1. Gases

2. Elements that exist as gases at 250C and 1 atmosphere
5.1

3. 5.1

4. Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
5.1

5. (force = mass x acceleration)
Area
Barometer
Pressure =
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mm Hg = 760 torr
= 101,325 Pa = 14.7 psi = in. Hg
5.2

6. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm
5.2

7. Boyle’s Law
P α 1/V
This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.
P1V1 = P2 V2
Robert Boyle ( ). Son of Earl of Cork, Ireland.

8. Charles’s Law If n and P are constant, then V α T
V and T are directly proportional.
V V2 =
T T2
If one temperature goes up, the volume goes up!
Jacques Charles ( ). Isolated boron and studied gases. Balloonist.

9. Gay-Lussac’s Law If n and V are constant, then P α T
P and T are directly proportional.
P P2 =
T T2
If one temperature goes up, the pressure goes up!
Joseph Louis Gay-Lussac ( )

10. No, it’s not related to R2D2
Combined Gas Law
The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V P2 V2 =
T T2
No, it’s not related to R2D2

11. And now, we pause for this commercial message from STP
OK, so it’s really not THIS kind of STP…
STP in chemistry stands for Standard Temperature and Pressure
Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0 deg C (273 K)
STP allows us to compare amounts of gases between different pressures and temperatures

12. Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2
Constant temperature
Constant pressure
V = constant x n
V1/n1 = V2/n2
5.3

13. Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT
5.4

14. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)
5.4

15. Solve for n, which = mass/molar mass
Density (d) Calculations
m is the mass of the gas in g m V = PM RT
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
Solve for n, which = mass/molar mass
dRT P
Or,
M =
d is the density of the gas in g/L
5.4

16. PV = nRT n = 0.0852 mol g n = M 2.10 L X 1 atm n = 0.0821 x 300.15 K
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas?
PV = nRT
2.10 L
X 1 atm
n =
L•atm
mol•K
x K
n = mol g
M =
54.6 g/mol
n = M
5.3

17. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L
5.5

18. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

19. PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
mole fraction (Xi) = ni nT
5.6

20. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
5.6

21. PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6
Bottle full of oxygen gas and water vapor
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2
5.6

22. 5.6

23. Scuba Diving and the Gas Laws
Chemistry in Action:
Scuba Diving and the Gas Laws
Depth (ft)
Pressure (atm) 1 33 2 66 3 P V
5.6

24. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
5.7

25. Kinetic theory of gases and …
Compressibility of Gases
Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
Charles’ Law
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
P a T
5.7

26. Kinetic theory of gases and …
Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
P a n
Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the presence of another gas
Ptotal = SPi
5.7

27. Deviations from Ideal Behavior
1 mole of ideal gas
Repulsive Forces
PV = nRT
n = PV RT
= 1.0
Attractive Forces
5.8

28. Effect of intermolecular forces on the pressure exerted by a gas.
5.8

29. ( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2
corrected
pressure }
corrected
volume
5.8

30.  Velocity of a Gas 3RT urms = M The distribution of speeds
of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
urms =
3RT M 
5.7

31. Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
5.7

32. Gas Diffusion relation of mass to rate of diffusion
HCl and NH3 diffuse from opposite ends of tube.
Gases meet to form NH4Cl
HCl heavier than NH3
Therefore, NH4Cl forms closer to HCl end of tube.

33. GAS DIFFUSION AND EFFUSION
diffusion is the gradual mixing of molecules of different gases.
effusion is the movement of molecules through a small hole into an empty container.

34. GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv2
Rate of effusion is inversely proportional to its molar mass.
Thomas Graham, Professor in Glasgow and London.

35. GAS DIFFUSION AND EFFUSION
Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is
proportional to T
inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T. He

36. Graham’s Law Problem 1
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees celsius according to the equation:
4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?

37. Graham’s Law Problem 2
What is the rate of effusion for H2 if ml of CO2 takes 4.55 sec to effuse out of a container?

38. Graham’s Law Problem 3
What is the molar mass of gas X if it effuses times as rapidly as N2(g)?