GENERAL GAS LAW
From the gas laws, we know that
the volume of a gas depends on
both its temperature and pressure.
The relationship between the three
variable; i.e. volume, temperature
and pressure can be summarized up
as follows:
If V ∝ P1 (Boyle’s law at constant
temperature) and V ∝ T (Charle’s law
at constant pressure)
V ∝ P1 × T (both temperature and
pressure may vary) or PVT = K (a
mathematical constant for a fixed
mass of gas)
PV
T =k is often known as the general
gas equation.
GENERAL GAS EQUATION
General gas equation states
that for fixed mass of a gas under
any set of conditions of V, P and T,
the value of PVT must remain
constant. If for a fixed mass of gas
V1 is the volume at pressure
P1 and absolute temperature T1 and
V2 is the volume at pressure P2 and
absolute temperature T2 it follows
that.
P1V 1
T = P T2 V2 2
The general gas equation can be
used to find the volume of a gas
when both its pressure and
temperature change. Thus;
V2 = P P1 V2 V1T2 2
The standard temperature and
pressure
The value of gases is sometimes
given in standard temperature and
pressure (S. T. P). These values are
standard temperature= 273k and
standard pressure = 760mmHg. The
S.I unit of standard pressure when
used is 1.01 × 105Nm-2
Examples
1. At S. T. P a certain mass of gas
occupies a volume of 790cm3, find
the temperature at which the gas
occupies 1000cm3 and has a
pressure of 720mmHg
P1V 1
T1 = P2V 2
T2
P1 = 760mmHg (at stp), V1=
790cm3
T1 = 273k (at stp), V2 =
1000cm3
P2 = 720mmHg
T2 = New Temperature
P 2 V 2T 1
∴ T2 = P1V 1
720× 1000 ×273
= 760× 790 = 327.4k
The new temperature of the gas is
327.4k given mass of gas occupies
850cm3at 320k and 0.92 × 103Nm-2
of pressure. Calculate the volume of
the gas at S.T.P.
P1V 1
T1 = P2V 2
T2
P1= 0.92 × 103Nm-2 T1=
320k
V1= 850cm3 P 2= SP =
1.01 × 103Nm-2
T2= 273k (at stp)
V2 = new volume of gas.
P 1 V 1T 2 0.92× 850 ×273
∴ V2 = P2T 1 = 3
1.01 ×10 ×320 = 660.5cm 3
EVALUATION
1. Explain the general gas equation.
2. If the volume of a given mass of a
gas at 298k and a pressure of 205.2
×10 Nm is 2.12dm . What is the volume
5 −2 3
of the gas S.T.P (standard
pressure= 101.3 ×103Nm-2, standard
temperature = 273k)?
PERIOD 4: IDEAL GAS LAW
The ideal gas: This is a gas sample
whose properties correspond, within
experimental error, to the
relationship PV =nRT. An ideal gas
must obey all the rules guiding
Boyle’s and Charles’s laws. Ideal
gas conforms to the kinetic theory
of gases. Four quantities are
important in all experimental work,
measurements or calculations
involving gases. They are: (i)
volume (ii) pressure (iii)
temperature and (iv) numbers of
moles
Ideal gas equation is given by PV =
nRT
Note: R is universal gas constant
The value of R for one mole of a
gas at 273K, 1atm and volume
22.4dm3 is 0.0821atmdm3K-1mol-
1
or 8.314JK-1mol-1
Examples:
1. Calculate the volume occupied
by 2.5moles of an ideal gas at -23oC
and 4.0atm. (R = 0.0821atmdm3K-
1
mol-1)
Solution:
Using PV = nRT where P = 4.0atm
n = 2.5 mole T = -23+273 = 250K
nRT 2.5 X 0.0821 X 250
Hence, V = P = 4
= 12.8dm3
NOTE: Pressure can also be
measured in other units.
760mmHg = 1atm = 101325Nm-
2
Ideal gases only exist at
experimental conditions of high
pressure and low temperature.
Basically, all gases are real
REASONS WHY REAL GASES
DEVIATE FROM IDEAL GAS
BEHAVIOUR
1. The forces of attraction in real
gases are not negligible.
2. The volume of real gases are not
negligible. Hence, real gases have
their own volume called excluded
volume.
3. Real gases undergo inelastic
collision
EVALUATION:
1. What is an ideal gas?
2. Write down the ideal gas
equation for n-mole of a gas.
GENERAL EVALUATION
OBJECTIVE TEST
1. A gas occupies 30.0dm at S.T.P.
3
What volume will occupy at 910C
and 52662.5 Nm . (a) 20.0dm (b) 40.0dm
−2 3 3
(c) 60.0dm (d) 76.96dm
3 3
2. Gases can be easily compressed
because. (a) the molecule are
relatively far apart (b) the molecule
are quite close together (c) the
molecule are very soft (d) the
molecules are in constant, rapid
motion
3. A give mass of gas occupies X cm 1
3
at Y1K. When the temperature is
changed to Y2K, the volume
becomes X2cm , the3
pressure
remaining constant. Which of the
following equations correctly
express the relationship between X 1
X2Y1andY2? (a) X Y1 = X2Y2 (b) YX 11 = YX 22
1
(c) X X2 =X Y2 (d)X = X2Y1Y2
1 1 1
4. Kelvin temperature can be
converted into Celsius temperature
by. (a) oC= K – 273 (b) k + 273 (c)
0 C+273
k (d) k +273
0C
5. What will be the new volume (v)
if the new pressure is halved and
the initial pressure remain the
same. (a) 2 p V = p V (b) p V =2 p V (c) P 12V 1
1 1 2 2 1 1 2 2
= P 22V 2 (d) p V = P 22V 2
1 1
ESSAY QUESTIONS
1. 130cm of a gas at 200C exert a
3
pressure of 750mmHg. Calculate its
volume is increased to 150cm at 350C 3
2. Draw the graphical
representation of both Boyle’s and
Charles’ law respectively.
3. Convert the following
temperature to K. (a) 150C (b)2750C
(c) 880C
4. The volume of gas at 250C (298k)
is 100cm . What will be the volume at
3
(a.) 750C (348k) (b). 500C (223)k,
pressure remaining constant?
WEEKEND ASSIGNMENT:
Read about Graham’s law,
Avogadro’s number and the mole
concept.
DALTON’S LAW OF PARTIAL
PRESSURE
Dalton’s law of partial pressure
states that for a mixture of gases
that do not react chemically, the
total pressure exerted by the
mixture of gases is equal to the sum
of the partial pressures of the
individual gases.
Mathematically, Dalton’s law of
partial pressure for a mixture of
gases can be expressed as:
Ptotal = P1 +P2+P3 +………..+ Pn
where Ptotal is the total pressure
exerted by the mixture of gases
that do not react, P1, P2, P3……Pn are
partial pressure of the individual
gases.
Example:
If 20.0dm3 of hydrogen were
collected over water at 17oC and
79.7Nm-2 pressure; Calculate the
(a) Pressure of dry hydrogen at this
temperature.
(b) Volume of dry hydrogen at s.t.p.
(Vapour pressure of water is 1.90
Nm-2 at 17OC)
Solution:
(a) PH2 = Ptotal – Pwater vapour
= 79.7 – 1.90
= 77.8 Nm-2
P1V 1 P2V 2
(b) T1 = T2
77.8 X 20 101.3V 2
290 = 273
V2 = 14.5dm3
Example2: see review chemistry for
examples
Assignment
Give three different between
evaporation and boiling
