Chemistry
States of matter – Session 1
Session Opener
Session Objectives
Session Objectives
1. Definition and differences between
solids, liquids and gases.
2. Measurable properties of gas: Mass,
volume and temperature
3. Boyle’s law
4. Charles’ law
5. Avogadro’s hypothesis
6. Ideal gas equation
7. Dalton’s law
8. Amagat’s law of partial volume
9. Molecular mass of mixture of gases
Solids, Liquids and Gases
Gaseous state-assumptions
(1)The molecules are very loosely packed
voids
(2) Intermolecular forces
negligible.
(3) Molecules move very rapidly in all directions
in a random manner.
(4) The molecules collide with one another and also
with walls of the container perfectly elastic
collisions.
(5) The mass, volume and pressure of gases can be
calculated.
Measurable Properties of Gases
Mass
Number of moles 
Mass of the gas
Molecular weight of the same gas
Volume
Volume of a gas is the space occupied by its
molecules and is equal to the volume of the container.
1m = 10 dm = 100 cm
1m3 = 103 dm3 = 106cm3
1L = 1 dm3 = 103cm3
Measurable Properties of Gases
Pressure
Pressure is measured in terms of
atmospheric pressure, which is
the pressure exerted by the atmospheric
gases on the surface of the earth.
1 atm = 76 cm of mercury = 760 mm of mercury
1 atm = 101.325 kPa
Measurable Properties of
Gases
Centigrade or
Celsius scale (°C)
Temperature
Fahrenheit Scale (°F)
Kelvin scale (K)
100° C = 180° F
–273.15° C = 0 K
or 0° C = 273.15 K and t° C = (t + 273.15) K
STP or NTP
Temperature = 0° C = 273.15 K = 273 K
Pressure = 1 atm = 76 cm = 760 mm = 101.325 kPa
Question
Illustrative example 1
The temperature which is same on
Fahrenheit and Celsius
scale is
(a) 40
(b) 90
(c) 30
(d) none
Solution:
Let that temperature be x
C F  32

5
9
9C  5(F  32)
9x  5x  160
9x  5x  160
4x  160
x
160
4
x  40
Boyle’s Law
Boyle’s law : P  1
V
(at constant temperature and fixed mass)
P1V1=P2V2=constant [at a constant temperature]
Questions
Illustrative example 2
A sample of gas occupies 2 L under a
pressure of 800 atmosphere. What will
be its volume if the pressure is decreased
to 500 atmosphere. Assume that the
temperature of the gas sample does not
change?
Solution:
V2=?
V1=2 L
P1=800 atm
P2=500 atm
According to Boyle’s law
So,
V2 
P1V1=P2V2
P1V1 800  2

 3.2L
P2
500
Illustrative example 3
A gas–filled freely collapsible balloon is
pushed from the surface level of a lake
to a depth of 100m. Approximately what
percent of its original volume will balloon
finally have? Assuming that the gas
behaves ideally.
Solution:
Volume of the balloon at the surface of the lake  V1
P1  1atm  76 cm of Hg
 76  13.6  981 dynes / cm2
 981  1033.6 dynes / cm2
Solution
Pr essure(P2 ) at adepth of 100metre
 76  13.6  981  100  100  1  981
 981(76  13.6  10000)
 981  1033.6 dynes / cm2
Applying, P1V1  P2V2
981  1033.6  V1  981  1033.6  V2
1033.6
V2 
 V1
11033
1033.6  V1

 100%  9.367%
11033.6  V1
Charle’s law
Charle’s law: V  T (at constant pressure
and fixed mass)
V1
V2

( at cons tan t pressure)
T1
T2
Avogadro’s law
V n
(at constant pressure and temperature)
“equal volumes of all gases contain
equal number of molecules under similar
conditions of temperature and
pressure”.
Do you know?
Loschmidt number.
It is the number of
molecules present in 1cc of a gas or
vapour at STP. Its value is
2.617 x 1019 per cc
Ideal gas equation
Boyle’s law,
1
P
V
Charles’ law,
VT
Avogadro’s hypothesis,
Combining these, we get
PV  nT or PV  nRT
V n
Ideal gas equation
The gas constant, R
Units of gas constant (R)
R = 0.0821 L atm K–1 mole–1
R = 8.314 107 ergs K–1 mole–1
= 8.314 JK–1 mole–1
R = 1.987 cal K–1 mole–1
R=
8.314 eV k-1 mol-1
1.602×10-19
= 5.1891019 eVK–1mol–1
Do you know?
Gas constant per molecule is known as
Boltzmann constant (K)
R
 1.38  1010 ergs / K / molecule
N
or 1.38  1023 JK 1 molecule1
K
Density and molar mass
relation
Number of moles of gas :
PV
n=
RT
If Wg of a gas of molecular mass M, occupies a volume
W
V, under pressure P at temperature T, then n 
M
W PV

M RT
W
P  
V
RT
M
PM
Density  
RT

RT
M
Vapour density
Wgas
PVM

RT
PV2
WH 
2
RT
Wgas
WH2
M

2
[Since molecular weight of H2 is 2]
= vapour density of a gas
Dalton’s Law
a moles of He
P
He
aRT

V
Dalton’s Law
b moles of O2
P
O2
bRT

V
Dalton’s Law
c moles of CO2
P
CO2
cRT

V
Dalton’s Law
Total gas pressure,
P
total
P
Total
Dalton’s law
of partial
pressure
RT
V
 (a  b  c)
P
He
P
O
2
P
CO
2
Question
Illustrative example 4
At 27° C a cylinder of 20 litres capacity
contains three gases He, O2 and N2
0.502 g, 0.250 g and 1.00 g
respectively. If all these gases behave
ideally, calculate partial pressure of each
gas as well as total pressure.
Solution:
Let number of moles of He, O2 and N2 be n1, n2, n3
respectively.
 n1 
0.502
 0.1255 moles of He
1
4.0 g mole
Solution
n2 = 0.25 g/32.0 g mole–1
= 0.0078 moles of O2
n3 = 1.00 g/28.0 mole–1
= 0.0357 moles of N2
Total number of moles in the gaseous mixture
n = 0.1255 + 0.0078 + 0.0357 = 0.169 moles
From gas equation,
nRT
P
V
(0.169 mole) (0.0821 L atm K 1 mole1 ) (300 K)

20 L
= 0.208 atm
Solution
Partial pressure of helium
0.1255
PHe 
 0.208
0.169
= 0.1545 atm
0.0078
PO 
 0.208
2
0.169
= 0.0096atm
PN
2
0.0357
 0.208
0.169
= 0.044 atm

Applications of Dalton’s law of
partial pressure
a. To determine the pressure of a dry gas
Pdry gas = pmoist gas – Aqueous Tension (at t° C)
b. To calculate partial pressure
In a mixture of non-reacting gases,
Partial pressure = mole fraction x total pressure
Question
Illustrative example 5
A certain quantity of a gas occupies
100 mL when collected over water
at 150C and 750mm pressure. It
occupies 91.9 mL in dry state at NTP.
Find the aqueous vapour pressure at
150C.
Solution:
P= aqueous vapour pressure
P1(dry gas)=750-p
P2=760 mm
V1=100 ml
V2=91.9 ml
T1=15+273=288 K
Solution
P1V1 P2V2

T1
T2
(750  p)  100 760  91.9

288
273
760  91.9  288
100  273
 736.8 mm
or 750  p 
p  736.8  750
 13.2 mm
Amagat’s Law of Partial Volume
V = VA + VB + VC + . . . . . . (in mixture of gases)
According to Amagat’s law,
VA
= mole fraction
V
Molecular mass of the mixture
of gases
For example, air contains 79%
nitrogen and 21% oxygen
approximately.Now mass of one
mole of air would be
(0.79 × 28) + (0.21 × 32) = 28.84 gm/mole.
 Mmixture 
 Mx
i i
= S Molecular mass of a gas × mole fraction of a gas
xi (mole fraction of a gas) =
ni
Moles of a gas

ni
Total moles of gas mixture
Class exercise
Class exercise 1
A gas at a pressure of 5 atm is
heated from 0° C to 546° C and
compressed to one-third of its
original volume. Hence final
pressure is
(a) 10 atm
(b) 30 atm
(c) 45 atm
(d) 5 atm
Solution:
we know,
P1V1 P2V2

T1
T2
5  V1
P2V1

273
3  819
 P2 = 45 atm.
Hence, the answer is (c).
Class exercise 2
What weight of CO2 at STP
could be contained in a vessel
that holds 4.8 g of O2 at STP
(a) 5.5 g
(b) 6.6 g
(c) 7.7 g
(d) 3.3 g
Solution
According to ideal gas equation
PV  nRT 
W
RT
M
PV WCO2

   (i)
RT
44
For O2,
PV 4.8

   (ii)
RT
32
Comparing both equations
WCO
44
2

4.8
32
 WCO  6.6 gm
2
Hence, the answer is (b).
Class exercise 3
The volume of helium is 44.8 L at
(a) 100° C and 1 atm
(b) 0° C and 1 atm
(c) 0° C and 0.5 atm
d) 100° C and 0.5 atm
Solution
At STP,
volume of 1 mole of helium = 22.4 L
New volume = 2 × 22.4 = 44.8 L
P1V1 P2V2

T1
T2
1  22.4 1  44.8

273
T2
(Keeping pressure constant)
T2 = 546 K = 273° C
1  22.4 P2  44.8

273
273
P2 = 0.5 atm
(Keeping temperature constant)
Hence, the answer is (c).
Class exercise 4
The density of O2 at NTP will be
(a) 1.43 g/L
(b) 1.45 g/L
(c) 1.55 g/L
(d) 1.59 g/L
Solution:
PV = nRT
PV 
w
RT
M
 PM = RT
PM
1  32
The density  

 1.43 g /L
RT 0.082  273
Hence, the answer is (a).
Class exercise 5
Which of the following gases
has the vapour density 14?
(a) O2
(b) CO2
(c) CO
(d) NO
Solution:
Molecular mass
Vapour density 
2
If vapour density = 14
 Molecular mass = 28
Hence, the answer is (c).
Class exercise 6
For a particular gas at NTP, the
pressure will be _______ at 100° C.
(a) 1.2 atm (b) 1.34 atm
(c) 1.4 atm(d) 1.9 atm
Solution:
We know,
P1
P
 2
T1 T2
P2
1

273 373
 P2 = 1.34 atm
Hence, the answer is (b).
Class exercise 7
100 ml of an ideal gas was
collected over water at 15° C and
750 mm of Hg pressure. The
volume of the dry gas is 91.9 ml
at NTP. Find the aqueous vapour
pressure at 15° C.
Solution:
Let the vapour pressure of water = p mm
Pressure of dry gas, P1 = (750 – p) mm
V1 = 100 ml
P2 = 760 mm
T1 = 288 K
V2 = 91.9 ml
T2 = 273 K
Solution
P1V1 P2V2
Since

T1
T2
750  p  100
288

p = 13.2 mm
760  91.9
273
Class exercise 8
The density of a mixture of CO
and CO2 is 1.5 g/L at 30° C and
730 torr. What is the composition
of the mixture.
Solution:
We know, PV = nRT
WRT
PV 
Mmix
RT
P
1.5  0.082  303  760
= 38.8 g

730
Mmix = 28x + (1 – x)44
Mmix 
38.8 = 28x + (1 – x)44
Solution
 x = 0.325
% of CO = 32.5
% of CO2 = 67.5
Thank you