45. Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?

46. What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C?

47. Calculate the density in g/L of O 2 gas at STP. From STP, we know the P and T.

48. How many L of O 2 are need to react 28.0 g NH 3 at 24°C and 0.950 atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

49. GAS DENSITY PV = nRT d and M proportional

50. Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT PM = d is the density of the gas in g/L