1. CHAPTER 13 – GASES
PRESSURE – Force per unit area
Due to the constant bombardment of the inside walls of the container by the gas molecules
6A-1 (of 34)

2. Atmospheric pressure is measured with a BAROMETER
STANDARD PRESSURE:
760.0 mm Hg
760.0 torr
1.000 atm
6A-2

3. Close-Ended Manometer
Gas pressure is measured with a MANOMETER
Close-Ended Manometer
125 mm more than 0
125 mm
Open-Ended Manometer
125 mm more than 765 mm
890. mm
6A-3

4. BOYLE’S LAW – The volume of a gas is inversely proportional to its pressure, provided the temperature and quantity of gas do not change
V  1
___ p
V = k
___ p
pV = k
p1V1 = k
p2V2 = k
p1V1 = p2V2
6A-4

5. 40. 0 mL of oxygen gas are collected with a pressure of 250. mm Hg
40.0 mL of oxygen gas are collected with a pressure of 250. mm Hg. Calculate the volume of the gas if the pressure changes to 300. mm Hg.
p1V1 = p2V2
p1 =
V1 =
250. mm
40.0 mL
p2 =
V2 =
300. mm
? mL
p1V1 = V2
______ p2
= (250. mm Hg)(40.0 mL)
_____________________________
(300. mm Hg)
= mL
6A-5

6. TEMPERATURE – A measure of the average kinetic energy of molecules

7. The lowest possible temperature is the temperature where molecular motion stops
ABSOLUTE SCALE – One with 0 as the lowest possible value
KELVIN SCALE (K) – A temperature scale where 0 is the lowest possible temperature
K = Cº
ABSOLUTE ZERO – The lowest possible temperature
6A-7

8. Change to K:
42 ºC
-22.0 ºC
42 ºC = K
-22.5 ºC = K
Change to ºC:
150. K
300.5 K
150. K = ºC
300.5 K = ºC
Standard temperature:
0.0 ºC
273.2 K
6A-8

9. T must be measured in Kelvin so that when T = 0, V = 0
CHARLES’ LAW – The volume of a gas is directly proportional to its temperature, provided the pressure and quantity of gas do not change
V  T
V = kT
V = k
___ T
V1 = k
___ T1
V2 = k
___ T2
V1 = V2
___ ___
T T2
T must be measured in Kelvin so that when T = 0, V = 0
6A-9

10. 15. 0 mL of hydrogen gas are collected at 10. ºC
15.0 mL of hydrogen gas are collected at 10.ºC. Calculate the volume of the gas if the temperature changes to 40.ºC.
V1 = V2
___ ___
T T2
V1 =
T1 =
15. 0 mL
283 K
V2 =
T2 =
? mL
313 K
V1T2 = V2
______ T1
= (15.0 mL)(313 K)
_____________________
(283 K)
= mL
6A-10

11. Combining Charles’ and Boyle’s Laws:
V1 = V2
___ ___
T T2
p1V1 = p2V2
p1V1 = p2V2
______ ______
T T2
6A-11

12. 30. 0 mL of a gas are collected at 22ºC and 750. torr
30.0 mL of a gas are collected at 22ºC and 750. torr. Calculate the volume of the gas at standard temperature and pressure.
273.2 K, torr
p1V1 = p2V2
______ ______
T T2
p1 =
V1 =
T1 =
750. torr
30.0 mL
295 K
p2 =
V2 =
T2 =
760.0 torr
? mL
273.2 K
p1V1T2 = V2
_________
p2T1
= (750. torr)(30.0 mL)(273.2 K)
____________________________________
(760.0 torr)(295 K)
= mL
6A-12

13. AVOGADRO’S LAW – The volume of a gas is directly proportional to the quantity of gas, provided the pressure and temperature of gas do not change
V  n
V = kn
6A-13

14. V  n Avogadro’s Law
V  nT
____ p or
pV  nT
V  1 Boyle’s Law
___ p
V  T Charles’ Law
6A-14

15. IDEAL GAS LAW EQUATION
pV = nRT
p = pressure
V = volume
n = quantity
T = temperature
R = Universal Gas Constant
(atm)
(L)
(mol)
(K)
( Latm/molK)
6A-15

16. pV = RTn
6A-16

17. Calculate the Celsius temperature of 0. 100 moles of a gas in a 700
Calculate the Celsius temperature of moles of a gas in a 700. mL container at a pressure of 2,500. torr.
pV = nRT
pV = T
____ nR
2,500. torr
x atm
_____________
760.0 torr
= atm 5
( atm)(0.700 L)
_________________________________________
(0.100 mol)( Latm/molK)
= K
281 K – = 8ºC
6A-17

18. Calculate the pressure of 2. 00 moles of nitrogen gas in a 5
Calculate the pressure of 2.00 moles of nitrogen gas in a 5.00 liter container at 20.ºC.
pV = nRT
p = nRT
_____ V
20.ºC
= K
(2.00 mol)( Latm/molK)(293 K)
________________________________________________
(5.00 L)
= atm
6A-18

19. PARTIAL PRESSURES
DALTON’S LAW OF PARTIAL PRESSURES – The total pressure of a mixture of gases is the sum of the pressures exerted by each gas
pN2 = torr
pO2 = 161 torr
pAr = 8 torr ___________
760. torr
PARTIAL PRESSURE – The pressure a gas would exert if it were alone in a container
6A-19

20. COLLECTING GASES IN LAB
To determine the quantity of gas collected in lab, you must measure the
(1) gas volume
(2) gas temperature
(3) gas pressure
pV = n
____ RT
EUDIOMETER – A gas collecting tube
6A-20

21. Atmospheric pressure: 765 mm Hg
Gas pressures?
1st – Hg level THE SAME in and out
∴ pressure inside eudiometer = pressure outside
∴ pressure of gas = atmospheric pressure
p = 765 mm Hg
6A-21

22. Atmospheric pressure: 765 mm Hg
Gas pressures?
2nd – Hg level HIGHER inside
∴ pressure inside eudiometer < pressure outside
∴ pressure of gas < atmospheric pressure
p = 765 mm Hg – 13 mm Hg = 752 mm Hg
6A-22

23. Atmospheric pressure: 765 mm Hg
Gas pressures?
3rd – Hg level LOWER inside
∴ pressure inside eudiometer > pressure outside
∴ pressure of gas > atmospheric pressure
p = 765 mm Hg + 6 mm Hg = 771 mm Hg
6A-23

24. COLLECTING GASES BY WATER DISPLACEMENT
Collecting O2 gas
When a gas is collected by water displacement, some water evaporates into the eudiometer
The eudiometer contains both oxygen gas and water vapor
If the total pressure is measured, the partial pressure of the water vapor must be subtracted to get the partial pressure of the oxygen gas
The partial pressure of the water vapor (called WATER VAPOR PRESSURE) depends only on temperature, and can be looked up
6A-24

25. Atmospheric pressure: 765 mm Hg
Temperature: 25ºC
Gas pressures?
1st – H2O level THE SAME in and out
∴ p of O2 (g) + p of H2O (g) = atmospheric pressure
∴ p of O2 (g) = atmospheric pressure – p of H2O (g)
Water Vapor Pressure at 25ºC = 24 mm Hg
p = 765 mm Hg – 24 mm Hg = 741 mm Hg
6A-25

26. Atmospheric pressure: 765 mm Hg
Temperature: 25ºC
Gas pressures?
2nd – H2O level HIGHER inside
∴ p of O2 (g) + p of H2O (g) < atmospheric pressure
Water level differences are converted to mercury level differences by dividing by the ratio of mercury’s density to water’s density: 13.6
27 mm H2O ÷ 13.6 = 2.0 mm Hg
p = 765 mm Hg – 2.0 mm Hg
– 24 mm Hg = 739 mm Hg
6A-26

27. Atmospheric Pressure = AP Water Vapor Pressure = WVP
1st – AP
2nd – AP – h
3rd – AP – WVP
4th – AP – (h÷13.6) – WVP
6A-27

28. VOLUME CALCULATIONS IN CHEMICAL REACTIONS
Calculate the mass of magnesium metal that would react with sulfuric acid to produce 250. mL of hydrogen gas at 23ºC and 1.05 atm.
Mg H2SO →
MgSO H2
x g
250. mL
1 mol
1 mol
mol
mol
pV = n
____ RT =
(1.05 atm)(0.250 L)
____________________________________
( Latm/molK)(296 K)
= mol H2 1
mol H2
x 1 mol Mg
_____________
1 mol H2
x g Mg
______________
1 mol Mg
= g Mg
6A-28

29. Calculate the mass of carbon that must be burned to produce 4
Calculate the mass of carbon that must be burned to produce 4.50 L of carbon dioxide gas at 29ºC and atm.
C O →
CO2
x g
4.50 L
1 mol
1 mol
0.179 mol
0.179 mol
pV = n
____ RT =
(0.987 atm)(4.50 L)
____________________________________
( Latm/molK)(302 K)
= mol CO2 2
mol CO2
x 1 mol C
_____________
1 mol CO2
x g C
____________
1 mol C
= g C
6A-29

30. Calculate the volume of hydrogen gas produced at STP by the reaction of 3.29 g of zinc metal with hydrochloric acid.
Zn HCl → 2
ZnCl H2
3.29 g
x L
1 mol
1 mol
mol
mol
3.29 g Zn
x 1 mol Zn
_____________
65.38 g Zn
x 1 mol H2
____________
1 mol Zn
= mol H2 2
V = nRT
_____ p =
( mol)( Latm/molK)(273 K)
_____________________________________________________
(1.00 atm)
= L H2
6A-30

31. Calculate the volume of oxygen gas at 25ºC and 745 torr needed to burn 5.00 g of propane, C3H8.
C3H O → 5 3
CO H2O 4
5.00 g
x L
1 mol
5 mol
0.113 mol
0.567 mol
5.00 g C3H8
x 1 mol C3H8
________________
44.11 g C3H8
x 5 mol O2
______________
1 mol C3H8
= mol O2 8
V = nRT
_____ p =
( mol)( Latm/molK)(298 K)
____________________________________________________
( atm)
= L O2
745 torr
x atm
____________
760.0 torr
= atm 3
6A-31

32. Calculate the volume of air at 25ºC and 745 torr needed to burn the 5
Calculate the volume of air at 25ºC and 745 torr needed to burn the 5.00 g of propane, if air is 20.9% oxygen by volume.
air =
20.9% O2
100 L
air =
20.9 L O2
14.14 L O2
x L air
___________
20.9 L O2
= L air
6A-32