B.Sc II sem III Chemistry

1. .SHRI R.R.LAHOTI SCIENCE
COLLEGE,MORSHI
E-LEARNING MATERIAL/MODULE
CLASS – BSc II SEM III
TOPIC – LIQUID STATE
Mr .G.D.Rawate (MSc,SET,NET)
DPARTMENT OF CHEMISTRY
SESSION-2017-18

3. Characteristic properties of liquids arise from
the nature and magnitude of intermolecular
force

4. Surface
tension(ɤ)
Defination
Force in Newton acting on the
liquid surface at right angle to
any line 1 meter in length
Unit
In CGS - dyne/centimeter
In SI - Newton/Meter

5. Determination of surface tension by drop
number method
The no.of drop formed for fixed volume of liquid is determined by
stalgmometer
i) The liquid under examination
is sucked upto the mark A
ii) Definite no.of drop receive in
weighing bottle
iii) From this weight of single
drop is calculated
iv) The radius of capillary is r
V)Surface tension of liquid is ɤ
vi) Weight W equal to force due
surface tension W= 2πr ɤ

6. For relative determination of no. of drop falling
between to fixed level are counted for water and
experimental liquid
Number of drops falling between two fixed marks
n1- no. of drops for water (reference liquid)
n2- no.of drops for experimental liquid
d1- density of water
d2 – density of experimental liquid
ɤ 1 – surface tension of water
ɤ 2 - surface tension of experimental liquid
Weight of drops = Force due to surface tension
W = mg = Vdg = 2πr ɤ

7. .
When drops fall
Wt of drops = Force due to surface tension
W = mg = V dg = 2πr ɤ -----------(1)
For Water
2πr ɤ 1 = V1d1g ---------------(2)
For Experimental liquid
2πr ɤ 2 = V2d2g ---------------(3)
Multiplying equation 2 &3 by n1 & n2 respectively
2πr ɤ 1 n1 = n1 V1d1g ---------------(4)
2πr ɤ 2 n2 = n2V2d2g ---------------(5)
But n1v1 = n2v2 =V
Volume are same for both liquid
2πr ɤ 1 n1 = Vd1g ---------------(6)
2πr ɤ 2 n2 = Vd2g ---------------(7)
Dividing equation 7 by 6
ɤ 2 n1d2
ɤ 1 n2d1
ɤ 2 - suface tension of experimental liquid

8. Effect of temperature on surface tension
for all liquids , surface tension decreases with
increase in temperature due to increased
molecular agitation.
S.No. Liquids Surface tension Nm-1 x 102
273 K 293K 313K
1 Water 7.56 7.28 6.96
2 Benzene 3.16 2.89 2.63
3 Toluene 3.07 2.84 2.61
4 Acetone 2.62 2.37 2.12
5 Ethyl Alcohol 2.40 2.23 2.06

9. Application of surface tension
 Important in the study of emulsion and colloid.
 It is an essential factor in the concentration of ore
by froth floation process.
 Important in biological science, particularly in bacteriology.
 The movement of moisture in soil and passage of sap in plants
involve surface tension.
 To determine parachor value, additive or constitutive property
used to investigate molecular structure.
 To compare cleaning power of detergent.
 Identifying presence of air bubble in blood stream and presence of blie
salt in urine.

10. Viscosity (η )
Resistance to flow of liquid is called viscosity
It is developed due to shearing effectof one layer of liquid
past another
A layer contact with stationary surface remain stationary, the
second layer move slowly, third faster than second and so on
this type of flow is called as laminar flow or stream line flow.
Force required to maintain steady velocity difference bet two
layers dv
Two layers separated by distance dx
Area of contact A
f α A . (dv/dx)
f = η A . (dv/dx)
η = Coefficient of viscosity
A = 1sqm ,dv = 1msec-I ,dx = 1m
η = f

11. Coefficient of Viscosity (η )
force that must be exerted between two parallel
layers 1 m2 in area and 1 meter apart in order to
maintain velocity difference of 1m/sec.
UNIT OF VISCOSITY
In CGS η is g cm-1 s-1 it is called poise
centipoise (10-2 poise)
millipoise (10 -3 poise)
SI unit Kgm-1s-1

12. Measurement of Viscosity by
using Ostwald Viscometer
 Direct measurement of viscosity is difficult.
 The relative viscosity is determined by using reference
liquid (water), The apparatus used for measurement
of viscosity is called as Ostwald viscometer (fig 1)
 Ostwald viscometer is cylindrical tube having U shape
and two arms, Provided with two bulb
for arm1 Small bulb A and for arm II large bulb B
 Fine capillary is present below small bulb through
which liquid is flow.
 Definite quantity of liquid is placed in larger bulb , It is
sucked above upper mark .
 Time flow is measured between upper mark C to
lower mark D when liquid flows through the
capillary.
 Time flow t is directly proportional to coefficient of
viscosity and inversely propotional to the density
t α η /d
η = k t d ------------------(1)

13. For water
ηw = k tw dw ------------------(2)
From equation 1 & 2
η t d
ηw tw dw
Where
η --- Viscosity coefficient for liquid
ηw -- Viscosity coefficient for water
t - time flow for liquid
tw- time flow for water
d – density of liquid
dw – density of water

14. Effect of temperature
on viscosity
 Viscosity of liquid decreases with increase
in temperature
The variation of viscosity with
temperature is given by the eq.
η = A eEa/RT
A &E are constants for given liquid
Taking logarithms
ln η =lnA +Ea/RT
log η =logA +Ea/ 2.303RT
 Plot η Vs 1/T is straight line.
 Viscosity decrease with temperature
because molecular agitation increases.

15. Application of viscosity measurement
 Gradation of lubricant oil , In watches special kind of lubricant
needed which should not change its viscosity with temperature.
Yield information regarding movement of liquid through pipes
To determine molecular weight of polymer.
 knowing the constitution of molecules through Rheochor R
R =(dv/dx) η 1/8
.Which is both additive as well as constitutive
property.
When CO2 accumulated in blood, breathing becomes difficult. blood
corpuscles swell, raise viscosity of blood. This process quickly leads
to heart attack.
In heart patient doctor prescribe medicine to lower the viscosity of
blood.

16. Numericals
1) Water required 120.5 sec to flow through a viscometer and same
volume of acetone required49.5 sec. if the density of water and
acetone are293 kare9.982x102kg m-3 and7.92x102
kgm3respectively and coefficient of viscosity of water at293k
is10.05 pascal second, calculate coefficient of viscosity of
acetone at this temperature.
Solution - ta =49.5 s tw =120.5 s da = 7.92x102 dw =9.982x10 2
nw =10.05 na= ‽
η a ta da
ηl tl dl
η a = 49.5x 7.92 x102 x10.05/120.5x9.982x102
= 3.257 pascal sec.

17. Problem 2
The coefficient of viscosity for two liquids at1.408x10-3 and1.594x10-3
and their densities are8.07x102 and 10.17x 102 respectively. If the time
of flow for first liquid is 100 sec. calculate time of flow for second liquid
Solution-
η 1= 1.408 x 10 -3, η2 = 1.594x10-3. d1 = 8.07102, d2 = 10.17 x102
t1 = 100sec flow time of second liquid = ‽
η 2 t2 d2 t2 = (η2d1/ η1d2)xt1
η1 t1 d1
t2 = 1.594x10-3 x8.07x102 x100/1.408x10-3x10.17x102
t2 = 89.83 sec