2. Gases

3. Compared to the mass of a dozen eggs, the
mass of air in an “empty refrigerator” is
Negligible
About a tenth as much
About the same
More

4. Why do we think of air as having no mass???
Answer:
One cubic meter of air at 0 degrees Celsius and normal atmospheric pressure has a mass of about 1.3 Kg. A medium-sized refrigerator has a volume of 0.6 cubic meters and contains 0.8 kilograms of air. A dozen large eggs weighs in at 0.75 kilograms.
Why do we think of air as having no mass???

5. Answer:
We don’t notice the weight of air on us because we are submerged in air. If someone handed you a bag of water while you were submerged in water, you wouldn’t notice its weight either.
Interesting Fact!
The average mass of the atmosphere is about 5,000 trillion metric tons (or 5 x 1018 kg) according to the National Center for Atmospheric Research.

6. The state of matter is determined by the motion of its particles.
Kinetic Theory
The state of matter is determined by the
motion of its particles.
 Phases of Matter

7. Kinetic Theory and Gases
Gases consist of tiny particles that are in constant motion
Collisions between particles of a gas and other objects are elastic-there is no loss of kinetic energy
Average kinetic energy of particles of a gas is directly proportional to the temperature of the gas

8. Nature of Gases Gases have no definite shape or volume
Gases have low density
Gases can be compressed
Gases are considered fluids
Gases are capable of diffusion-they can spread out with the absence of circulating currents

9. Ideal Gas
An imaginary gas that conforms perfectly to all of the assumptions of kinetic theory

10. Real Gas
A gas that does not obey all of the assumptions of kinetic theory
Gases will deviate from ideal gas behavior under conditions of high temperature and pressure.

11. Pressure The force per unit area on a surface Pressure = Force / Area
Gas molecules exert pressure on every surface with which they collide-the pressure exerted depends on:
Number of molecules present
Temperature
Volume of container

12. What is normal atmospheric pressure?
At sea level, the normal atmospheric pressure is 1 atmosphere (atm)
1 atm = 760 mm Hg
1 atm = 760 torr
1 atm = kPa

13. Gas Laws

14. Standard Temperature and Pressure (STP)
(STP) is defined as:
1 atm and 0 degrees Celsius

15. Boyle’s Law
Relates changes in pressure and volume assuming constant temp. and molecules of gas.
States that pressure and volume are inversely proportional (If you ↓V, you ↑ P)
P1V1=P2V2

16. Charles’ Law
Relates changes in volume and temp. assuming constant pressure and molecules of gas.
States that volume and temp. are directly proportional (increase temp, increase volume)
V1 = V2
T1 T2
Varies directly with the Kelvin scale so all temps must be in Kelvin
Remember…
K= C + 273

17. Gay Lussac’s Law
Deals with changes in pressure and temperature assuming volume and amount of gas are held constant.
Pressure and temperature are directly proportional (↑Temp, ↑Pressure)
P1 = P2
T1 T2

18. Combined Gas Law
Relates changes in pressure, temp. and volume assuming the amount of gas is held constant.
P1V1 = P2V2
T T2

19. Example Problems:
Look at what information is given to decide which gas law must be used.
Check to make sure that units are the same on both sides of the equation and that all temps. are in Kelvin

20. A sample of nitrogen occupies a volume of 250 mL at 25 C
A sample of nitrogen occupies a volume of 250 mL at 25 C. What volume will it occupy at 95 C?
A sample of carbon dioxide gas occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume was decreased to 2.00 L?
A helium filled balloon has a volume of 50 L at 25 C and 820 mmHg. What volume will it occupy at 650 mm Hg and 10 C?

21. A sample of nitrogen occupies a volume of 250 mL at 25 C
A sample of nitrogen occupies a volume of 250 mL at 25 C. What volume will it occupy at 95 C?
Convert temps to Kelvin: K = °C + 273
V1 = 250 mL
V2 = ?
T1 = 25°C = 298 K
T2 = 95°C = 368 K
V1 = V mL = V2
T1 T K K
***Cross multiply, and solve for V2
(250)x(368) = (298)x(V2) V2 = mL

22. A sample of carbon dioxide gas occupies a volume of 3. 50 L at 125 kPa
A sample of carbon dioxide gas occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume was decreased to 2.00 L?
V1 = 3.50 L
V2 = 2.00 L
P1 = 125 kPa
P2 = ?
P1V1=P2V (125)x(3.50) = (P2)x(2.00)
P2 = kPa

23. A helium filled balloon has a volume of 50 L at 25 C and 820 mmHg
A helium filled balloon has a volume of 50 L at 25 C and 820 mmHg. What volume will it occupy at 650 mm Hg and 10 C?
V1 = 50 L
V2 = ?
T1 = 25°C = 298 K
T2 = 10°C = 283 K
P1 = 820 mmHg
P2 = 650 mm Hg
P1V1 = P2V2 (820)(50) = (650)(V2)
T T K K
V2 = 59.9 L

24. Dalton’s Law of Partial Pressure
The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Partial pressure- the pressure of an individual gas in a mixture of gases
Pt = P1 + P2 + P3 ……
Where Pt = total pressure and P1, P2, etc. = partial pressures of individual gases

25. Partial Pressures are used to:
Calculate water vapor pressure when a gas is collected over water.
Because water molecules at a liquid surface always evaporate, gases collected over water are not pure….they are always mixed with water vapor….so Dalton’s Law is used to calculate the partial pressure of the gas.

26. Example Problem-Partial Pressure
Oxygen from the decomposition of potassium chlorate was collected over water. The barometric pressure and the temp. during the experiment were 731 mm Hg and 20 C. What was the partial pressure of oxygen collected?
(Use the water vapor pressure to solve)
PressureTotal = PressureOxygen + Pressure Water vapor
731 mm Hg = Pressure Oxygen mm Hg
Pressureoxygen = mm Hg

27. Example Problem-Partial Pressure
A sample of gas was collected over water at 740 mm Hg and 23 C. The collecting tube is left in place, and the volume is not measured until the next day when the dry gas was found to have a pressure of 745 mm Hg and a temp. of 20 C, at which time the volume is found to be 15.3 mL. What is the original volume (from yesterday)?
PressureTotal = PressureGas + Pressure Water vapor
740 mm Hg = Pressure Gas mm Hg
PressureGas = mm Hg (pressure from yesterday)
P1 = mm Hg, V1 = ?, T1 = 23°C = 296 K
P2 = 745 mm Hg, V2 = 15.3 mL, T2 = 20°C = 293 K

28. …Continued Solving for the original volume: Use the equation:
P1 = mm Hg, V1 = ?, T1 = 23°C = 296 K
P2 = 745 mm Hg, V2 = 15.3 mL, T2 = 20°C = 293 K
Use the equation:
P1V1 = P2V2
T T2
(718.9)(V1) = (745)(15.3)
V1 = L

29. Ideal Gas Law
Expresses a relationship between pressure, volume, temp. and moles of gas
PV= nRT
(where P= pressure, V= volume, n= moles, R= ideal gas constant, and T= temp)

30. Ideal Gas Constant
The constant’ value depends on the units chosen for pressure, volume, and temp.
If 1 mole of a gas occupies 22.4 L at STP (0 degrees C and 1 atm) and these values are substituted into the ideal gas law, the value for R can be calculated:
R = PV = (1atm)( 22.4 L) = L . atm
nT (1 mol) (273 K) mol . K
Since R has units of L, atm, mol, and K, all pressure, volumes, temps., and amounts must also be expressed in these units.

31. Examples:
What is the pressure in atm exerted by 0.50 mol sample of nitrogen in a 10 L container at 298 K?
PV = nRT
(P)(10) = (0.5)(0.0821)(298)
P = 1.2 atm

32. Examples:
What is the volume in liters occupied by 0.25 mol of oxygen at 20 degrees C and 740 mm Hg?
Convert all units to K, atm, and L!!
740 mm Hg x atm = atm
760 mm Hg
20°C = 293 K
PV = nRT
(0.974)(V) = (0.25)(0.0821)(293)
V = 6.17 L

33. More Examples:
At 28 C and atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of the gas? (Hint.. Remember that molar mass is in units of g/mol)
PV = nRT °C = 301 K
(0.974)(1.00) = (n)(0.0821)(301)
n = mol
Molar mass: grams/mol g/ mol
Molar mass = g/mol

34. More Examples:
What is the density of a sample of ammonia (NH3), if the pressure is atm and the temp. is 63.0 C and there is 2 L of gas?
(Density = mass / volume)
PV = nRT 63°C = 336
(0.928)(2) = (n)(0.0821)(336)
n = mol NH3 (Change to grams!!)
mol NH3 x g = g NH3
1 mol NH3
Density = mass/volume = g/2L = g/L

35. Effusion and Diffusion
Effusion- the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container
Diffusion- the spreading out of a gas due to the constant motion of gas molecules

36. Graham’s Law of Effusion
Rates of effusion and diffusion depend on the velocities of gas molecules.
The velocity of a gas varies inversely with its mass (lighter molecules move faster).

37. Suppose you wanted to compare the velocity of two gases:
Graham’s Law states that the rates of effusion of gases at the same temp. and pressure are inversely proportional to the square roots of their molar masses.
Rate of effusion of A = √Mb
Rate of effusion of B √Ma

38. Examples:
Compare the rate of effusion of carbon dioxide with that of hydrochloric acid at the same temp. and pressure.
CO2 has a molar mass of g
HCl has a molar mass of g
Since mass and velocity are inversely related, the smaller molar mass would have the greater velocity
Therefore, HCl has a higher rate of effusion

39. Rate of effusion of A = √Mb
Examples:
If a molecule of neon gas travels at an average of 400 m/s at a given temp., estimate the avg. speed of a molecule of butane (C4H10) at the same temp.
Assign Ne as “A” and C4H10 as “B”
Molar mass of C4H10 = g = Mb
Molar mass of Ne = g = Ma
400 m/s = √58.12
Rate C4H √20.18
Rate of effusion of C4H10 = m/s
Rate of effusion of A = √Mb
Rate of effusion of B √Ma

40. Rate of effusion of A = √Mb
More Examples:
A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Assign H2 as “A” and Unknown as “B”
Assign rate of Unknown as “x” so rate of H2 would be “9x”
9x = √MB
x √2.016
(√2.016)(9x) = (x)(√MB) The “x” cancels on both sides.
(√2.016)(9) = (√MB)
12.78 = (√MB)
MB = g
Rate of effusion of A = √Mb
Rate of effusion of B √Ma

41. Stoichiometry of Gases
Remember… 1 mol of any gas at STP will occupy 22.4 L -this is called molar volume

42. When solving stoichiometry problems involving gases the following will be used:
Given
Unknown
Mol ratio –compares given and unknown
Molar mass-if given is in grams or unknown is asked for in grams
Molar volume- Only if at STP (used to convert moles to liters or liters to moles)

43. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g)
Example:
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g)
What volume in liters of oxygen would be required for the complete combustion of L of propane at STP?
0.350 L C3H8 x 1 mol C3H8 x 5 mol O2 x L O2 = 17.5 L O2
22.4 L mol C3H8 1 mol O2
b) What volume of CO2 will be produced at STP?
0.350 L C3H8 x 1 mol C3H8 x 3 mol O2 x L O2 = 1.05 L O2
22.4 L mol C3H mol O2

44. CaCO3 (s)  CaO (s) + CO2 (g)
How many grams of calcium carbonate must be decomposed to produce 5 L of carbon dioxide at STP?
CaCO3 (s)  CaO (s) + CO2 (g)
5 L CO2 x 1 mol CO2 x 1 mol CaCO3 x g CaCO3 = g CaCO3
22.4 L mol CO mol CaCO3

45. WO3 (s) +3H2 (g)  W (s) + 3 H2O (l)
How many liters of hydrogen at 35 C and 745 mm Hg are needed to react completely with 875 g of tungsten oxide?
(Because we are not at STP, 1 mol of the gas will no longer occupy 22.4 L..so molar volume can not be used but we need a way to convert mol into liters to use in stoichiometry….we will have to use the Ideal Gas Law)

46. WO3 (s) +3H2 (g)  W (s) + 3 H2O (l)
How many liters of hydrogen at 35 C and 745 mm Hg are needed to react completely with 875 g of tungsten oxide?
875 g WO3 x 1 mol WO3 x 3 mol H2 = mol H2
g WO3 1 mol WO3
745 mm Hg x 1 atm = atm °C = 308 K
760 mm Hg
PV = nRT
(0.9802)(V) = (11.32)(0.0821)(308)
V = L H2

47. To decide which to use first:
Write down the given and unknown
Write down the mole ratio
Write down any molar masses needed
Write down the variables for the gas law, using in pressures, volumes, temps., that are given in the problem.
If there is enough info to solve the gas law 1st, do so. You will be solving for moles of whatever the volume amount is for.

48. Summary: What to do for problems that are NOT at STP!
If there is not enough info to solve the gas law first, use the given to find the moles of what you are trying to find the volume of…plug moles back into the gas law and solve for volume.

49. 2Na (s) + Cl2 (g)  2NaCl (s)
How many grams of Na should be used to react with 3.54 L of chlorine at 38 C and 1.63 atm?
***Note: This is not at STP so we can’t use the 1 mol = 22.4 L conversion. Therefore, we use the Ideal Gas Law and stoichiometry.
PV = nRT
(1.63)(3.54) = (n)(0.0821)(311)
n = mol Cl2
0.226 mol Cl2 x 2 mol Na x g Na = g Na
1 mol Cl mol Na