كيمياء عامة1

1. CChhaapptteerr 55
GGaasseess

2.  Matter exist in three physical states: gas, liquid, solid.
 Few substances exist in gaseous state under typical conditions.
 Gases are very important…we live at the bottom of an ocean of air
solution (atmosphere) whose composition by volume is roughly
78% N2, 21% O2, and 1% other gases including CO2.
 The atmosphere both support life and act as a waste receptor and
also shield us from harmful radiation from the sun.
 In this chapter we will look carefully at the properties of gases
which lead to various types of laws.

3.  The following table shows some substances that exist as gases

4. Gases differ from solids and liquids in several aspects:
 fill any container uniformly and move freely within it,
so the volume of the gas = the volume of the container
 mixed completely with any other gases. when there
are several gases in a mixture, the volume of each
component is the same as the volume occupied by the entire mixture
highly compressible, when a gas is subjected to pressure, its volume
decreases.
The attraction forces between the molecules of gases are weak which
allow rapid, independent movement of the molecules and cause the
physical behavior of a gas to be nearly independent of its chemical
composition.
The behavior of a gas controlled by its volume, pressure, temperature,
and the number of moles.

5. PPrreessssuurree ooff aa GGaass::
Gases exert pressure on any surface with which they come in contact, because
gas molecules are constantly in motion.
Pressure is one of the most readily measurable properties of gas
Pressure define as force acting on an object per unit area.
P = F/A
F = mass x acceleration = kg. m/s = newton
P = (kg. m/s) / m2 = N/m2 = pascal
The SI unit of pressure is the pascal, which define as one newton per square meter.
The atmosphere surround the earth is a mixture of gases, it exert a pressure
uniformly on everything on earth.
The pressure of the atmosphere is measured with a device called a barometer.
It was invented in 1643 by Torricilli.
The barometer is constructed by filling a glass tube (about 1m long) with
liquid mercury (Hg) and inverting it in a dish of mercury as shown in the
following figure.

6. BBaarroommeetteerr
The pressure of the atmosphere
at sea level will hold a column
of mercury 760 mm Hg.
1 atm = 760 mm Hg =760torr
The SI unit of pressure is pascal
(pa): define as 1 Newton per
square meter
1 pa = 1 N/m2
1 atm.= 101325pa=101.325kpa
1 atm
Pressure
760 mm
Hg
Vacuum
Pressure
of Hg atm. Pressure = pressure of Hg
The Hg will stop get out from the
column

7. Measuring pressure of trapped gases
It is often desirable to measure the pressure of a gas
present in closed system, the instrument used is
called manometer.
Two types of manometers are used:
1) Open end manometer: it is simply a U tube
containing Hg, one arm of the tube is connecting to a
gas whose pressure is to be measured, while the
other arm is remain open to the atmosphere.
2) Closed end manometer
it is used to measure low pressure gases, it consist
of U-shaped tube with one arm sealed at the top.

8. Gas
Patm. Patm.
h
P gas = Patm. P gas Patm. P gas Patm.
Patm.
Gas Gas
h
Open end manometer
P gas = Patm - h P gas = Patm + h

9. gas
gas
P gas = Patm.
P = 0
P gas Patm.
PHg
Closed end manometer
vaccume

10. 10
EExxaammpplleess::
aa)) CCoonnvveerrtt 00..552277 aattmm ttoo ttoorrrr
PP ((ttoorrrr)) == 00..552277 aattmm xx
bb)) CCoonnvveerrtt 773300 ttoorrrr ttoo kkPPaa
PP ((kkPPaa)) == 773300 ttoorrrr xx
cc)) CCoonnvveerrtt 114477..22 kkPPaa ttoo ((11)) aattmm aanndd ((22)) ttoorrrr
11)) PP ((aattmm..)) == 114477..22 KKppaa xx
22)) PP ((ttoorrrr)) == 114477..22 KKppaa xx
760 torr = 400.52 torr
1 atm.
101.235 Kpa = 97.24 Kpa
760 torr
1 atm. = 1.45 atm.
101.235 Kpa
760 torr = 1105.1 torr
101.235 Kpa

11. 11
Using liquids other than Hg in barometer and manometer:
Lab. Manometers contain Hg rather than other liquids because its high
density allows the manometer to be convenient size.
If low pressures are to be measured, liquids other than Hg is used. This
is because Hg with its large densities gives very small differences in the
height of the columns in manometers.
If liquids of lower densities is used, the difference in the heights can be
much larger.
1 torr Hg = 1 mmHg = 13.6 mmHg
There is a very simple relationship between the height of the columns
of fluids in manometer (or barometer) and their densities
(height x density)liquid = (height x density)Hg
(h x d) liquid = (h x d)Hg

12. 12
Example:
A long glass tube, sealed at one end has an inner diameter
of 10 mm. the tube is filled with water and inverted into a
pail of water. If the atmospheric pressure is 760 mmHg.
How high (in mmH2O) is the column of water in the tube.
(density H2O = 1g/ml , density Hg = 13.5 g/ml)
(h x d) water = (h x d)Hg
h x 1 g/ml = 760 mmHg x 13.5 g/ml
h = 10260 mmH2O
h = 1026.0 cmH2O
h = 10.26 mH2O

13. 13
The Gas Laws:
Physical behavior (or state) of a gas can be describe by four
variables will affect the state of a gas:
- Temperature, (T) - Volume, (V) -Pressure, (P)
- Quantity of gas present, n (moles)
These variables are interdependent : any one of them can
be determine by measuring the other three.
We will consider several mathematical laws that relate
these properties of gases which known as gas laws.
These laws derived from experiments involving careful
measurements of the relevant gas properties.
From these experimental results, the mathematical
relationship among the properties can be discovered.

14. 14
Boyle’s Law(1662): P-V relationship
He use J-shape tube closed at one end, he study the
relationship between the pressure of the trapped gas and
its volume. Representative values from Boyle’s law are
given in the following table:
(Volume (in3 (P (inch Hg P x V
117.5 12.0 x 102 14.1
87.2 16.0 x 102 14.0
70.7 20.0 x 102 14.1
58.8 24.0 x 102 14.1
44.2 32.0 x 102 14.1
35.3 40.0 x 102 14.1
29.1 48.0 x 102 14.0

15. 15
These data show that the P x V of the trapped gas is
constant:
PV= k , k: constant depend on (n, T)
Boyle’s Law: the volume of a fixed quantity of gas is
inversely proportional to its pressure.
•Mathematically:
V = constant x
1
P
PV = constant
P1V1 = P2V2
•A plot of V versus P is a hyperbola.
•Similarly, a plot of V versus 1/P must be a straight line
passing through the origin.
•The Value of the constant depends on the temperature and
quantity of gas in the sample.

16. 16
Slope = k

17. 17
PV
P (at constant T)
Ideal
CO2
22.41 L atm
Ne
O2
CO2
Boyle’s law holds precisely at very law pressures.
Measurement at higher pressures reveals that PV is not constant,
but varies as the pressure varies as shown in this figure.
A gas that obey Boyle’s law is called ideal gas

18. 18
Example (1):
20.5 L of nitrogen at 25ºC and 742 torr are compressed to 9.8
atm at constant T. What is the new volume?
9.8 atm. = 9.8 atm. x
760 torr = 7448 torr
PV= PV1 atm.
11 22
742 torr x 20.5 L = 7448 torr x VV= 2.04 L
2 2 Example (2):
30.6 mL of carbon dioxide at 740 torr is expanded at constant
temperature to 750 mL. What is the final pressure in kPa?
P1V1 = P2V2 P =
P (kPa) = 30.19 torr x
30.6 ml x 740 torr
750 ml
= 30.19 torr
101.235 kPa = 4.02 kPa
760 torr

19. 19
Charles’s law (1787): T-V relationship
he found that the volume of a gas at constant pressure increase
linearly with the temperature of the gas.
A plot of the volume of a gas (at constant pressure) versus temp. (ºC)
gives a straight line.
A very interesting feature of this plot is that the volume of all gases
decrease until the gas occupy a theoritical zero volume at -273.15
ºC. This point represents the temp. at which all gases if they did not
condense would have a volume zero
Kelvin used this linear relationship between gas volume and temp. to
device the absolute temp. scale.
In this scale, the absolute zero (0 K or -273ºC) is the temp. at which an
ideal gas would have zero volume.
K = ºC + 273
0 K is called absolute temp., it has never been reached

20. A plot of volume versus temp.
20 V (L) T (ºC)
He
CH4
H2O
H2
-273.15ºC

21. 21
Charles’s law: the volume of a gas is directly proportional
with the absolute temperature at constant pressure.
V = kT (T is in Kelvin)
V1
T1
= V2
T2

22. 22
EExxaammppllee ((11))::
WWhhaatt wwoouulldd tthhee ffiinnaall vvoolluummee bbee iiff 224477 mmLL ooff ggaass aatt 2222ºCC iiss
hheeaatteedd ttoo 9988ºCC ,, iiff tthhee pprreessssuurree iiss hheelldd ccoonnssttaanntt..
=
V1
T1
V2
T2
Example(2):
V2 = 247 ml x 371 K
295 K
= 310.6 ml
At what temperature would 40.5 L of gas at 23.4ºC have a volume of
81.0 L at constant pressure?
V1 = V2
40.5 L
TT=
1 2 296.4 K
T2
80.0 L
T2 = 80.0 L x 296.4 K
40.5 L
= 585.5 K

23. 23
Avogadro’s law:
Avogadro postulate that: equal volumes of gas at the same
temperature and pressure will contain the same number
of molecules.
V α n (n is the number of moles)
V = k n (k is constant)
Avogadro’s Law: the volume of gas at a given temperature
and pressure is directly proportional to the number of
moles of gas.
V1
V2
= n1
n2
This relationship is obey closely by gases at low pressure

24. 24
The ideal gas law
We have considered three laws that describe the behavior of gases.
Boyle’s law: V α 1/P (at constant T and n)
Charles’s law: V α T (at constant P and n)
Avogadro’s law: V α n (at constant T and P)
These relationships can be combine as follows:
T n
P
V α V= R ( )
R is the combine proportionality constant called the universal gas
constant. R = 0.0821 L. atm./K mol.
The preceding equation can be rearranged to the more familiar form
of the ideal gas law:
PV = nRT
T n
P

25. 25
The ideal gas law is an equation of state for a gas, where the
state of the gas is its condition at a given time.
The gas that obeys this equation is said to behave ideally.
Ideal gas is a hypothetical gas that would obey this equation
over all ranges of T and P, real gases deviate from ideal
behavior but many of them come quite close to being ideal
at low P( 1atm.) and high T.
 Knowledge of any three of these properties is enough to
define the state of a gas, since the fourth property can be
determined from the equation.
 The ideal gas law is also used to calculate the changes that
will occur when the conditions of the gas are changed

26. 26
Example (1):
A 47.3 L container containing 1.62 mol of He is heated
until the pressure reaches 1.85 atm. What is the
temperature
PV = nRT 1.85 atm. T x 47.3 = PV/ L
nR
= 658 K
T =
1.62 mol x 0.0821 (L. atm./mol K)
Example (2):
Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at
24.8ºC What is the mass of Kr (molar mass = 83.8
g/mol).
8.61 atm. x 18.5 L
= 6.93 mol
PV = 0.0821 nRT (L. atm. /mol K) n x = 279.8 PV / K
RT
n =
Mass of K (g) =6.93 mol x 83.8 g/mol = 581 g

27. 27
Example (3):
Calculate the number of moles of a gas that occupy1.50 L
at 37 oC and 725 torr;
P(atm) = 725 torr. x
1 atm. = 0.954 atm.
760 torr
T (K) = 37 + 273 = 310 K
n = P V
R T =
0.954 atm. x 1.5 L
0.0821 (L. atm./mol K) x 310 K
n = 0.056 mol

28. 28
The ideal gas law is also used to calculate the changes that
will occur when the condition of the gas are changed.
In general, if we have aa ggaass uunnddeerr ttwwoo sseettss ooff ccoonnddiittiioonnss,,
tthheenn::
P1V1
P2V2
= n1T1
n2T2
Example (1):
A sample of gas has a volume of 4.18 L at 29ºC and 732
torr. What would its volume be at 24.8ºC and 756 torr?
n: is constant (the same sample of a gas)
P1V1
P2V2
T1
T2
= V2
=
P1V1T1
P2T2
V2 = 732 torr. x 4.18 L x 303 K
756 torr. x 297.8 K
= 4.12 L

29. 29
We define STP (standard temperature and pressure) = 0°C,
273.15 K, 1 atm.
Volume of 1 mol of gas at STP is:
V = n R T
P =
1 mol x 0.0821 L. atm./mol K) x 273 K
1 atm.
V = 22.42 L
This volume is called the molar volume of ideal gas
because it is the volume of 1.0 mol under these
conditions

30. 30
The ideal gas equation can be used to determine:
 the density of a gas.
 the molar mass of the gas
 the volume of gases formed or consumed in chemical reaction.
Molar mass and the density of a gas:
One very important use of ideal gas law is in the calculation of the
molar mass of a gas from its density.
n (number of moles) = mass (g)
Molar mass (g/mol)
=
m
Molar mass (g/mol)
Substituting into the ideal gas equation
V = (m/molar mass) RT
P = n RT
V
= m R T
V x molar mass
Density (d) = mass (g)
Volume (L)= M (g)
V(L)

31. 31
d R T
molar mass
P = d =
P x molar mass
R T
d RT
P
molar mass = or molar mass =
Example (1):
What is the density of ammonia, NH(molar mass = 17g/mol)
3at 23ºC and 735 torr.
P(atm.) = 735 torr. x
1 atm.
760 torr. = 0.967 atm.
T (K) = 273 + 23ºC = 296 K
d =
P x molar mass
R T
= 0.967 atm. x 17 g/mol
0.0821 (L.atm./mol K) x 296 K
= 0.676 g/L
m RT
P V

32. 32
Example (2):
A compound has the empirical formula CHCl. A 256 mL
flask at 100.ºC and 750 torr contains 0.80 g of the gaseous
compound. What is the molecular formula?
P= 750 torr. x T(K) =273+100
Molar mass =
1 atm.
760 torr.
= 0.987 atm.
m RT
P V
0.8 g x 0.0821 L.atm/mol K x 373 K
=
0.987 atm. x 0.256 L
Molar mass = 97 g/mol
X = molar mass compound
Molar mass E.F.(CHCl)
= 97
48.5
= 2
Molecular formula is 2(CHCl) , C2H2Cl2

33. 33
Gas stoichiometry
Many chemical reactions involve gases, by assuming ideal behavior
for these gases we can carry out stoichiometric calculations if P, V,
T of the gases are known.
Reactions happen in moles
At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1
mole of gas occupies 22.42 L.
If not at STP, use the ideal gas law to calculate moles of reactant or
volume of product.
For the reaction:
2 H2(g) + O2(g) 2 H2O(g)
2 mol 1 mol 2 mol
2 x 22.4 L 1 x 22.42 L 2 x 22.42 L
2 x volume H2 1 x volume O2 2 x volume H2O
2 : 1 mol ratio of H2 : O2 mean 2 : 1 volume ratio of H2 : O2

34. 34
Example(1):
Consider the following reaction
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
A) What volume of H2O at 1.0 atm and 1000ºC can be
produced from 10.0 L of NH3 and excess O2 at the same
temperature and pressure?
B) What volume of O2 measured at STP will be consumed
when 10.0 kg NH3 (molar mass = 17g/mol) is reacted?
Volume of H2O = 10.0 L NH3
x
A)
6 L H2O = 15.0 L H2O
4 L NH3
B) Mol of NH3 = 10 x 103 g NH3
17 g/mol
= 588.2 mol NH3
V(NH3) =588.2 mol x 0.0821 (L. atm/mol K) x 273 K
1 atm. = 1.32x104 L
Volume of O2= 1.32x104 L NH3 x 5 L O2
4 L NH3
= 1.65 x 104 L

35. 35
Example(2):
Mercury can be achieved by the following reaction
Heat
2HgO 2Hg(ℓ) + O2(g)
A) What volume of oxygen gas can be produced from 4.10 g
of mercury (II) oxide (molar mass = 216.5 g/mol) at STP.
B) At 400.ºC and 740 torr.
mol of HgO = 4.1 g
216.5 g/mol = 0.019 mol HgO
V(O2) = 0.019 mol x 0.0821 (L. atm/mol K) x 273 K
1 atm. = 0.426 L
A)
B) V(O2) =
0.0095 mol x 0.0821 (L. atm/mol K) x 673 K
(740/760) atm.
= 0.539 L
mol O2 = 0.019 mol HgO x 1 mol O2
2 mol HgO
= 0.0095 mol O2

36. 36
Dalton’s Law
When two or more gases do not react chemically are placed
in the same container, the pressure exert by each gas in the
mixture is the same as it would be if it were the only gas in
the container.
The pressure exerted by each gas in the mixture is called its
partial pressure and observed by Dalton and called Dalton’s
law of partial pressure
The total pressure in a container is the sum of the pressure
each gas would exert if it were alone in the container.
The total pressure is the sum of the partial pressures.
PTotal = P1 + P2 + P3 + P4 + P5 ...
For each: P = nRT/V

37. 37
PT = n1 RT + n2 RT + n3 RT + n4 RT V V V V
PT = (n1 + n2 + n3 + n4+….)
+ ………
RT
V
PT = nT RT
V
nT :sum of the number of moles of all gases
The pressure exert by an ideal gas is not affected by the identity
(composition) of the gas particles.
The partial pressure of a gas is related quantitatively to the total
pressure by its mole fraction.
Mol fraction (x): the ratio of the number of moles of a given
compound in a mixture to the total number of mol of mixture
x = n1 nT
= n1 n1 + n2 + n3 + n4+….)
x = P1V/RT
(P1V/RT) + (P2V/RT) + (P3V/RT) +….

38. 38
x =
P1(V/RT)
(P1 + P2+ P3 + P4..)(V/RT)
x = P1
P1 + P2 + P3 + P4…. = P1
PT
P1 = x PT
The partial pressure of a particular component of a
gaseous mixture is the mole fraction of that
component times the total pressure

39. 39
Example:
A) The partial pressure of nitrogen in air is 592 torr. Air
pressure is 752 torr, what is the mole fraction of nitrogen?
B) What is the partial pressure of nitrogen if the container
holding the air is compressed to 5.25 atm?
A) X=
= 0.787
N2 PN2
PT
= 592 torr.
752
torr.
B) PN2 = XN2 PT = 0.787 x 5.25 atm. = 4.133 atm.

40. Example:
The partial pressure of CH4 is 0.175 atm. and that of O2 is 0.25 atm. In a
mixture of two gases, calculate:
1. The total pressure of the mixture.
2. The mol fraction of each gas in the mixture.
3. The mass of each gas in the mixture if it occupy 10.5 L at 65ºC.
1. PT = PCH4 + PO2 = 0.175 atm. + 0.25 atm. = 0.425 atm.
2. XCH4 = (0.175/0.425)= 0.412 XO2 = (0.25/0.425)= 0.588
3. nT = P V
40
R T =
0.425 atm. x 10.5 L
0.0821 (L. atm./mol K)x 338 K = 0.161 mol
XCH4 =
nCH4 nT
nCH4 = 0.161 x 0.412 = 0.0663 mol
mass CH4 = 0.0663 mol x 16 g/mol = 1.061g
nO2 = 0.161 x 0.588 = 0.0947 mol
mass O2 = 0.0947 mol x 32 g/mol = 3.03 g

41. 41
3.50 L
O2
1.50 L
N2
example
4.00 L
CH4
2.70 atm
4.58 atm 0.752 atm
When these valves are opened, what is each partial pressure
and the total pressure?
(Pi Vi)CH4 = (Pf Vf)CH4
2.7 atm. x 4 L = P x 9 L
PfCH4 = 1.2 atm.
Pf N2 = 0.76 atm.
Pf O2 = 0.29 atm.
PT = 1.2 atm.+0.76 atm.+ 0.29 atm. = 2.25 atm.

42. 42
Collecting gas over water:
Water-insoluble gases prepared in lab. Are collected by
displacement of water as shown

43. 43
A gas collected in this manner become “contaminated” with
water molecules that evaporate into the gas, so the gas in
the bottle is a mixture of water vapor and the gas evolved
from the reaction.
Water vapor is present because water molecules escape
from the surface of liquid, these water molecules also exert
a pressure called the vapor pressure of water which depends
on the temperature of liquid water.
So PT = P(gas) + P(H2O)

44. 44
Example:
Helium is collected over water at 25ºC and 1.0 atm. total
pressure, what volume of gas must be collected to obtain
0.586 g He (molar mass= 4g/mol) . (PH2O = 23.8 torr.=
0.0313 atm. at 25ºC)
PT = P(He) + P(H2O)
P(He)= P(T) - P(H2O) P(He)= 1- 0.0313= 0.967 atm.
mol of He = (0.586g/4 g/mol) = 0.1465 mol
VHe = n RT/P
VHe =
0.1465 mol x 0.0821 (L. atm./mol K) x 298 K
= 3.69 L
0.967 atm

45. 45
Example:
N2O can be produced by the following reaction
heat
NH4NO3(s) N2O (g) + 2 H2O(ℓ)
what volume of N2O collected over water at a total pressure
of 94 kPa and 22ºC can be produced from 2.6 g of
NH4NO3(molar mass = 80g/mol)? ( the vapor pressure of water
at 22ºC is 21 torr)
PT = 94 kPa = 0.93 atm. PH2O =21 torr. = 0.028 atm.
PN2O = 0.93 – 0.028 = 0.902 atm.
mol NH4NO3 = (2.6 g/80g/mol) = 0.0325 mol
mol N2O = 0.0325 mol
V N2O = (n RT/P) = 0.0325 mol x 0.0821 (L. atm./mol K) x 295 K
0.902 atm
V N2O = 0.873 L

46. 46
EEffffuussiioonn aanndd DDiiffffuussiioonn
If two gases are placed in the same container their molecules
gradually mixed until the composition of the gas become
uniform. This process of mixing is called Diffusion.
Diffusion: is the term used for mixing gases.
A process similar to diffusion called effusion
Effusion: Passage of gas under pressure through a small
hole, into a vacuum.
The effusion rate measures how fast this happens.
Graham study the rate of effusion of various gases.
He found that the rate of effusion is inversely proportional to
the square root densities of gases (d) at the same T P .

47. 47
Rate of effusion α √ d
If you have two gases A and B, then
Rate of effusion of gas A
Rate of effusion of gas B = √
dB
dA
Molar mass = d R T
P
( molarmass )
B
( )
molarmass A
Rate of effusion for gas A =
Rate of effusion for gas B

48. 48
Example:
Which gas will effuse faster NH3 or CO2 , what are their
relative rates of effusion.
molar mass NH3 = 17 g/mol, molar mass CO2 = 44 g/mol
NH3 effuse faster (smaller molar mass)
Rate of effusion of NH3
Rate of effusion of CO2 = √
44/17 =1.6

49. 49
Example:
An unknown gas effuse at a rate that is 0.355 times that of
O2 at the same T P, what is the molar mass of this
unknown gas
rate of effusion of gas A = 0.355 rate of effusion of O2
Rate of effusion of gas A
Rate of effusion of gas O2
= 0.355 = √
32 g/mol
Molar mass (A)
Molar mass (A) = 254 g/mol