All matters exist in three states: gas, liquid and solid. Of the three states of matter, the gaseous state is the one most studied and best understood.
In a gaseous state, molecules remain separated wide apart in empty space. The molecules are free to move about throughout the container in which they are placed.
General Characteristics of Gases
1. Expansibility
Gases have limitless expansibility because of little intermolecular attraction among the gas molecules. They expand to fill the entire vessel they are placed in.
2. Compressibility
Due to large intermolecular space, gases can be easily compressed by the application of pressure to a movable piston fitted in the container.
3. Diffusibility
Gases can diffuse rapidly through each other to form a homogeneous mixture.
4. Pressure
Gases exert pressure on the walls of the container in all directions.
5. Effect of Heat
When a gas, confined in a vessel is heated, its pressure increases. Upon heating in a vessel fitted with a piston, volume of the gas increases.
Parameters of a Gas (Measurable Properties)
A gas sample can be described in terms of four parameters:
(1) The volume, V of the gas
(2) Its pressure, P
(3) Its temperature, T
(4) The number of moles, n, of gas in the container
The Volume, V
The volume of the container is the volume of the gas sample. It is usually given in litre (l or L) or mililitres (ml or mL).
The Pressure, P
The pressure of a gas is defined as the force exerted by the impacts of its molecules per unit surface area in contact.
The pressure of a gas sample can be measured with the help of a mercury manometer. Similarly, the atmospheric pressure can be determined with a mercury barometer.
The pressure of air that can support 760 mm Hg column at sea level is called one atmosphere (1 atm). The unit of pressure, millimetre of mercury, is also called torr.
Thus, 1 atm = 760 mm Hg = 760 torr
The SI unit of pressure is the Pascal (Pa). The relation between atmosphere, torr and pascal is:
1 atm = 760 torr = 1.013 × 105 Pa
The unit of pressure ‘Pascal’ is not commonly used today.
Temperature, T
The temperature of a gas may be measured in Centigrade degrees (°C) or Celsius degrees. The SI unit of temperature is Kelvin (K) or Absolute degree. The centigrade degrees can be converted to kelvins by using the equation-
K = °C + 273
The Kelvin temperature (or absolute temperature) is always used in calculations of other parameters of gases. Remember that the degree sign (°) is not used with K.
The Moles of a Gas Sample, n
The number of moles, n, of a sample of a gas in a container can be found by dividing the mass, m, of the sample by the molar mass, M (molecular mass).
Moles of gas (n) = mass of gas sample (m)molecular mass of gas (M)
The Gas Laws
The volume of a given sample of gas depends on the temperature and pressure applied to it. Any change in temperature or pressure will affect the volume of the gas. As results of experimental studies from 17th to 19th century, scientists derived the relationships among the pressure, temperature and volume of a given mass of gas.
These relationships, which describe the general behaviour of gases, are called the gas laws.
Boyle’s Law
In 1660 Robert Boyle found out experimentally the change in volume of a given sample of gas with pressure at room temperature. From his observations he formulated a generalization known as Boyle’s Law.
It states that: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved.
Fig. Boyle’s law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved.
The Boyle’s Law may be expressed mathematically as-
V 1/P (T, n are constant)
Or, V = k × 1/P
Where, k is proportionality constant.
So, PV = k
If P1, V1 are the initial pressure and volume of a given sample of gas and P2, V2 the changed pressure and volume, we can write;
P1V1 = k = P2V2
Or, P1V1 = P2V2
This relationship is useful for the determination of the volume of a gas at any pressure, if its volume at any other pressure is known.
Fig. Graphical representation of Boyle's law. (left ) a plot of V versus P for a gas sample is hyperbola; ( right) a plot of V versus 1/P is a straight line.
যখন বেলুনের ওপর চাপ বাড়বে তখন তার আয়তন কমবে। উল্টা কোরে বললে যখন বালুনের আয়তন কমান হচ্ছে কিন্তু মোল সংখ্যা ঠিক আছে তারমানে ক্রমাগত চাপ বাড়ছে আর বাড়তে বাড়তে একসময় বেলুনটা ফেটে যাবে।যদি V বাড়ে P কমবে এখন ওই P কে inverse করলে (1/P) (pressure) value বাড়বে সুতরাংV increase means 1/P increase, in a constant way giving us a straight line পাওয়া যাবে।y=mx+c (V=1/P).
Charles’s Law
In 1787 Jacques Charles investigated the effect of change of temperature on the volume of a fixed amount of gas at constant pressure. He established a generalization which is called the Charles’ Law.
It states that: At constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature or absolute temperature. If the absolute temperature is doubled, the volume is doubled.
Fig. Charles law state that at constant pressure, the volume of a fixed mass of gas is directly proportional to the absolute temperature.
Charles’ Law may be expressed mathematically as-
VT (P, n are constant)
Or, V = kT Where, k is a constant.
If V1, T1 are the initial volume and temperature of a given mass of gas at constant pressure and V2, T2 be the new values, we can write-
Using this expression, the new volume V2 can be found from the experimental values of V1, T1 and T2.
The Combined Gas Law
Boyle’s Law and Charles’ Law can be combined into a single relationship called the Combined Gas Law.
Boyle’s Law, V ∝ 1/P (T, n constant)
Charles’ Law, V T (P, n constant)
Therefore, V T/P (n constant)
The combined law can be stated as: For a fixed mass of gas, the volume is directly proportional to kelvin temperature and inversely proportional to the pressure.
If k be the proportionality constant,
V = kT/P (n constant)
Or, PV/T = k (n constant)
If the pressure, volume and temperature of a gas be changed from P1, V1 and T1 to P2, T2 and V2, then-
This is the form of combined law for two sets of conditions. It can be used to solve problems involving a change in the three variables P, V and T for a fixed mass of gas.
Gay Lussac’s Law
In 1802 Joseph Gay Lussac as a result of his experiments established a general relation between the pressure and temperature of a gas. This is known as Gay Lussac’s Law or Pressure-Temperature Law. It states that: at constant volume, the pressure of a fixed mass of gas is directly proportional to the Kelvin temperature or absolute temperature.
The law may be expressed mathematically as-
P T (Volume, n are constant)
Or, P = kT
Or, P/T = k
For different conditions of pressure and temperature
Knowing P1, T1, and T2, P2 can be calculated.
Avogadro’s Law
Let us take a balloon containing a certain mass of gas. If we add to it more mass of gas, holding the temperature (T) and pressure (P) constant, the volume of gas (V) will increase. It was found experimentally that the amount of gas in moles is proportional to the volume. That is,
V n (T and P constant)
Or, V = A n
Where, A is constant of proportionality.
Or, V/n = A
For any two gases with volumes V1, V2 and moles n1, n2 at constant T and P,
If V1 = V2 then, n1 = n2
Thus, for equal volumes of the two gases at fixed T and P, number of moles is also equal. This is the basis of Avogadro’s Law which may be stated as: Equal volumes of gases at the same temperature and pressure contain equal number of moles or molecules. If the molar amount is doubled, the volume is doubled.
Fig. Avogadro’s law states that under equal conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules.
The Molar Gas Volume
It follows as a corollary of Avogadro’s Law that one mole of any gas at a given temperature (T) and pressure (P) has the same fixed volume. It is called the molar gas volume or molar volume. In order to compare the molar volumes of gases, chemists use a fixed reference temperature and pressure. This is called standard temperature and pressure (abbreviated, STP). The standard temperature used is 273 K (0°C) and the standard pressure is 1 atm (760 mm Hg). At STP we find experimentally that one mole of any gas occupies a volume of 22.4 litres. To put it in the form of an equation, we have
1 mole of a gas at STP = 22.4 litres
The Ideal Gas Equation
We have studied three simple gas laws:
Boyle’s Law, V 1/P (T, n constant)
Charles’ Law, V T (n, P constant)
Avogadro’s Law, V n (P, T constant)
These three laws can be combined into a single more general gas law:
V nTP …………………………………………………………………...(1)
This is called the Universal Gas Law. It is also called Ideal Gas Law as it applies to all gases which exhibit ideal behavior i.e., obeys the gas laws perfectly.
The ideal gas law may be stated as: The volume of a given amount of gas is directly proportional to the number of moles of gas, directly proportional to the temperature, and inversely proportional to the pressure.
Introducing the proportionality constant R in the expression (1) we can write-
V = R nTP
Or, PV = nRT ..........................................................................................................................(2)
The equation (2) is called the Ideal Gas Equation or simply the general Gas Equation. The constant R is called the Gas constant. The ideal gas equation holds fairly accurately for all gases at low pressures. For one mole (n = 1) of a gas, the ideal-gas equation is reduced to
PV = RT ............................................................................................................................(3)
The ideal-gas equation is called an Equation of State for a gas because it contains all the variables (T, P, V and n) which describe completely the condition or state of any gas sample. If we know the three of these variables, it is enough to specify the system completely because the fourth variable can be calculated from the ideal-gas equation.
The Numerical Value of R
From the ideal-gas equation, we can write
R = PVnT
We know that one mole of any gas at STP occupies a volume of 22.4 litres. Substituting the values in the above expression, we have
R = 1 atm × 22.4 litres1 mole ×273 K
= 0.0821 atm. litre mol–1 K–1
It may be noted that the unit for R is complex; it is a composite of all the units used in calculating the constant.
Mathematical Problems
1) A gas occupies 100 mL volume at 150 kPa. Find its volume at 200 kPa. (P1V1 = P2V2)
2) A gas’s pressure is 765 torr at 23°C. At what temperature will the pressure be 560 torr? (P1/T1 = P2/T2)
3) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. ( P1V1T1 = P2V2T2)
4) 25.8 litre of a gas has a pressure of 690 torr and a temperature of 17°C. What will be the volume if the pressure is changed to 1.85 atm and the temperature to 345 K?
( P1V1T1 = P2V2T2)
5) Calculate the pressure in atmospheres of 0.412 mol of He gas at 16°C that is occupying 3.25 L. (PV = nRT)
Dalton’s Law of Partial Pressures
John Dalton visualized that in a mixture of gases, each component gas exerted a pressure as if it were alone in the container. The individual pressure of each gas in the mixture is defined as its Partial Pressure. Based on experimental evidence, in 1807, Dalton enunciated what is commonly known as the Dalton’s Law of Partial Pressures. It states that: The total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases present.
Fig. Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by each gas. The pressure of the mixture of O2 and N2 (Tanks) is the sum of the pressures in O2 and N2 tanks.
Mathematically the law can be expressed as Ptotal = P1 + P2 + P3 ... (V and T are constant) where, P1, P2 and P3 are partial pressures of the three gases 1, 2 and 3; and so on.
Dalton’s Law of Partial Pressures follows by application of the ideal-gas equation
PV= nRT separately to each gas of the mixture.
Thus, we can write the partial pressures P1, P2 and P3 of the three gases-
P1 = n1 (RT/V) P2 = n2 (RT/V) P3 = n3 (RT/V)
Where, n1, n2 and n3 are moles of gases 1, 2 and 3. The total pressure, Pt, of the mixture is
Pt = (n1 + n2 + n3) RTV
Or, Pt = nt RTV
In the words, the total pressure of the mixture is determined by the total number of moles present whether of just one gas or a mixture of gases.
Another derivation
Let us consider a container with the volume V . This container has n1,n2 ,n3….. moles of non-reacting gases 1,2 3 and so on .
At a fix temperature T , the individual pressure of the gas molecules will be
P1 + P2 + P3 …….
Here these values of individual pressure are the part or portion of the total pressure.
Thus the total pressure will as Ptotal = P1 + P2 + P3 ... (V and T are constant) . This is the mathematically expression for the law of partial pressure.
Derivation:
Dalton’s Law of Partial Pressures follows by application of the ideal-gas equation
PV= nRT separately to each gas of the mixture.
Thus, we can write the partial pressures P1, P2 and P3 of the three gases-
P1 V= n1 RT P2 V= n2 RT P3 V= n3 RT
Or, P1 = n1 (RT/V) -------(i) P2 = n2 (RT/V)-------(ii) P3 = n3 (RT/V) --------(iii)
Where, n1, n2 and n3 are moles of gases 1, 2 and 3.
And for the total pressure the equation will be, Ptotal = ntotal RTV -------------------------(iv)
Now, the total pressure,, of the mixture is will be,
P1 + P2 + P3 = (n1 + n2 + n3) RTV (Combining the equations i, ii and iii)
Or, P1 + P2 + P3 = ntotal RTV (Since, n1 + n2 + n3 =ntotal)
Or, P1 + P2 + P3 = Ptotal
In the words, the total pressure of the mixture is determined by the total number of
moles present whether of just one gas or a mixture of gases.
Mathematical Problems
1. What pressure is exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273K in a
10 litre vessel?
2. A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? Vapour pressure of water at 35°C is 42.2 torr.
Graham’s Law of Diffusion
When two gases are placed in contact, they mix spontaneously. This is due to the movement of molecules of one gas into the other gas. This process of mixing of gases by random motion of the molecules is called Diffusion. Thomas Graham observed that molecules with smaller masses diffused faster than heavy molecules.
Fig. A light molecule diffuses quicker than a heavy molecule.
In 1829, Graham formulated what is now known as Graham’s Law of Diffusion. It states that:
Under the same conditions of temperature and pressure, the rates of diffusion of different gases are inversely proportional to the square roots of their molecular masses.
Derivation
Let us consider there are two gases, Gas-1 and Gas-2
The rate of effusion for the Gas-1 is =r1 and molecular weight =M1
On the other hand the rate of diffusion for the Gas-2 is = r2 and molecular weight =M2
Now according to the Law of diffusion for the Gas-1,
r1∝ 1√M1
Or, r1=k1 1√M1---------(i)
Similarly for the Gas-2,
r2∝ 1√M2
Or, r2=k2 1√M2---------(ii)
Dividing equation iby iiwe get
r1r2 = √M2M1
This is the mathematical expression of the law
When a gas escapes through a pin-hole into a region of low pressure or vacuum, the process is called Effusion. The rate of effusion of a gas also depends on the molecular mass of the gas.
Dalton’s law when applied to effusion of a gas is called the Dalton’s Law of Effusion. It may be expressed mathematically as-
(P, T constant)
The determination of rate of effusion is much easier compared to the rate of diffusion. Therefore,
Dalton’s law of effusion is often used to find the molecular mass of a given gas.
Mathematical Problems
1. An unknown gas diffuses 4.0 times faster than O2. Find its molecular mass.
2. 50 ml of gas ‘A’ effuse through a pin-hole in 146 seconds. The same volume of CO2 under identical conditions effuses in 115 seconds. Calculate the molecular mass of A.
Kinetic Molecular Theory of Gases
Maxwell and Boltzmann (1859) developed a mathematical theory to explain the behaviour of gases and the gas laws. It is based on the fundamental concept that a gas is made of a large number of molecules in perpetual motion. Hence, the theory is called the kinetic molecular theory or simply the kinetic theory of gases (The word kinetic implies motion). The kinetic theory makes the following assumptions.
Assumptions of the Kinetic Molecular Theory
(1) A gas consists of extremely small discrete particles called molecules dispersed throughout the container.
The actual volume of the molecules is negligible compared to the total volume of the gas. The molecules of a given gas are identical and have the same mass (m).
(2) Gas molecules are in constant random motion with high velocities.
They move in straight lines with uniform velocity and change direction on collision with other molecules or the walls of the container.
(3) The distances between the molecules are very large and it is assumed that Van Der Waals attractive forces between them do not exist. Thus, the gas molecules can move freely, independent of each other.
(4) All collisions are perfectly elastic. Hence, there is no loss of the kinetic energy of a molecule during a collision.
(5) The pressure of a gas is caused by the hits recorded by molecules on the walls of the container.
(6) The average kinetic energy (1/2 mv2) of the gas molecules is directly proportional to absolute temperature (Kelvin temperature). This implies that the average kinetic energy of molecules is the same at a given temperature.
How Does an Ideal Gas Differ from Real Gases?
A gas that confirms to the assumptions of the kinetic theory of gases is called an ideal gas. It obeys the basic laws strictly under all conditions of temperature and pressure.
The real gases as hydrogen, oxygen, nitrogen etc., are opposed to the assumptions (1), (3) and (4) Stated above.
Thus:
(a) The actual volume of molecules in an ideal gas is negligible, while in a real gas it is appreciable.
(b) There are no attractive forces between molecules in an ideal gas while these exist in a real gas.
(c) Molecular collisions in an ideal gas are perfectly elastic while it is not so in a real gas.
For the reasons listed above, real gases obey the gas laws under moderate conditions of temperature and pressure. At very low temperature and very high pressure, the clauses (1), (3) and (4) of kinetic theory do not hold. Therefore, under these conditions the real gases show considerable deviations from the ideal gas behaviour.
Different Kinds of Velocities
In our study of kinetic theory we come across two different kinds of molecular velocities:
(1) The Average velocity (V)
(2) The Root Mean Square velocity (µ)
Average Velocity
Let, there be n molecules of a gas having individual velocities v1, v2, v3 ..... vn. The ordinary average velocity is the arithmetic mean of the various velocities of the molecules,
Root Mean Square Velocity
If v1, v2, v3 ..... vn are the velocities of n molecules in a gas, µ2, the mean of the squares of all the velocities is-
µ is thus the Root Mean Square velocity or RMS velocity. It is denoted by u.
The value of the RMS of velocity u, at a given temperature can be calculated from the Kinetic Gas Equation.
By substituting the values of R, T and M, the value of u (RMS velocity) can be determined. RMS velocity is superior to the average velocity considered earlier. With the help of u, the total Kinetic energy of a gas sample can be calculated.
Derivation of Kinetic Gas Equation
Starting from the postulates of the kinetic molecular theory of gases we can develop an important equation. This equation expresses PV of a gas in terms of the molecular mass, number of molecules and molecular velocity. This equation which we shall name as the Kinetic Gas Equation may be derived by the following clauses.
Let us consider a certain mass of gas enclosed in a cubic box at a fixed temperature.
Suppose that:
The length of each side of the box = l cm
The total number of gas molecules = N
The mass of one molecule = m
The velocity of a molecule = v
The kinetic gas equation may be derived by the following steps:
(1) Resolution of Velocity ‘v’ of a Single Molecule Along X, Y and Z Axes
According to the kinetic theory, a molecule of a gas can move with velocity ‘v’ in any direction.
Velocity is a vector quantity and can be resolved into the components vx, vy, vz along the X, Y and Z axes. These components are related to the velocity v by the following expression.
v2 = vx2 + vy 2+ vz2
Now we can consider the motion of a single molecule moving with the component velocities independently in each direction.
(2) The Number of Collisions per Second on Face ‘A’ Due to One Molecule
Consider a molecule is moving in OX direction between opposite faces A and B. It will strike the face A with velocity vx and rebound with velocity – vx
(since the collision is assumed to be elastic nothing would be changed i.e. the backward velocity would be the same as before.).
To hit the same face again (That means to hit the A face again) , the molecule must travel l cm to collide with the opposite face B and then again l cm to return to face A
( This is because the gas molecule will not be able to hit A until it is directed towards it. And for this it must be strike in B and B will and that will direct it to the A)
Therefore, the time between two collisions on face A = 2l/ vx seconds
[Speed of a body is measured by the distance travelled per unit time. i.e Speed = distance ÷time
or, time=distance ÷ speed (velocity). If a moving body travels a distance `s’ in time `t’ then the speed v is v =s ÷ t or, t=s ÷ v
Here, the gas molecule travelled l distance (from B-A) for the first strike on the face A.
And for striking the face B (from A-B) it must to travel the same distance l.
Therefore the total distance travelled by the molecule for 2 consecutive collisions (1st on Face A and again on Face A) =l+l=2l i.e. s=2l and the velocity of the gas molecules were Vx. i.e. v=Vx
That means the time between two successive collisions (only on Face A) will be= 2l÷Vx)]
The number of collisions per second on face A = vx /2l (This is the frequency)
(3) The Total Change of Momentum on All Faces of the Box Due to One Molecule Only
Each impact of the molecule on the face A causes a change of momentum
(mass × velocity):
The momentum before the impact = mvx
The momentum after the impact = m (–vx)
∴ The change of momentum = mvx – (– mvx)
= 2mvx (Alternative)☺
But, the number of collisions per second on face A due to one molecule = vx /2l
Therefore, the total change of momentum per second on face A caused by one molecule,
The change of momentum on both the opposite faces A and B along X-axis would be double i.e.,
2mvx2/l similarly, the change of momentum along Y-axis and Z-axis will be 2mvy2/l and 2mvz2/l respectively.
Hence, the overall change of momentum per second on all faces of the box will be
(v2= vx2+ vy2+ vz2)
☺ [We know that acceleration, a=(change of velocity ÷time). If the initial velocity of a body is u and its final velocity after time t is v.Then change of velocity in time t = v – u . Hence, the change of velocity in unit time = (v-u) ÷t
∴ Rate of change of velocity, i.e. acceleration, a = (v-u) ÷t {(final velocity – initial velocity) ÷ unit time or distance/time2}
We also know that, Momentum = (mass× velocity)
or, Momentum = mv & the rate of change of momentum in time t = mv – mu
Since, Force = mass× acceleration
Thus, F=ma or, F= m (v-u/t) or, F=(mv-mu)/t or, F=(change of momentum)÷t .
Now the change of momentum on the face A = mvx – (– mvx) = 2mvx.
Putting the values of change in momentum for the two successive collision on the face A we get= {(2mvx)÷(2l/vx)}=mv2x÷l
The change of momentum on both the opposite faces A and B along X-axis would be= 2mv2x÷l
The change of momentum along Y-axis and Z-axis will be 2mv2y/l and 2mv2z/l respectively.]
(4) Total Change of Momentum Due to Impacts of All the Molecules on All Faces of the Box
Suppose, there are N molecules in the box each of which is moving with a different velocity v1, v2, v3, etc. The total change of momentum due to impacts of all the molecules on all faces of the box
Multiplying and dividing by n, we have
(5) Calculation of Pressure from Change of Momentum; Derivation of Kinetic Gas Equation
Since force may be defined as the change in momentum per second, we can write
Since l3 is the volume of the cube, V, we have
This is the fundamental equation of the kinetic molecular theory of gases. It is called the kinetic gas equation. This equation although derived for a cubical vessel, is equally valid for a vessel of any shape. The available volume in the vessel could well be considered as made up of a large number of infinitesimally small cubes for each of which the equation holds.
Collision Properties
In the derivation of kinetic gas equation we did not take into account collisions between molecules. The molecules in a gas are constantly colliding with one another. The transport properties of gases such as diffusion, viscosity and mean free path depend on molecular collisions.
The Mean Free Path
At a given temperature, a molecule travels in a straight line before collision with another molecule. The distance travelled by the molecule before collision is termed free path. The free path for a molecule varies from time to time. The mean distance travelled by a molecule between two successive collisions is called the Mean Free Path. It is denoted by 𝜆.
If l1, l2, l3 are the free paths for a molecule of a gas, its mean free path-
Where, n is the number of molecules with which the molecule collides. Evidently, the number of molecular collisions will be less at a lower pressure or lower density and longer will be the mean free path.
The Collision Diameter
When two gas molecules approach one another, they cannot come closer beyond a certain distance.
The closest distance between the centres of the two molecules taking part in a collision is called the Collision Diameter. It is denoted by .
Whenever the distance between the centres of two molecules is σ, a collision occurs.
The collision diameter is obviously related to the mean free path of molecules.
The smaller the collision or molecular diameter, the larger is the mean free path.
The Collision Frequency
The collision frequency of a gas is defined as: The number of molecular collisions taking place per second per unit volume (c.c.) of the gas.
Let a gas contain N molecules per c.c. From kinetic consideration it has been established that the number of molecules, n, with which a single molecule will collide per second, is given by the relation
Where v = average velocity; σ = collision diameter.
If the total number of collisions taking place per second is denoted by Z, we have
Since each collision involves two molecules, the number of collision per second per c.c. of the gas will be Z/2.
Hence the collision frequency
Evidently, the collision frequency of a gas increases with increase in temperature, molecular size and the number of molecules per c.c.
Deviations from Ideal Behaviour
An ideal gas is one which obeys the gas laws or the gas equation PV = RT at all pressures and temperatures. However, no gas is ideal. Almost all gases show significant deviations from the ideal behaviour. Thus, the gases H2, N2 and CO2 which fail to obey the ideal-gas equation are termed nonideal or real gases.
Compressibility Factor
The extent to which a real gas departs from the ideal behaviour may be depicted in terms of a new function called the compressibility factor, denoted by Z.
It is defined as, Z = PVRT
For an ideal gas, Z = 1 and it is independent of temperature and pressure. The deviations from ideal behaviour of a real gas will be determined by the value of Z being greater or less than 1.
The difference between unity and the value of the compressibility factor of a gas is a measure of the degree of non-ideality of the gas.
For a real gas, the deviations from ideal behaviour depend on pressure and temperature. This will be illustrated by examining the compressibility curves of some gases discussed below with the variation of pressure and temperature.
Effect of Pressure Variation on Deviations
The following figure shows the compressibility factor, Z, plotted against pressure for H2, N2 and CO2 at a constant temperature.
At very low pressure, for all these gases Z is approximately equal to one. This indicates that at low pressures (upto 10 atm), real gases exhibit nearly ideal behaviour. As the pressure is increased, H2 shows a continuous increase in Z (from Z = 1). Thus, the H2 curve lies above the ideal gas curve at all pressures.
For N2 and CO2, Z first decreases (Z < 1).
It passes through a minimum and then increases continuously with pressure (Z > 1). For a gas like CO2 the dip in the curve is greatest as it is most easily liquefied.
Effect of Temperature on Deviations
Figure shows plots of Z or PV/RT against P for N2 at different temperatures.
It is clear from the shape of the curves that the deviations from the ideal gas behaviour become less and less with increase of temperature. At lower temperature, the dip in the curve is large and the slope of the curve is negative. That is, Z < 1. As the temperature is raised, the dip in the curve decreases.
At a certain temperature, the minimum in the curve vanishes and the curve remains horizontal for an appreciable range of pressures.
At this temperature, PV/RT is almost unity and the Boyle’s law is obeyed. Hence this temperature for the gas is called Boyle’s temperature. The Boyle temperature of each gas is characteristic e.g., for N2 it is 332 K.
Conclusions
From the above discussions we conclude that:
(1) At low pressures and fairly high temperatures, real gases show nearly ideal behaviour and the ideal-gas equation is obeyed.
(2) At low temperatures and sufficiently high pressures, a real gas deviates significantly from ideality and the ideal-gas equation is no longer valid.
(3) The closer the gas is to the liquefaction point the larger will be the deviation from the ideal behaviour.
Explanation of Deviations – Van Der Waals Equation
Van Der Waals (1873) attributed the deviations of real gases from ideal behaviour to two erroneous postulates of the kinetic theory. These are:
(1) The molecules in a gas are point masses and possess no volume.
(2) There are no intermolecular attractions in a gas.
Therefore, the ideal gas equation PV = nRT derived from kinetic theory could not hold for real gases. Van Der Waals pointed out that both the pressure (P) and volume (V) factors in the ideal gas equation needed correction in order to make it applicable to real gases.
Volume Correction
The volume of a gas is the free space in the container in which molecules move about. Volume (V) of an ideal gas is the same as the volume of the container. The dot molecules of ideal gas have zero volume and the entire space in the container is available for their movement. However, Van Der Waals assumed that molecules of a real gas are rigid spherical particles which possess a definite volume.
Fig. Volume of a Real gas.
The volume of a real gas is, therefore, ideal volume minus the volume occupied by gas molecules.(Fig.)
If b is the effective volume of molecules per mole of the gas, the volume in the ideal gas equation is corrected as: (V – b)
For n moles of the gas, the corrected volume is: (V – nb)
Where, b is termed the excluded volume which is constant and characteristic for each gas.
Pressure Correction
A molecule in the interior of a gas is attracted by other molecules on all sides. These attractive forces cancel out. But a molecule about to strike the wall of the vessel is attracted by molecules on one side only. Hence, it experiences an inward pull (Figure). Therefore, it strikes the wall with reduced velocity and the actual pressure of the gas, P, will be less than the ideal pressure. If the actual pressure P, is less than Pideal by a quantity p, we have
P = Pideal – p
Or, Pideal = P + p
Fig. (a) A molecule about to strike the wall has a net inward pull; (b) A molecule in the interior of gas has balanced attractions.
p is determined by the force of attraction between molecules (A) striking the wall of container and the molecules (B) pulling them inward.
The net force of attraction is, therefore, proportional to the concentration of (A) type molecules and also of (B) type of molecules. That is,
Where, n is total number of gas molecules in volume V and ‘a’ is proportionality constant characteristic of the gas.
Thus, the pressure P in the ideal gas equation is corrected as: for n moles of gas.
Van Der Waals Equation
Substituting the values of corrected pressure and volume in the ideal gas equation, PV = nRT, we get-
This is known as Van Der Waals equation for n moles of a gas. For 1 mole of a gas (n = 1), Van Der Waals equation becomes
Constants a and b in Van Der Waals equation are called Van Der Waals constants. These constants are characteristic of each gas.
Liquefaction of Gases – Critical Phenomenon
A gas can be liquefied by lowering the temperature and increasing the pressure.
At lower temperature, the gas molecules lose kinetic energy. The slow moving molecules then aggregate due to attractions between them and are converted into liquid. The same effect is produced by the increase of pressure. The gas molecules come closer by compression and coalesce to form the liquid.
Andrews (1869) studied the P–T conditions of liquefaction of several gases. He established that for every gas there is a temperature below which the gas can be liquefied but above it the gas defies liquefaction. This temperature is called the critical temperature of the gas.
The critical temperature, Tc, of a gas may be defined as that temperature above which it cannot be liquefied no matter how great the pressure applied.
The critical pressure, Pc, is the minimum pressure required to liquefy the gas at its critical temperature.
The critical volume, Vc, is the volume occupied by a mole of the gas at the critical temperature and critical pressure.
Tc, Pc and Vc are collectively called the critical constants of the gas. All real gases have characteristic critical constants.
At critical temperature and critical pressure, the gas becomes identical with its liquid and is said to be in critical state. The smooth merging of the gas with its liquid is referred to as the critical phenomenon.
Table: The critical constants of some common gases
Methods of Liquefaction of Gases
If a gas is cooled below its critical temperature and then subjected to adequate pressure, it liquefies. The various methods employed for the liquefaction of gases depend on the technique used to attain low temperature. The three important methods are:
Faraday’s method in which cooling is done with a freezing mixture
Linde’s method in which a compressed gas is released at a narrow jet (Joule-Thomson effect)
Claude’s method in which a gas is allowed to do mechanical work
Faraday’s Method
Faraday (1823) used freezing mixtures of ice with various salts for external cooling of gases. The melting of ice and dissolution of salts both are endothermic processes. The temperature of the mixture is lowered up to a temperature when the solution becomes saturated.
Fig. Faraday's method for the liquefaction of gases.
Faraday succeeded in liquefying a number of gases such as SO2, CO2, NO and Cl2 by this method. He employed a V-shaped tube in one arm of which the gas was prepared. In the other arm, the gas was liquefied under its own pressure. The gases liquefied by this method had their critical temperature above or just below the ordinary atmospheric temperature. The other gases including H2, N2 and O2 having low critical points could not be liquefied by Faraday’s method.
Linde’s Method
Linde (1895) used Joule Thomson effect as the basis for the liquefaction of gases. When a compressed gas is allowed to expand into vacuum or a region of low pressure, it produces intense cooling. In a compressed gas the molecules are very close and the attractions between them are appreciable. As the gas expands, the molecules move apart. In doing so, the intermolecular attraction must be overcome. The energy for it is taken from the gas itself which is thereby cooled and become liquid.
Figure: Linde’s Method for Liquefaction of Air
Linde used an apparatus worked on the above principle for the liquefaction of air. Pure dry air is compressed to about 200 atmospheres. It is passed through a pipe cooled by a refrigerating liquid such as ammonia. Here, the heat of compression is removed. The compressed air is then passed into a spiral pipe with a jet at the lower end. The free expansion of air at the jet results in a considerable drop of temperature. The cooled air which is now at about one atmosphere pressure passed up the expansion chamber. It further cools the incoming air of the spiral tube and returns to the compressor. By repeating the process of compression and expansion, a temperature low enough to liquefy air is reached. The liquefied air is collected at the bottom of the expansion chamber.
Claude’s Method
This method for liquefaction of gases is more efficient than that of Linde. Here, also the cooling is produced by free expansion of compressed gas. But in addition, the gas is made to do work by driving an engine. The energy for it comes from the gas itself which cools.
Figure: Claude’s Method for Liquefaction of Air
Thus, in Claude’s method the gas is cooled not only by overcoming the intermolecular forces but also by performance of work. That is why, the cooling produced is greater than in Linde’s method.
Pure dry air is compressed to about 200 atmospheres. It is led through a tube cooled by refrigerating liquid to remove any heat produced during the compression. The tube carrying the compressed air then enters the ‘expansion chamber’. The tube bifurcates and a part of the air passes through the side-tube into the cylinder of an engine. Here it expands and pushes back the piston. Thus, the air does mechanical work whereby it cools. The air then enters the expansion chamber and cools the incoming compressed air through the spiral tube. The air undergoes further cooling by expansion at the jet and liquefies. The gas escaping liquefaction goes back to the compressor and the whole process is repeated over and over again.
Uses of Liquefied Gas
To produce low temperature
Commercial preparation of Oxygen and Nitrogen
Liquid He can be used for intense cold
Liquid Cl2 can be used for bleaching purpose
Liquid O2 and He can be used for welding purpose
Determination of Molecular Weight
The ideal gas equation is-
PV = nRT
= (w/M) RT
So, P = (w/V) . (RT/M)
= d (RT/M)
Where, d is the density of the gas.
Thus, if a liquid can conveniently be converted into the vapour state then the measurement of the density of vapour will yield the molecular weight of the liquid in the vapour state.
There are some methods to determine the molecular weight of gas such as-
Regnault’s method
Duma’s method
Hofmann’s method
Victor Meyer’s method
The Buoyancy method
Victor Meyer’s Method
This is perhaps the most common method for the determination of the vapour density of substances which are liquid or solid at ordinary temperature. In this method the volume of a known weight of the substance is determined by measuring the volume of air displaced by it.
Figure: Victor Meyer’s Apparatus
When the condition becomes saturated temperature is noted on a thermometer inside the vapour jacket. The volume of vapour is noted in the graduated tube upon the mercury level. The difference of mercury level before and after introduction of vapour is the volume of the vapour. Since the volume of the vapour may be easily read from the graduated tube and the weight of liquid introduced is known as well as the density so the molecular weight of the vapour (the substance being examined) may be calculated.
Mathematical problems
Moles: http://mmsphyschem.com/moleWks.pdf
7) 8.00 g NaOH 8) 5.657 g H2SO4 9) 32 g KNO3 = ? mole//
7) n = 8.00 g NaOH x 1 mol NaOH/40.00 g NaOH = 0.200 mol NaOH
8) n = 5.657 g H2SO4 x 1 mol H2SO4/98.08 g H2SO4 = 0.05768 mol H2SO4
9) n = 32 g KNO3 x 1 mol KNO3/101.11 g KNO3 = 0.32 mol KN
Boyles: http://mmsphyschem.com/boyleL.pdf
1) A container holds 500. mL of CO2 at 20.° C and 742 torr. What will be the volume of the CO2 if the pressure is increased to 795 torr?
2) A gas tank holds 2785 L of propane, C3H8, at 830. mm Hg. What is the volume of the propane at standard pressure?
3) A balloon contains 7.2 L of He. The pressure is reduced to 2.00 atm and the balloon expands to occupy a volume of 25.1 L. What was the initial pressure exerted on the balloon?
4) A sample of neon occupies a volume of 461 mL at STP. What will be the volume of the neon when the pressure is reduced to 93.3 kPa?
5) 352 mL of chlorine under a pressure of 680. mm Hg are placed in a container under a pressure of 1210 mm Hg. The temperature remains constant at 296 K. What is the volume of the container in liters?
1. 467 mL CO2 2. 3040 L C3H8 3. 7.0 atm 4. 501 mL Ne 5. 198 mL Cl2
Charles: http://mmsphyschem.com/chuckL.pdf
1) A container holds 50.0 mL of nitrogen at 25° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?
2) A sample of oxygen occupies a volume of 160 dm3 at 91° C. What will be volume of oxygen when the temperature drops to 0.00° C?
4) 568 cm3 of chlorine at 25° C will occupy what volume at -25° C while the pressure remains constant?
1. 55.9 mL N2 2. 120 dm3 O2 4. 473 cm3 Cl2
Ideal: http://mmsphyschem.com/idealGas.pdf
1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?
2) Calculate the mass of 15.0 L of NH3 at 27° C and 900. mm Hg.
4) Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg.
1. 3.48 x 1013 N2 molecules 2. 12.3 g NH3 4. 2.82 g/L
Partial pressure: http://mmsphyschem.com/daltonL.pdf
1) The volume of hydrogen collected over water is 453 mL at 18° C and 780. mm Hg. What is its volume dry at STP?
2) A 423 mL sample of dry oxygen at STP is transferred to a container over water at 22° C and 738 mm Hg. What is the new volume of the oxygen?
3) Calculate the mass of 400. mL of carbon dioxide collected over water at 30.° C and 749 mm Hg.
4) 50.0 mL of dry fluorine at 20.0° C and 795 mm Hg will occupy what volume over water at the same temperature and pressure?
1. 436 mL H2 2. 484 mL O2 3. 0.669 g CO2 4. = 51.1 mL
