F.Sc. Part 1 Chemistry.Ch.03.Test (Malik Xufyan)

th
Maliks Chemistry -9 Class
(Board Paper-wise Test Series)
th
(Board Paper -wise Test Series)
Maliks Chemistry - F.Sc. l
Maliks Chemistry- F.Sc. ll
5
6
7
8
th
(Chapter-wise Test Series)
th
Maliks Chemistry - F.Sc. l
Maliks Chemistry- F.Sc. ll
1
2
3
4
Msc.Chemistry
CHEMISTRYCHEMISTRYCHEMISTRY
Chapter-wise
Test Series
Malik Xufyan
JIAS ACADEMY
Malik
Jhang Ins tute for Advanced Studies
0313-7355727
Our Other Publica ons
Jhang Ins tuteJhang Ins tuteJhang Ins tute
for Advance Studies
Jhang Sadar

CHEMISTRYCHEMISTRYCHEMISTRYCHEMISTRYCHEMISTRYCHEMISTRY
Malik Xufyan
JIAS ACADEMY
Malik
‫ﺧﻮﺷﺨﺒﺮی‬
Jhang Ins tute for Advanced Studies
Contact:
0313-7355727

Chapter # Topic Page #
Basic concepts – Unsolved paper1
2
3
4
5
6
7
8
9
10
11
5
6
16
17
24
25
36
37
45
46
54
55
64
65
73
74
83
84
93
94
103
104
Experimental Techniques – Solved Paper
Gases – Unsolved Paper
Gases – Solved Paper
Liquid – Unsolved Paper
Liquid – Solved Paper
Atomic Structure – Unsolved Paper
Atomic Structure – Solved Paper
Chemical Bonding –Unsolved Paper
Chemical Bonding – Solved Paper
Thermochemistry – Unsolved Paper
Thermochemistry – Solved Paper
Chemical Equilibrium – Unsolved Paper
Chemical Equilibrium – Solved Paper
Solu on – Unsolved Paper
Solu on – Solved Paper
Electrochemistry – Unsolved Paper
Electrochemistry – Solved Paper
Chemical Kine cs – Unsolved Paper
Chemical Kine cs – Solved Paper
Basic concepts – Solved Paper
Experimental techniques – Unsolved Paper
INDEX

Chapter#3:Gases Malik Chemistry
Jhang Institute for Advanced25
Name : CH # Gases Class 11th
Chemistry Test # 03 , CH #03
(Complete)
Marks : 85
Time : 2 & Half Hour Objective & Subjective JIAS Test System
Section 1: Objective
1. Choose the correct a nswer
Sr # Statement A B C D
1. Plasma is conductor of electricity Poor Bad Good None
2. The value of R is 8.214 8.314 Both None
3. The law which correlates the volume and
temperature of a gas is called
Boyle law Charles law Avogadro
law
Graham law
4. The temperature at which the volume of a
gas theoretically becomes zero is called
Absolute
temperature
Critical
temperature
Transition
temperature
Normal
temperature
5. Both Celsius and Fahrenheit scales are
intercontertable by using the formula
F=1.8o
C+32 F=1.8o
C-40 F=1.8o
C+212 None
6 Which of the following matters does not
show diffusion?
Solids Liquids Gas All
7. Diffusion of different species is due to
difference of
Potential energy Density Temperature All
8. ……. of known universe is in the plasma
state
59 % 99 % 5 % 2 %
9. Which of the following gas is ideal at
-200o
C?
nitrogen Helium Both None
10. Real gases deviate from the ideal behavior
at very
High pressure Low
temperature
Low pressure Both a and b
11. Which of the following expression is right? V= m/d PM = dRT PV = nRT All
12. Plasma is known as the 3rd
state of
matter
4th
state of
matter
2nd
state of
matter
None
13. The phenomenon in which sudden
expansion of a gas causes cooling is called
Joule Thomson
effect
evaporation Cooling sublimation
14. Density of a substance varies ……to
pressure at given temperature
Inversely Directly No change Can be any
15. Which of the following represents minimum
temperature
10 Centigrade 10 Fahrenheit 10 Kelvin All are equal
16. The colour of nitrogen dioxide is Yellow Green Brown Blue
17. The deviation of a gas from ideal behavior
is maximum at
-10o
C and 5atm -10o
C and
2atm
100o
C and
2atm
Oo
C and 2atm

Jhang Institute for Advanced Studeies 26
Section-ll: Short questions
2. Attempt only EIGHT questions 8 x 2= 16
i. Explain SO2
is comparatively non-ideal at 273 K but behaves
ideally at 327o
C.
ii. Explain water vapours do not behave ideally at 273 K.
iii. Explain the plot of PV versus P is a straight line at constant
temperature and with a fixed number of moles of an ideal gas.
iv. Hydrogen and helium are ideal at room temperature, but SO2
and Cl2
are non-ideal. How will you explain this?
v. How does kinetic molecular theory of gases explain the
Charles’s law?
vi. What do you know about aqueous tension?
vii. What is quantitative definition of Charles law also write formu-
la.
viii. Why Charles law is not being applied when temperature is mea-
sured on Celsius scale .also prove mathematically.
ix. Why -273.16 o
C are unattainable for real gases?
x. Why do we feel uncomfortable breathing at higher altitudes?
xi. Differentiate between diffusion and effusion of gas/
xii. What are the different values of ideal gas constant R? Derive its
value in SI units.
3. Attempt only 8 questions 8x2=16
i. Derive a relation for density of an ideal gas?
ii. What is an Avogadro LAW? Also explain graphically.
iii. What is Dalton’s la of partial pressure? Write mathematically.
iv. Why the mole fraction of mixture of anyone of gas is less than
unity?
v. The process of respiration can be explained by Dalton’s law
.how?
vi. State graham’s law of diffusion? Write formula.
vii. Write four points of kinetic molecular theory of gases.
viii. Which of the two points of kinetic theory of gases are faulty?
ix. Explain Avogadro’s law from kinetic theory of gases?
x. How heat flows from one body to another?
xi. What is the general principal of liquefaction?
4. Attempt any 6 questions 6x2=12
i. Define critical temperature and also write critical temperature
of some gases.
ii. What is joule Thomson effect?
iii. Why hydrogen and helium cannot be liquefied by Lind’s meth-
od of liquefaction?
iv. What is compressibility factor?
v. Why real gases deviate from ideality?
vi. How the various scales of thermometry can be inter- converted?
vii. Give some properties of gases?
viii. What is pressure? Give its different units.
ix. What are gas laws also state and derive Boyle’s law.
x. What is isotherm? Draw a graph of PV=K
Section-lll: Long Questions.
Attempt any three questions. 8 x 3 = 24
5) A) Write comprehensive note on plasma in detail?
b) Calculate the number of molecules and atoms
of methane gas taken at 0 O
c and 700mmHg
having volume 20 cm3
6) a) Derive van der waals equation for pressure correction.
b) 250 cm3
of the sample of hydrogen effuses four
times as rapidly as 250 cm3
of an unknown gas
.calculate the molar mass of an unknown gas?
7) a) Explain Lind’s method of liquefaction of gases also
draw diagram.
b) what pressure is exerted by a mixture of 2.00g of
H2
and 8.00g of N2
At 273K in a 10dm3
8) (a) What do you know about kinetic interpretation of
temperature?
(b) Calculate the masses of 1020
molecules of H2
, O2
and CO2
at STP. What will happen to the masses
of these gases, when the temperatures of these
gases are by 100 o
C and the pressure is decreased
by 100 torr.
9) (a) Write expression for pressure of an ideal gas By
R.J Clausius.
(b) 25 cm3
of the sample of hydrogen effuses four
times as rapidly as 250 cm3
of an unknown gas.
Calculate the molar mass of unknown gas
10) (a) What is Charles’s law? Explain it with experiment
and graphical explanation.
(b) A sample of krypton with a volume of 6.25 dm3
,
a pressure of 765 torr and a temperature of 20o
C
is expanded to a volume of 9.55 dm3
a pressure
of 375 torr. What will be its final temperature?

Section-Il: Short questions
2) Attempt only EIGHT questions 8 x 2= 16
i. Explain SO2
is comparatively non-ideal at 273 K but behaves ideally at 327o
C.
Ans:
At low temperature, the molecules of SO2
possess low kinetic energy. They come close to each other.
The intermolecular attractive forces become very strong. So, it behaves non-ideally at 273K.
At high temperature, the molecules of SO2
have high kinetic energy. The molecules are at larger dis-
tances from one another. The intermolecular attractive forces become very weak. So, it behaves ideally
at 327 K.
ii. Explain water vapours do not behave ideally at 273 k.
Ans:
Water vapors present at 273K do not behave ideally because polar water molecules exert force of attrac-
tion on one another. The intermolecular forces become strong.
iii. Explain the plot of PV versus P is a straight line at constant temperature and wit a fixed number of
moles of an ideal gas.
Ans:
At constant temperature and with a fixed number of moles of an ideal gas, when the pressure of the gas
is varied, its volume changes, but the product PV remains constant. Thus,
P1
V1
= P2
V2
= P3
V3
Hence, for any fixed temperature, the product PV when plotted against P. A straight line parallel to
P-axis is obtained. This straight line indicates that PV remains constant quantity.
iv. Hydrogen and helium are ideal at room temperature, but SO2
and Cl2
are non-ideal. How will you
explain this?
Ans:
Hydrogen (B.P: 253 o
C) and helium (B.P: 269 o
C)have a very low boiling points. They are far away
from their boiling points at room temperature. Also, they have smaller number of electrons in their
molecules and smaller molecular sizes, i.e., molecular weight. So, intermolecular forces are negligible
at room temperature. Hence, they behave as an ideal gases at room temperature.
On the other hand, SO2
(B.P:10 o
C) and C12
(B.P: 34 o
C) have boiling points near to room temperature.
They are not far away from their boiling points at room temperature. Also, they have larger number of
electrons in their molecules and larger molecular sizes. So, sufficient intermolecular attractive forces
are present at room temperature. Hence, they behave as non-ideal at room temperature.
v. How does kinetic molecular theory of gases explain the Charles’s law?
Consider the equation which has just been derived
vi. What do you know about aqueous tension?
Ans:

Aqueous tension:
Some gases are collected over water in the laboratory. The gas during collection gathers water vapours
and becomes moist. The pressure exerted by this moist gas is, therefore, the sum of the partial pressures
of the dry gas and that of water vapours. The partial pressure exerted by the water vapours is called
aqueous tension.
Formula:
Pmoist
= pdry
+ p w.vapour
Pmoist
= pdry
+ aqueous tension
Pdry
= Pmoist
- aqueous tension
vii. What is quantitative definition of Charles law also write formula.
It is quantitative relationship between temperature and volume of a gas and was give by French Scien-
tist J.Charles in 1787. According to this law, the volume of the given mass of a gas is directly propor-
tional to the absolute temperature when the pressure is kept constant.
Formula:
viii. Why Charles law is not being applied when temperature is measured on Celsius scale .Also prove
mathematically.
Ans:
Charles Law was defined on the basis of Kelvin scale. Because the value of k will not remain constant
on different temperatures if temperature is measured on Celsius scale.For example, if hypothetical gas
is warmed on different scale of temperature.
For Kalvin Scale For Celsius Scale
That’s why Charles law is not being applicable when temperature is measured on Celsius scale due to
above example in column.
ix. Why -273.16 o
C are unattainable for real gases?
Ans:

This temperature will be attained when the volume becomes zero. But for a real gas the zero volume is
impossible which shows that this temperature cannot be attained for a real gas. Actually, all the gases
are converted into liquids above this temperature. If real gas is converted into liquid at this temperature
then volume of gas cannot be zero.
x. Why do we feel uncomfortable breathing at higher altitudes?
Ans:At higher altitude, the pilots feel uncomfortable breathing because the partial pressure of oxygen in
the un-pressurized cabin is low as compared to 159 torr, where one feels comfortable breathing.
xi. Differentiate between diffusion and effusion of gas.
Ans:
Sr # Diffusion Effusion
1. Spontaneously intermixing of
molecules of one gas with another at
a given temperature and pressure is
called diffusion.
Effusion of a gas is movement through an
extremely small opening into region of low
pressure.
2. NO2
and O2
are intermixed
homogeneously after reacting them.
Effusion is when molecules escape from
their container into the space only when
they happen to hit the hole.
xii. What are the different values of ideal gas constant R? Derive its value in SI units.
Ans: If the pressure is expressed in mm of mercury or torr and the volume of the gas in cm3
then values
of R are :
R = 0.0821 dm3
atm K-1
mol-1
= 0.0821 x 760 dm3
mm Hg K-1
mol-1
= 62.4 dm3
mm Hg K-1
mol-1
since (1 mm of Hg =1 torr)
= 62.4 dm3
torr K-1
mol-1
= 62400 cm3
torr K-1
mol-1
since (1dm3
= 1000 cm3
)
Using SI units of pressure, volume and temperature in the general gas equation, the value of R is calcu-
lated as follows. The SI units of pressure are Nm-2
and of volume are m3
. By using Avogadro’s principle
1 atm = 7600 torr = 101 Nm-2
1 m3
= 1000 dm3
n = 1 mole
T = 273.16 K
P = 1 atm = 101325 Nm-2
Putting their values, alongwith units
3) Attempt only 8 questions 8 X 2 = 16
i. Derive a relation for density of an ideal gas?
Ans: for idea gas:

This equation is another form of general gas equation that may be employed to calculate the mass of a gas
whose P,T,V and molar mass are known. Rearranging equation:
Hence the density of an ideal gas is directly proportional to its molar mass.
ii. What is an Avogadro LAW? Also explain with example.
Ans:
Avagadro’s law : it states that
‘’ equal volume of the ideal gases at the same temperature and pressure contain equal number of mol-
ecules’’.
Example:
22.414 dm3
of a gas at 273.16 K and one atmospheric pressure has number of molecules = 6.02 x 1023
iii. What is Dalton’s law of partial pressure? Write mathematically.
Ans:
Dalton law of partial pressure:
John Dalton studied the mixture of gases his law of partial pressure. It states that:
‘’The total pressure exerted by a mixture of non-reacting gases is equal to the sum of their individual
partial pressures.’’
Mathematically:
Let the gases are designed as 1,2,3 and their partial pressure are p1
,p2
,p3
. The total pressure (P) of the
mixture of gases is given by
Pt
= p1
+ p2
+ p3
iv. Why the mole fraction of mixture of anyone of gas is less than unity?
Ans:
Because mole fraction of a component is the ratio of mole of component to the total number of moles
and we know that total number of moles are always greater than any of its component. So the formula
of mole fraction tells us the small value (of component) is divided by large value (total no .of moles) an
answer comes always equal to less than 1.
v. The process of respiration can be explained by Dalton’s law .how?
Ans:
Dalton law finds its applications during the process of respiration. The process of respiration depends
upon the difference in partial pressure. When animals inhale air then oxygen moves into lungs as the
partial pressure of oxygen in the air is 149 torr, while the partial pressure of oxygen in the lungs 116
torr, CO2
produced during respiration moves out in the opposite directions, as it’s partial pressure is

more in the lungs than that in air.
vi. State graham’s law of diffusion? Write formula.
Ans.
Thomas Graham, an English scientist, found that the rate of diffusion or effusion of a gas is inversely
proportional to the square root of its density at constant temperature and pressure.
Formula:
vii. Write four points of kinetic molecular theory of gases.
Ans:
Points of KMT:
1) The actual volume of molecules of a gas is negligible as compared to the volume of the gas
2) The molecules of a gas have not forces of attraction for each other
3) The molecules of a gas are very widely separated from one another and the sufficient empty spaces
among them.
4) Every gas consists of a large number of very small particles called molecules. Gases like He , Ne, Ar
have mono-atomic molecules
viii. Which of the two points of kinetic theory of gases are faulty?
Ans:
Faulty Points of KMT:
• The molecules of a gas have not forces of attraction for each other
• The actual volume of molecules of a gas is negligible as compared to the volume of the gas
ix. Explain Avogadro’s law from kinetic theory of gases?
Ans:
Avogadro’s law from KMT:
Consider two gases 1 and 2 with N1
& N2
(number of molecules), m1
& m2
are masses respectively.

According to kinetic equation:
Equalizing

when the temperature of both gases is the same, their mean kinetic energies per molecules will also be
same, so
Divide the equation (a) by (b)
N1
= N2
Hence , equal volume of all the ideal gases at the same temperature and pressure contain equal number
of molecules which is Avogadro’s law.
x. How heat flows from one body to another?
Ans:
When heat flows from one body to another, the molecules in the hotter body give up some of their
kinetic energy through collision to molecules in the colder body. This process of flow of heat continues
until the average translational kinetic energies of all the molecules become equal.
xi. What is the general principal of liquefaction?
Ans:
The conversion of a gas into a liquid requires high pressure and low temperature. High pressure brings
the molecules of a gas close to each other. Low temperature deprives the molecules from kinetic energy
and attractive forces start dominating.
For every gas there exist temperatures above which the gas cannot be liquefied, no matter how much
pressure is applied. The highest temperature at which a substance can exist as liquid is called its critical
temperature.
4) Attempt any 6 questions 6 x 2 =12
i. Define critical temperature. Also write critical temperature of some gases.
Ans:
Critical temperature:
The highest temperature, at which a substance can exist as a liquid, is called its critical temperature (Tc
).
Critical temperature of gases:
Water : 374.44 o
C
Oxygen : 118.75 o
C
Nitrogen : 147.06 o
C
Ammonia : 132.44 o
C
ii. What is joule Thomson effect?
Ans:
Definition:
When a compressed gas is allowed to expand into a region of low pressure it gets cooled.
The molecules of the compressed gas are very close to each other and appreciable attractive forces a
represent among term. When a gas is allowed to undergo sudden expansion through the nozzle of a

jet, then the molecules move apart. In this way energy is needed to overcome the intermolecular attrac-
tions. This energy is taken from the gas itself, which is cooled.
iii. Why hydrogen and helium cannot be liquefied by Lind’s method of liquefaction?
Ans:
Gases can be liquefying with the help of critical temperature. The value of critical temperature of Rea-
son a gas depends upon its size, shape and intermolecular forces present in it.
Helium and Hydrogen are small sizes gases.
Due to their very very small size as compared to other gases, these gases exist almost no force of attrac-
tion between their molecules .Hence they cannot be liquefied by Lind’s method.
Lind’s method is based on the Joule-Thomson’s effect. The liquefaction temperature of H2
is -252.7 o
C
and that of He is -268 o
C. These temperatures are very close to -273.16 o
C which is difficult to attain.
That’s why H2
and He cannot be liquefied by the Lind’s method.
iv. What is compressibility factor?
Ans:
Compressibility factor:
A graph is plotted between pressure on x-axis and on y-axis for an ideal gas. The factor is called
compressibility factor. Its value is unity under all conditions for an ideal gas. Since the increase of pres-
sure, the volume in such a way that remains constant at a constant temperature so straight line is
obtained parallel to the pressure axis.
i. Why real gases deviate from ideality?
Ans:
There are two basic reason:
a) When the pressure on a gas is high and the temperature is low then the attractive forces among the
molecules significant, so the ideal gas equation PV = nRT does not hold, actually under this condition
gas does not remain ideal.
b) The actual volume of the molecules of a gas usually very small as compared to the volume of the vessel
and hence it can be neglected. However, this volume does not remain negligible when the gas is sub-
jected to high pressure.
ii. How the various scales of thermometry can be inter- converted?
Ans:
Conversions: These are three conversion of scale of temperature
K= o
C + 273.16
o
C = [o
F – 32 ]
o
F = (o
C) + 32
iii. Give some properties of gases?
Ans:
Properties of gases:
a) Gases don’t have definite volume and occupy all the available space. The volume of a gas is the volume

of the container.
b) They don’t have definite shape and take the shape of the container just like liquids.
c) Due to low densities of gases as compared to those of liquids and solids, the gases bubble through liq-
uids and tend to rise up.
d) Gases can diffuse and effuse. This property is negligible in solids but operates in liquids as well.
e) The intermolecular forces in gases are very weak.
iv. What is pressure? Give its different units.
Ans:
Pressure:
It is the force exerted by 760 mm or 760 cm long column of mercury on an area of 1 cm2
at 0 o
C.
Units:
1) Atmosphere 2) torr
3) pascal 4) millibar
5) pounds inch-2
6) kilopascal
7) pounds per square 8) Nm-2
v. What are gas laws also state and derive Boyle’s law.
Gas laws:
The relationship between volume of a given amount of a gas and the prevailing conditions of tempera-
ture and pressure are called gas laws. Gas laws describe uniform behavior of gases.
Boyle’s law:
the volume of a given mass of a gas at constant temperature is inversely proportional to the gas.

V=
PV = k (when T and n are constant)
For initial and final condition
P1
V1
= k
P2
V2
= k
P1
V1
= P2
V2
Where
P1
V1
= initial condition
P2
V2
= final condition
vi. What is isotherm? Draw a graph of PV=K.
Ans:
The P-V curves obtained at constant temperat are called isotherms. These curves are ob-
tained by plotting a graph between pressure on the x-axis and volume on the y-a

Pressure (atm)
Volume (dm3
)
Isotherm at Oo
C
X
Y
Section-ll: Long Questions.
Attempt any three questions 8 x 3 = 24
5) a) Write comprehensive note on plasma in detail?
b) Calculate the number of molecules and atoms of methane gas taken at 0O
c and 700 mmHg having
volume 20 cm3
6) a) Derive van der waals equation for pressure correction.
b) 250 cm3
of the sample of hydrogen effuses four times as rapidly as 250 cm3
of an unknown gas.
Calculate the molar mass of an unknown gas?
6) a) Explain Lind’s method of liquefaction of gases also draw diagram.
b)what pressure is exerted by a mixture of 2.00g of H2
and 8.00g of N2
at 273 K in a 10dm3
7) (a) What do you know about kinetic interpretation of temperature?
(b) Calculate the masses of 1020
molecules of H2
, O2
and CO2
at STP. What will happen to the masses
of these gases, when the temperature of these gases are by 100 o
C and the pressure is decreased by
100 torr.
8) (a) Write expression for pressure of an ideal gas By R.J Clausius.
(b) 25 cm3
of the sample of hydrogen effuses four times as rapidly as 250 cm3
of an unknown gas.
Calculate the molar mass of unknown gas
9) (a) What is Charles’s law? Explain it with experiment and graphical explanation.
(b) A sample of krypton with a volume of 6.25 dm3
, a pressure of 765 torr and a temperature of 20o
C
is expanded to a volume of 9.55 dm3
a pressure of 375 torr. What will be its final temperature?

F.Sc. Part 1 Chemistry.Ch.03.Test (Malik Xufyan)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to F.Sc. Part 1 Chemistry.Ch.03.Test (Malik Xufyan)

Similar to F.Sc. Part 1 Chemistry.Ch.03.Test (Malik Xufyan) (20)

More from Malik Xufyan

More from Malik Xufyan (20)

Recently uploaded

Recently uploaded (20)

F.Sc. Part 1 Chemistry.Ch.03.Test (Malik Xufyan)