Test Questions on
Chapter # 4, Gaseous State
Q1. State Avogadro’s law and Charles law.
Q2. Explain pressure and its various units.
Q3. What is Atmospheric
Pressure? On what factor Atmospheric Pressure depends upon? Give its effect on
weather. Explain low and High atmospheric
pressure system
Q4. What
is liquid air? Mention its uses.
Q5. State
the basic postulates of Kinetic Molecular Theory. Explain diffusion and
compressibility in three states of matter on the basis of KMT.
Q6. derive the General gas equation. Deduce the value of General gas constant (R) in:
(i) S.I. System (J/mol.K)
(ii) Non S.I. System (atm dm3/mol.K)
(iii) calorie
Q7. State and explain
Dalton’s law of partial pressure with its applications.
Q8. State and explain
Avogadro’s law.
Q9. What
is meant by diffusion and effusion? State and explain Graham’s law of diffusion
with its applications. Compare the rate
of diffusion of H2 and D2.
Q10. Differentiate
between ideal and non-ideal gas. What are the causes of deviation of real gas
from ideal behaviour? Write down Graphical Explanation
of Deviation of Real Gases. Discuss the deviation of Ideal behaviour of gases at low temperature and high
pressure.
Q11. Derive van der Waal’s equation. Deduce Units
for van der Waal’s Constants ‘a’ and ‘b’
Q12. Write down faulty
assumptions of KMT. Define following
STP, absolute zero, compressibility
factor, Mean free path, gas constant R, elastic collisions and pressure, Liquefaction,
Critical Temperature, Critical Pressure, Critical
Volume, Joule-Thomson
Effect, SCUBA
Q13. What
is plasma? Give its significance in daily life.
Q14. Give reasons for
following:
i) Why the process of diffusion occurs more rapidly in
gases, less rapidly in liquids and very slow in solids.
ii)100 cm3 of O2 and 100 cm3
of NH3 contain the same number of molecules at stp.
iii) The rates of diffusion of CO2 and C3H8
gases are the same.
iv) Real gases deviate from ideal behavior at high
pressure and low temperature.
v) air pressure decreases due to increase in altitude
vi) How Atmospheric Pressure is
originated?
Important Numericals
Q1. The pressure of gas filled in automobile tire is generally measured in psi. Convert 32.8 psi into
(i) atmosphere (2.32 atm)
(ii) kPa (226.085 kPa)
(iii) torr (1695.8 torr)
Q2. Gorakh Hill station is coldest area of Sindh province and has 83 kPa barometer pressure. What will be the pressure of this area in psi and atm units?
(Answer; 12.04 psi, 0.819 atm)
Q3. A cylinder contains 2.2 moles of oxygen gas at STP. When more oxygen gas is pumped into the cylinder, the volume of gas is changed from 2.0 dm3 to 3.4 dm3. Calculate how many moles of the oxygen gas are added to the cylinder?
(Book question; page no 72, Example # 4.2)
[Answer; n2 =
3.74, moles added = 1.54 moles]
Q4. At standard temperature and pressure 26.4 dm3 of a gas contains 1.26 moles. If 0.25 moles are added to the gas, what will be the new volume of the gas?
(Book question; page no 72, Self-Assessment)
[Answer; 31.63 dm3
Q5. Laughing gas (N2O) at 30°C and 820 torr pressure occupies a volume of 10.32 dm3. Calculate the volume that it will occupy at standard temperature and pressure.
(Page 83; Example 4.3)
[Answer;
V2= 10.03 dm3]
Q6. An steel gas cylinder has a capacity of 15.8 dm3 and filled with 785 g of helium gas at a temperature of 20°C. Calculate the pressure of helium in the cylinder (molar mass of helium is 4 g/mol).
(Self-assessment; Page 83).
[Answer; P = 298.79 atm]
Q7. Calculate the volume occupied by 8g of methane gas at 40°C and 842 torr pressure?
(Assignment)
[Answer; P = 11.58
atm]
Q8.At 35°C, oxygen gas in a cylinder has 456 cm3 volume and 0.85 atm pressure. Calculate the pressure when this oxygen gas is transferred to 10 dm3 cylinder and cooled to 20°C.
(Assignment)
[Answer; P2 = 0.037 atm]
Q9. Four containers of equal volume are
filled as follows: (Assignment) [Hint: Suppose Volume = 1 dm3].
(i) 2.0
g of H2 at 0°C
(ii) 1.0
g of H2 at 273°C
(iii) 24g
of O2 at 0°C
(iv) 16
g of CH4 at 273°C
(a) Which
container is at the greatest pressure?
(b) Which container is at the lowest pressure?
[Answer; Container (iv) is at highest pressure, Container (iii) is at lowest pressure]
(Ans: P1
= 22.4 atm; P2 = 22.4 atm; P3 = 16.8 atm; P4 =
44.8 atm)
Q10. 40dm3 of hydrogen gas was collected over water at 831 torr pressure at 23oC. What would be the volume of dry hydrogen gas at standard conditions? The vapour pressure of water at 23oC is 21 torr of Hg.
(Assignment)
[Answer; V2 = 39.32 dn3]
Q11. A 500cm3 vessel contains H2 gas at 400 torr pressure and another 1 dm3 vessel contains O2 gas at 600 torr pressure. If under the similar condition of temperature these gases are transferred to 2dm3 empty vessel, calculate the pressure of the mixture of gases in new vessel.
(Assignment)
[PT =
100 + 300 = 400 torr]
Q12.Calculate the density of oxygen gas at 45°C when the gas is confined in cylinder at 1.54 atmospheric pressure.
(Example 4.4; Page 76)
[Answer; 1.889 g/dm3]
Q13. One
mole of ammonia gas is kept in a cylinder of 5.5 dm3 at 27°C.
Assuming ammonia gas as a real gas, determine its pressure. The van der Waals
constant for ammonia are
a = 4.17 atm dm6 mol–2 and b = 0.0371 dm3 mol–1.
[Answer; P = 4.365 atm]
Q14. Two
moles of oxygen gas is kept in a vessel of 15.5 dm3 at a temperature
of 37°C. Calculate the pressure
exerted by the gas if
(a) gas behaves as ideal
(b) gas behave is non-ideal
a = 1.36 atm dm6 mol–2
and b = 0.0318 dm3 mol–1.
[Answer; Pideal = 3.28 atm, Preal = 3.27
atm]
Q15. Compare the rates of diffusion of helium
(He) and methane (CH4) gases. (Example 4.9; Page 95)
[Answer; He diffuses two times faster than CH4]
Q16. Compare the rates of diffusion of the
following pairs of gases: (Assignment book; Page 95)
(a) H2 and
D2 [Answer; H2 diffuses 1.414
times faster than D2]
(b) He and SO2 [Answer; He diffuses 4 times
faster than O2]
(b) SF6
and SO2 [Answer;
SO2 diffuses 1.51 times faster than SF6]
Q17. The ratio of rates of diffusion of two gases A and B is 1.5:1. If the relative molecular mass of gas A is 16, find out the relative molecular mass of gas B?
(Example 4.10; Page 95)
[Answer;
MB= 36 amu]
Q18. At a specific temperature and pressure, it takes 290 s for a 1.5 dm3 sample of He to effuse through a porous membrane. Under similar conditions, if 1.5 dm3 of an unknown gas “X” takes 1085 s to effuse, calculate the molar mass of gas “X”.
(Example 4.11; Page 95)
[Mx = 55.96 ≈ 56 g/mol]
Q19. If it takes 8.5 seconds for 5 cm3 of CO2 gas to effuse through a porous material at a particular temperature and pressure. How long would it take for 5 cm3 of SO2 gas to effuse from the same container at the same temperature and pressure?
(Self-assessment; Page 97)
[Answer; 10.43 seconds]
Q20. If 16 cm3 of hydrogen effuses in 30 sec, from a porous material, what volume of SO2 will effuse in the same time (30 sec) under similar conditions?
(assignment; Page)
[Answer; 2.82 cm3]
Q21. A 20 dm3 cylinder is filled with 4.25 moles of oxygen gas and 12 moles of helium gas at 25°C. Calculate the total pressure of gas mixture and partial pressures of oxygen and helium.
(Example 4.7; Page 83)
[Answer; Pt= 19.88 atm, PO2 = 5.19
atm PHe = 14.68 atm]
Q22. Oxygen gas is produced by heating
potassium nitrate
2KNO3 → 2KNO2 + O2
The gas is
collected over water. If 225 cm3 of gas is collected at 25oC
and 785 mm Hg total pressure, what
is the mass of O2 gas collected? (Pressure of vapours at 25oC
is 23.8 mm Hg)
(Example 4.8;
Page 84)
[Answer; nO2
= 9.2 x 10−3 mole, Mass = 0.294 g O2
Q23. The mole fraction of oxygen in air is 0.2093, determine the partial pressure of oxygen in air if the atmospheric pressure is 760 torr.
(Self-Assessment; Page 85)
[Answer; PO2 = 159.068 torr]
Answers of Test Questions on Chapter # 4, Gaseous State
Q1. State and explain Avogadro’s law and Charles
law.
Answer
Avogadro’s Law
Introduction and Statement
In 1811, Amadeo Avogadro (an Italian scientist) advanced a
brilliant relationship (hypothesis) regarding the relationship between the volume
and number of moles (or molecules) of gases at fixed temperature. This
relationship or hypothesis, now called Avogadro’s Law.
Under the similar conditions of
temperature and pressure, equal volumes of all contain the same number of
moles. Thus volume of a gas is directly proportional to the number of molecules
of the gas at constant temperature and pressure.
Thus under similar conditions of temperature and pressure, 1 dm3
of any gas contains same number of molecules as same number of moles contains
same number of molecules. 1 mole of any gas at standard temperature (0oC)
and pressure (1 atm) occupies 22.4 dm3 (molar volume) and contains
6.02 x 1023 molecules. It shows that two or more gases having same
volume must have same number of molecules and moles but different masses.
Mathematical Expression
V a n (at constant temperature and pressure).
V = Kn [K is constant of proportionality which
depends upon pressure and temp]
V/n = K
It shows that the ratio of volume to number of moles of a gas remains
constant.
Suppose a gas of n1 moles is enclosed in a vessel of V1
volume. If we add more gas, the volume increases to V2, hence
V1/n1 = K ……………………. Initial State
V2/n2 = K ……………………. Final State
V1/n1
= V2/n2 = K
Charles Law (Effect of Absolute temperature on gas
volume)
Importance
Gases exhibit enormous changes in volume due to expansion and contraction
due to presence of large intermolecular spaces. The effect of temperature
changes on volume of gases at constant pressure were studied by French
Scientist Jacques Charles (1746-1823). He explained that the volume of given
mass of a gas increases or decreases by 1/273 times of its original volume at 0oC for every
degree rise or fall of temperature at given pressure.
Statement
The volume of a fixed mass of a gas is directly
proportional to the absolute (kelvin) temperature at a given pressure.
According to Charles law, when absolute temperature
of a gas increases, its volume also increases because rise of temperature
increases the kinetic energy of gas molecules, which makes the molecules to
move freely resulting in the expansion of gas.
Charles law can be expressed as:
V a T (Pressure constant) and V
= KT
Graphical
Explanation of Absolute zero
When the graph is plotted between temperature (T) on ox-axis and
volume (V) of gas on y-axis, a straight line is obtained showing the direct
relation between volume and temperature. This straight line in upward direction
shows that volume increases with the increase of temperature.
If this straight line is further extended downward, it will
intercept the temperature axis at –273.15oC. According to Charles, at –273.15°C, the
volume of a gas should be zero. Actually no real gas can achieve this lowest
possible temperature and before –273.15°C all gases are condensed into liquids.
Absolute zero
The temperature of –273.15°C is referred to as Absolute Zero or
Zero Degree Absolute or Zero kelvin. The hypothetical temperature of –273.15°C
at which volume of an ideal gas becomes zero and all molecular motion (all
gases) ceases to exist is called Absolute Zero or Zero Degree kelvin.
The concept of absolute zero cannot be applied to real gases.
Q2. Explain pressure and its various units.
Answer
Definition
The magnitude of force that is applied on the surface of an object
per unit area is called pressure. Pressure of a gas is due to the elastic
collisions of its molecules with the walls of container.
Pressure (P)
= Force (F)/Area (A)
Units of
Pressure
1. The SI unit of
pressure of N/m2 which is known as ‘pascal’ (Pa). A
pascal is defined as the pressure exerted by a force of one newton acting on
area of one square meter i.e. 1 Pa is equal to N–m–2 or 1 kg–m–s–2.
1
Pa =
1 newton/m2 = 1
kg/m-sec2 (1N
= kg.m/sec2)
2. Pressure
may be measured in Pound per square inch (Psi), Standard Atmospheric
Pressure (atm), mm Hg, cm Hg, inch Hg, torr,
kPa, bar, millibar, dyne/cm2.
101325 Pa = 1 atm = 760 mm Hg = 14.7 psi
Q3. What is Atmospheric
Pressure? On what factor Atmospheric Pressure depends upon? Give its effect on
weather. Explain low and High atmospheric
pressure system
Answer
Atmospheric Pressure
The pressure exerted by the air at sea level (14.7 Psi) which will
support a column of mercury 760 mm Hg high (approximately 30 inch) is called standard
barometric or atmospheric pressure abbreviated as atm which is equal
to 14.7 Psi.
Factor Affecting Atmospheric Pressure
The size, number and nature of the
molecules determine the density and temperature of the air. As the number and
motion of the particles increases, atmospheric pressure increases and vice
versa.
Low and High atmospheric pressure
system
Low atmospheric pressure system is also
called a depression. An area which has low pressure as compared to the
surrounding is generally warmer where moist warm air rises up and cools
down. This area has suffocating weather and invites winds, clouds and
precipitation. Here the wind blows in anticlockwise direction.
During day time, there is comparatively moderate temperature while
nights are warmer due to trapping of solar radiations in cloudy
weather.
The area which experiences a high air
pressure as compared to surrounding is said to be in high pressure system.
In such area, dry cold and dense air moves downward to the
ground. Here the wind blows in clockwise direction. Thus, clear sky and calm
weather are developed.
Q5. State
the basic postulates of Kinetic Molecular Theory. Explain diffusion and
compressibility in three states of matter on the basis of KMT.
Answer
Basic Postulates of Kinetic Molecular Theory
1. Molecule and Intermolecular Distance
All gases consist of a very large number of
extremely small, tiny and discrete particles called molecules, which are far apart from one another. Thus gas molecules are very small in comparison to
the distances between them. Molecules may be
monoatomic (He, Ne, Ar, Kr, Xe, Rn), diatomic (H2, O2, N2,
F2, Cl2) or polyatomic (CH4, C4H10).
2. Negligible Molecular Volume (faulty
assumption)
Gas molecules are far away from each
other and occupy negligible volume as compared to the total volume of
the container. In other words, ideal
gases have no volume). This
assumption explains the great compressibility of gases.
3. Random, Linear Molecular Motion and Mean
Free Path
The gas molecules are in
constant continuous state of rapid random haphazard motion in straight line. The distance covered by the molecules
between two successive collisions is called free path. The average distance
travelled by a molecule between two consecutive collisions (i.e. one collision
and the next) is called Mean Free Path (MFP).
4. Elastic Molecular Collisions
The
collisions exhibited by gas particles are completely elastic involving no
energy loss; when two molecules collide, total kinetic energy is conserved.
5. Gas Pressure Results from Molecular
Collisions
Gas pressure is the
result of elastic collisions (impacts) of gas molecules with the walls of
container. The greater is the
number of collision exerted by gas molecules, the greater is the pressure of
gas.
8. No Attractive or Repulsive Force (faulty
assumption)
There are no appreciable
attractive or repulsive forces among molecules in an ideal gas. Thus each molecule acts as quite
independently. (In real gases, there
actually is attraction between their molecules which is negligibly weak and
will be ignored).
6.Average Kinetic energy Proportional to
Absolute temperature
The molecules possess kinetic energy due to their continuous motion
which depends on their mass and velocity (K.E = ½ mv2).
The average kinetic energy of gas molecules is directly
proportional to the absolute (kelvin) temperature of gas. all
gas molecules (regardless of their molecular mass) have the same average
kinetic energy at the same kelvin temperature. Thus average kinetic energy of
molecules is independent of the kind of molecules or nature of the gas.
KE
𝛂 a ½ mv2 a T (Absolute temperature) Or
v2 a T
Kinetic Equation
On the basis of above
postulates, R. J Clausius derived a kinetic equation; PV = 1/3 mN
Where
P =
V =
m = mass of a single molecule of a gas
N = number of moles of gas molecules
C2= mean square velocity of the gas
molecules
Under given conditions, the gas
molecules do not possess the same velocities, instead mean square velocity
is taken for the molecules in the above equation. If n1 molecules
possess c1 velocity, n2 molecules with velocity c2,
and so on, then mean or average square velocity (
Q6. derive the General gas equation. Deduce the value of General gas constant (R) in:
(i) S.I. System (J/mol.K)
(ii) Non S.I. System (atm dm3/mol.K)
(iii) calorie
Answer
Definition
The combined form of Boyle’s law, Charles’s law and Avogadro’s law
in which neither temperature nor pressure is constant is called General Gas
Equation.
Boyle’s law, Charles’s law and Avogadro’s law can be combined into
a single statement known as ideal gas law or ideal gas equation which
describes the behaviour of a gas on the basis of relationship between volume
and other variables like pressure, temperature and number of moles. When the
values of any three of the variables P, V, T and n are known, the value of
fourth can be calculated by using ideal gas equation.
Derivation of an Ideal Gas
Equation (Equation of State)
According to
Charles’s law…………..V a T (P is constant)
Avogadro’s
law………………………… V a n (P and T are constants)
According to
Boyle’s law…………… V a 1/P (T is
constant)
Combining
above these laws …….. V a nT/P OR V a nT/P
V = R nT/P
PV= nRT (Ideal
gas equation)
Where R is constant of proportionality and is called General gas
constant or Universal Gas Constant. The above equation is also known as Ideal
Gas Equation. (This equation is called Equation of State because when we
specify the four variables – pressure, temperature, volume and number of moles,
we define the state of a gas).
Second Form of Ideal Gas
Equation (Equation of State)
Since for
one mole of a gas; n = 1
\ PV =
RT
PV/T = R
If pressure changes from P1 to P2 and
temperature changes from T1 to T2 then volume also
changes from V1 to V2.
Then,
P1V1/T1
= R (For initial state) –––––––––– (1)
Similarly,
P2V2/T2
= R (For
final state) –––––––––– (2)
Value of R for 1 mole of
a gas at STP when pressure is in atmosphere and volume in dm3
P = 1 atm
T = 273 K
n = 1 mole
V = 22.4 dm3
(molar volume)
R =
0.0821 atm dm3K−1.mole−1
Value of R for 1 mole of
a gas at STP when pressure is in N/m2 and volume in m3
P = 101300 N/m2 (equal
to 1 atm)
T = 273 K
n = 1
V = 0.0224 m3 (
Q 1 dm3 = 10¯3 m3)
R = 8.3143 N-m K−1 mole−1 or J K−1 mole−1 (Since 1Nm = 1 J)
Since 1 cal = 4.18 J, therefore the value of R may
also be written as
R = 1.99 cal K−1
mole−1
Q7. State and explain
Dalton’s law of partial pressure with its applications.
Answer
Partial Pressure
The pressure exerted by each individual
gas in a mixture of gases on the wall of the container is called partial
pressure of that gas.
Introduction
John Dalton formulated a law in 1801 about the partial pressures and
total pressure exerted by non-reacting mixture of gases.
Statement
The total
pressure exerted by a mixture of non-reacting gases (that do not react
chemically) in a closed vessel (fixed volume) is always equal to the sum of
their partial pressures of the individual gases in the mixture at constant
temperature.
Mathematical Form
If the total
pressure is Pt and partial pressures of gases in the mixture is
denoted by PA, PB and PC and so on, then
according to Dalton’s Law total pressure of the mixture is written as.
Pt = PA + PB + PC +
…………… Pn
Example and
Experimental Proof
Suppose we have three empty cylinders of equal capacity
(1dm3), O2 and N2 gases are
filled separately in first two cylinder at constant temperature
Let the pressure exerted by the O2 gas and N2
gas are 500 torr and 700 torr respectively. If now these gases
are transferred into third empty container under the same conditions of
temperature and pressure; the pressure exerted by the mixture of these two
gases is found to be 1200 torr which is exactly equal to the sum of
partial pressures of O2 and N2 thereby proving Dalton’
law
According to
Dalton’ law
Pt = PO2
+ PN2 = 500 + 700 = 1200 torr
Relationship between Partial Pressure and Mole (Mole Fraction)
To derive the relationship between partial pressure and mole,
Consider a gaseous mixture of two gases denoted as 1 and 2 confined in a
container of volume V at T kelvin. The respective partial pressures are P1
and P2 and their respective moles are n1 and n2
Calculation of Partial Pressures of Each
Gas Using Ideal Gas Equation
Calculation of Total Pressure Using
Dalton’s Law
Dividing Each Partial Pressure by Total Pressure:
Since mole fraction is the ratio of number of moles of individual gas and the total number of moles of all gases present in the mixture. Hence above equation becomes reduced to
From above equation, it is conclude that partial pressure of any
gas of the mixture is equal to the product of mole fraction of that gas and
total pressure of the mixture. Mole fraction of any gas in the mixture is less
than one but the sum of the mole fractions is always equal to one.
Applications of Law
1. Collection of gases over water
Dalton’s law helps in calculating the pressure of dry gas which is
collected over water by downward displacement of water in a gas jar.
Ptotal = Pdry gas + Pwater vapours
Pdry gas = Ptotal – Pwater vapours
2. Maintenance of
Oxygen pressure at high Altitudes
Normally, respiration depends upon the difference between the partial
pressure of oxygen in the air (159 torr) and in the lungs (116 torr). At higher
altitudes, due to low partial pressure of oxygen causes the problem in the
process of respiration.
3. Maintenance of Oxygen pressure for deep
sea divers
Opposite to altitude, as distance increases downward in the sea, partial
pressure of oxygen increases. At the depth of 40 meters, pressure increases to
five times. This increased pressure also causes problem in respiration.
Therefore deep sea divers use the SCUBA (self contained underwater
Breathing Apparatus), breathing tank for respiration. Scuba contains 96% helium gas
and 4% oxygen gas.
Q8. What is meant by diffusion and effusion?
State and explain Graham’s law of diffusion with its applications. Compare the rate
of diffusion of H2 and D2.
Answer
Effusion
and Diffusion
effusion is the movement of gas molecules through a tiny hole into
the region of low pressure. The diameter of tiny hole is considerably smaller
than mean free path. Slow escaping of air from a tire pinhole is an example of
effusion.
Diffusion is the process of homogeneous
intermixing of molecules of different gases with each other due to their
motion. Spreading of fragrance of a rose flower or a perfume is a common
example of diffusion. Lighter gases diffuse more rapidly than heavier gases.
Introduction and Statement
In 1881, Thomas Graham established a quantitative relationship
between the rates of effusion or diffusion of gases and their densities (and
molar masses) by maintaining the temperature and pressure and is termed as
Graham’s Law of effusion or Diffusion.
The
rate of effusion or diffusion of a gas is inversely proportional to the square
root of its density or molar mass under the same conditions of temperature and
pressure.
Graham also studied the comparative rates of diffusion of two
gases. Therefore, the law is also stated as:
The relative rates of diffusion or effusion of two gases,
under the same conditions of temperature and pressure, are inversely proportional
to the square roots of their densities or molecular masses
Mathematical Expressions
1. For a single
gas
2. For Two Gases
If there are
two gases 1 and 2, having rates of diffusion r1 and r2
and densities d1 and d2, then law may be expressed as:
Since the
density of a gas is proportional to its molecular mass, we can write the above
equation as:
Relation between Diffusion Time &
Volume of gas diffused with Molar Mass
Applications
of Graham’s Law
1. It helps to determine
densities and molar masses of different gases.
2. It is used to
determine comparative rates of diffusion or effusion of gases.
3. It helps in separating
one gas from another gas.
4. It
helps in separating different isotopes of gases due to their different
diffusion rate owing to their different masses.
5. In diluting the
poisonous gases and minimizing their toxic effects by diffusing them into air.
Comparing the rate of
diffusion of H2 and D2.
Q9. Differentiate between ideal and
non-ideal gas. What are the conditions of deviation of real gas from ideal
behaviour? Write down Graphical
Explanation of Deviation of Real Gases. Discuss the deviation of Ideal behaviour of gases at low
temperature and high pressure.
Answer
Graphical Explanation of Deviation of Real Gases
To understand the attitude of read
gases graphically, the general gas equation should be amended as
PV = z (nRT) or z = PV/nRT
Where ‘z’ is called as compressibility
factor, its value is unity for an ideal gas, less than 1 for real
gases showing negative deviation and more than unity for real gases
showing positive deviation.
The graph plotted between z and P for
an ideal gas and various real gases provides the following information:
(i) All gases reaches a value of z =1 ,
when the pressure approached to zero. This reveals that all gases tend to act
like ideal gas at very low pressure.
(ii) The extent of deviation of real gases
from ideality is based on pressure, temperature and the nature of gases.
Conditions
of Deviation of Real Gases from Ideal Behaviour
There are
two conditions under which gases show considerable deviation from their ideal
behaviour:
1. At very high pressure.
2. At very low temperature.
Causes of Deviation
To analyze why real gases deviate from ideal behaviour, we should
know the basics of formulation of the ideal gas equation which was obtained
from certain faulty assumptions of kinetic molecular theory. The deviation of
real gases from the ideality is due to following to faulty assumptions of
kinetic molecular theory:
1. Actual volume of gas
molecules is negligibly small as compared to total volume of gas.
2. Gas Molecules have
neither attractive nor repulsive forces.
Contrary to these faulty assumptions, molecules of all gases do
exert some attractive force on one another and they do occupy some space in the
total volume of gas. The assumption 1 is valid at low pressure while assumption
2 is effective at high temperature.
The reason is that the attractive
forces diminish rapidly as the distance between molecules increases. Thus at
high pressure and low temperature intermolecular forces becomes significant
because molecules tend to be close together.
Q10. Derive van der Waal’s equation. Deduce
Units for van der Waal’s Constants ‘a’ and ‘b’.
Answer
Derivation of van Der Waal’s Equation
Definition
van der Waal Equation is the modification
of the ideal gas law and updated version of the ideal gas equation to take into
account molecular size and molecular interaction forces dealing with the two
conflicting postulates of the kinetic molecular theory regarding volumes and
intermolecular attraction of gas molecules.
Volume Correction in van der Waal’s Equation
Therefore, the available volume for free movement of the gas
molecules becomes less than the original molar volume or volume of the
container (Vvessel) due to the fact that gas molecules have definite
volume. The volume available for the gas molecules is less than the volume of
the container, V.
Keeping in view the definite volume of gas molecules, van der Waal calculated the actual volume (available volume) of a gas by subtracting excluded volume of ‘n’
moles of gas, ‘nb’ from the volume of the container (Vvessel). The
volume of a real gas is, therefore, ideal volume minus the volume occupied by
gas molecules.
Available volume = V = Vvessel – b
(for 1 one molecule)
V = Vvessel – nb
(for n mole )………………………… (i)
Where
V = free volume/actual volume (molar volume of ideal gas)
Vvessel = Volume of the vessel in which gas molecules
are present
‘n’ = number of moles
‘b’ = volume correction factor/excluded volume/theoretical volume/Incompressible Volume of gas molecules per mole in highly
compressed gaseous state. It is a constant and characteristic of gas, depends upon size of molecule.
The ideal gas equation can be written after correcting for this
as:
P(V– nb ) = nRT
Excluded volume (‘b’) is not equal to
actual volume of gas molecule. In fact, it is 4 times of actual volume of gas molecule.
‘b’ = 4Vmolecule
Pressure correction in van der
Waal’s Equation
The pressure of the real gas is less
than the expected pressure due to attractions between the molecules. These
attractions slow down the motion of gas molecules and result in:
i) reduction of frequency of collisions
over the walls and
ii) reduction in the force with which the
molecules strike the walls.
It means
that pressure produced on the wall would be little bit lesser than pressure of
an ideal gas molecule. Therefore, if observed pressure is simply
indicated by P, ideal pressure Pi and pressure-correcting
term PL (which is the pressure drop due to backward pull of striking
molecules) then equation will be
Pobserved/real or P = Pideal – Pless (∴ Pideal > Preal)
P = Pi – PL
Pideal = Preal +
Pless
Pi = P + PL ------------------------- (ii)
However, the reduction in pressure
depends upon the number of particles (A and B) per unit
volume i.e. is proportional to the square of molar concentration; n/V (one
factor for reduction in frequency of collisions and the second factor for
reduction in strength of their impulses on the walls).
Where ‘a’ is a proportionality
constant called van der Waals constant of attraction and is characteristic
of a gas. Higher values of ‘a’ indicate greater attraction between gas
molecules. The easily compressible gases like ammonia, HCl possess higher ‘a’
values. Greater the value of ‘a’ for a gas easier is the liquefaction.
The corrected pressure and volume is now put in ideal gas equation
to modify it, we get
This is van der Waal’s equation. Here ‘a’ and ‘b’ are van der
Waal’s constants and contain positive values. The constants are the
characteristic of the individual gas. When gas is ideal or that it
behaves ideally then both the constant will be zero. Generally, ‘a’ constant
help in the correction of the intermolecular forces while the ‘b’ constant
helps in making adjustments for the volume occupied by the gas particles.
Units for van der Waal’s Constants ‘a’ and ‘b’
From the pressure correction
expression, the value of ‘a’ is calculated. If the pressure is expressed in
atmospheres and volume in liters,
Since P =an2/V2, hence a = PV2/n
But substituting the units of P (atm), V (dm3) and n
(mol), we get
Unit of a = atm dm6 mol−2
Since ‘nb’ is
excluded volume for n moles of gas, ‘b’ is expressed b is expressed in litre
mol–1 units if volume is taken in litres,
b = volume/n =
liter/mol or litre mol–1
Unit of b = dm3/mol, since ‘b’ represent the volume per
mol of gas.
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