Articles by "PROPERTIES OF GASES"

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In 1873, Van der Waals modified the Ideal Gas Equation,
PiVi = RT
    By incorporating the size effect and inter molecular attraction effect of the real gases. These above two effects are discussing under the Volume Correction and Pressure Correction of the Ideal Gas Equation.

Volume Correction:

    Molecules are assumed to be a hard rigid spheres and in a real gas, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1 mole gas molecules,
Vi = (V - b)
    where V is the molar volume of the gas and b is the volume correction factor.
    Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.
    Let us take r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
What is the van der waals equation?
Encounter of the gas molecules.
    Then it can be shown that the volume correction term or effective volume of one mole gas molecules,
    b = 4 × N0 (4/3) π r3
    When two molecules encounter each other the distance between the centers of the two molecules would be σ. They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.
    This excluded volume = 4/3 π σ3 for a pair of molecules.
    Thus a single molecule,
    = 1/2 × 4/3 π σ3
    The effective volume for Avogadro number of molecules (present in 1 mole gas),
b = N0 × 2/3 × π σ3
or, b = N0 × 2/3 × π (2r)3
or, b = 4 N0 × 4/3 × π (r)3
Thus, b = N0 × 2/3 × π (2r)3
∴ b = 2/3 × N0 × π σ3
    Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule. The equation becomes,
    Pi (V - b) = RT

Pressure Correction:

    Pressure of the gas is developed due to the wall collision of the gas molecules. But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas Pi.
    Thus, Pi 〉P
    or, Pi = P + Pa
    Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular attraction in a gas, greater is the magnitude of Pa.
    Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules.
    Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
    Pa ∝ 1/V2
    Since, density ∝ 1/V
    ∴ Pa = a/V2
    Where a = constant for the gas that measures the attractive force between the molecule.
    Thus, Pi = P + (a/V2)
    Using the two corrections we have Van der Waals equation for 1 mole Real Gas is,
(P + a/V2 )(V - b) = RT
    To convert the equation for n moles volume has to change as it is the only extensive property in the equation.
    Let V be the volume of n moles gas hence, V = v/n. Putting this term in the Van der Waals equation, we have,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas
    Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 270C. Given, a = 1.4 atm lit2 mol-2 and b=0.04 lit mol-1 using Van der Waals equation.
    Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.
    P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:

    (a)For the real gas a=0 (that is no inter molecular attraction )
    but b≠0 (size considers):
    We have Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT but a=0
    Hence, P = nRT/(V - nb) 〉Pi
    Since, Pi = nRT/V only.
    (b)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):
    It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume.
    We have Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT but b=0(no size).
    Hence, P = (nRT/V - an2/V2)ㄑPi
    Since, Pi = nRT/V only.
    Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of Van der Waals Constant a and b:

    From the Van der Waals Equation,
    (P+ an2/V2)(V - nb) = nRT
    where, Pa = an2/V2
    Where Pa is the pressure correction term, called often the internal pressure of the gas.
    Then, a = Pa × (V2/n2)
Thus the unit of a = atm lit2 mol-2
    Again, nb = Unit of volume (say liter)
Hence, the unit of b = lit mol-1
    Find the Units and Dimensions of Van dar Walls constant 'a' and 'b' from Van der walls Equation.
    In SI system,unit of 'a' = N m4 mol-2
    In CGS system,unit of 'a' =dyne cm4 mol-2
    In SI system,unit of 'b' = m3 mol-1
    In CGS system,unit of 'b' = cm3 mol-1

Significance of Van der Waals constants 'a' and 'b':

    a’ term originates from the inter molecular attraction and Pa = an2/V2. Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules. Higher the value of ‘a’ grater is the inter molecular attraction and more easily the gas could be liquefied.
    Thus, a of CO2 = 3.95 atm lit2 mol-2 while, a of H2 = 0.22 atm lit2 mol-2.
    Another constant, ‘b’ measures the molecular size and also a measure of repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ). The grater the value of ‘b’ larger the size of the gas molecule.
    Thus, b of CO2 = 0.04 lit mol-1 while, b of H2 = 0.02 lit mol-1.

For a Van der Waals gas the expression for Boyle's temperature:

    Mathematical condition for Boyle’s temperature is,
    TB = [d(PV)/dP]T When P→0
    From the Van der Waals Equation for 1 mole gas,
    (P + a/V2)(V - b) = RT
    or, P = RT/(V - b) - a/V2
    Hence, PV = {RTV/(V - b)} - a/V
    Thus, TB = [d(PV)/dP]T
    = [RT/(V-b)-{RTV/(V-b)2}+a/V2][(dV/dP)]T
    = [{RT(V - b) - RTV
    }/(V - b)2 + a/V2] [(dV/dP)]T
    =[{- RTb/(V - b)2} + a/V2] [(dV/dP)]T
    But when T = TB,
    [d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0
    Thus, We have, RTBb/(V - b)2 = a/V2
    or, TB = (a/Rb) {(V - b)/V}2
    Since P → 0, V is large.
    Thus, (V - b)/V ≃ 1
Hence,TB = a/Rb
    Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1.
    TB = 427 K
    Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.
  1. When, a = 0, Van der Waals Equation, P (V - b) = RT or, PV = RT + Pb Differentiating with respect of pressure at constant temperature, [d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0. The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]T ≠ 0.
  2. Again when a = b = 0, Van der Walls equation becomes PV = RT; or, [d(PV)/dP]T = 0.
    That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.

Explanation of Amagat’s Curves:

Amagat’s Curves.
Van der Waals equation for 1 mole real gas,
(P + a/V2)(V - b) = RT
or, PV - Pb + (a/V) + (a/V2) = RT
Neglecting the small term (a/V2), we get,
PV = RT + Pb - (a/V)
Using ideal gas equation for small term,
a/V = aP/RT and taking Z = (PV/RT),
We have, Z = 1 + (1/RT){b - (a/RT} P
This Shows that, Z = (T,P)
    This equation can be uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO2 ‘a’ is very high since the gas is easily liquefied.
    Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve .
    That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
(b) For H2 ‘a’ is very small and It is not easily liquefied.
    So, a/RT〈 b and slope of Z vs P curve for H2 becomes (+) ve and the value of Z increases with pressure.
(i) When T〈 TB
or, T〈 a/Rb
or, b〈 a/RT
and {b−(a/RT)} = (-)ve.
    That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO2 , Z〈 1 and more compressible.
(ii) When T = TB = a/Rb
or, b = a/RT
hence, {b - (a/RT)} = 0
    That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region.
(iii) When T 〉TB
or, T 〉a/Rb
or, b 〉a/RT
hence {b - (a/RT} = (+)ve.
    That is value of Z increases with increase of P when T 〉TB.
    The size effect dominates over the effect due to inter molecular attraction. For hydrogen and helium, 0°C is grater then their TB values and Z vs P slope becomes (+)ve.
    At very low pressure P→0 and high temperature,the volume is very large and both the size effect and attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys ideal behavior.

A gas can be liquefied by lowering temperature and increasing of pressure. But influence of temperature is more important. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.
This temperature of the gas is called its Critical Temperature (Tc). A gas can only be liquefied when the temperature is kept below Tc of the gas.

Definitions of Critical Constants:

    Critical Temperature (Tc):
    Critical Temperature (Tc) is the maximum temperature at which a gas can be liquefied, that is the temperature above which a liquid cannot exist.
    Critical Pressure (Pc):
    Critical Pressure (Pc) is the maximum pressure required to cause liquefaction at the temperature (Tc).
    Critical Volume (Vc):
    Critical Volume (Vc) is the volume occupied by one mole of a gas at critical temperate (Tc) and Critical Pressure (Pc).

Andrews Isotherms:

    In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Andrews Isotherms.

Following are observed from this graph:

  1. At high temperature, such as T4, the isotherms look like those of an Ideal Gas.
  2. At low temperatures, the curves have altogether different appearances. Consider, for examples, a typical curve abcd. As the pressures increases, the volume of the gas decreases (curve a to b).
  3. At point b liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure P.
  4. At the point C, liquefaction is complete and thus the cd is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that ab represents the gaseous state, bc, liquid and vapour in equilibrium, and cd shows the liquid state only.
  5. At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.
  6. At temperatures Tc the horizontal portion is reduced to a mere point. At temperatures higher then Tc there is no indication of qualification at all.
Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

Continuity of State:

    It appears from the Amagat curve at T there is discontinuity or break during the transformation of gas to liquid. But it is not so.
    The continuity of the states from the gas to liquid can be explains from the following isotherm pqrs at T.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Continuity of State.
    The gas at A is heated to B at constant volume (V) along AB. Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant V until the point D is reached.
    No where in the process liquid would appear. At D, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state.
    Hence there is hardly a distinction between the liquid state and the gaseous state. There are no line of separation between the two phases. This is known as the principle of continuity of state.

Critical Phenomena and Van der Walls Equation:

(P + a/V2)(V - b) = RT
V3 - (b + RT/P)V2 + (a/P)V - ab/P = 0
    This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Critical Phenomena and Van der Walls Equation.

The main characteristics of the above isotherm are as follows:

  1. At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
  2. At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V1, V2 and V3 at pressure P1.
    The section AB and ED of the Van der walls curve at T1 can be realized experimentally. ED represents Supersaturated or Super cooled vapour and AB represents super heated liquid. Both these states are meta-stable.
    These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
  3. The section BCD of the Van der Walls isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease of pressure. The line BCD represents the meta- stable state.

Expression of Critical Constants in Terms of Van der Waals Constants:

    Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.
    Thus the mathematical condition of critical point is,
    (dP/dV)T = 0 and (d2P/dV2)T = 0
    Van der Waals equation for 1 mole gas is,
    (P + a/V2)(V - b) = RT
    or, P = RT/(V - b) - a/V2
    Differentiating Van der Waals equation with respect to V at constant T,
    We get Slope,
    (dP/dV)T = - {RT/(V - b)2} + 2a/V3
    And Curvature,
    (d2P/dV2)T = {2RT/(V - b)3} - 6a/V4
    Hence at the critical point,
    - {RTc/(Vc - b)2} + 2a/Vc3 = 0
    or, RTc/(Vc - b)2 = 2a/Vc3
    and {2RTc/(Vc - b)3} - 6a/Vc4 = 0
    or, 2RTc/(Vc -b)3 = 6a/Vc4
    Thus, (Vc - b)/2 = Vc/3
∴ Vc = 3b
    Putting the value of Vc = 3b in RTc/(Vc - b)2 = 2a/Vc3.
    We have, RTc/4b2 = 2a/27b3
∴ Tc= 8a/27Rb
    Again the Van der Walls equation at the critical state,
    Pc = RTc/(Vc - b) - a/Vc2
    Putting the value of Vc and Tc,
∴ Pc = a/27b2
    The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm3mol-1. Calculate the value of Van der Waals constants a and b.
    We have Tc = 647 K,
    Pc = 22.09 Mpa = 22.09 × 103 kPa
    and Vc = 0.0566 dm3 mol-1
    Thus, b = Vc/3 = (0.0566 dm3 mol-1)/3
    ∴ b = 0.0189 dm3 mol-1
    a = 3 Pc Vc2
    = 3 (22.09 × 103 kPa)(0.0566 dm3 mol-1)2
    ∴ a = 213.3 kPa mol-2

Compressibility Factor at the Critical State (Zc):

    The Critical coefficient is defined as,
RTc/PcVc
    Thus the value of Critical Coefficient,
    = {R × (8a/27Rb)}/{(a/27b2) × 3b}
    = 8/3
    = 2.66
    Thus the value of Compressibility factor at the critical state (ZC)
    = PCVC/RTC
    = 3/8
    = 0.375

Van der Waals constants in terms of critical constants:

    Van der Waals constants can be determined from critical constants Tc and Pc of the gas. Vc in the expression is avoided due to difficulty in its determination.
    We have, b = Vc/3
    but, PcVc/RTCc = 3/8
    or, Vc = (3/8) × (RTc/Pc)
∴ b = (1/8)(RTc/Pc)
    Again, a = Pc × 27b2
    = 3 × Pc × (3b)2
    = 3 PcVc2
    = 3Pc × (3RTc/8Pc)2
∴ a = (27/64)(R2Tc2/Pc)
    Calculate Van der Waals constants for Ethylene. (TC = 280.8 K and PC = 50 atm). 
    a = 0.057 lit mol-1 and b = 4.47 lit2 atm mol-2
    Argon has (TC = - 122°C, PC = 48 atm). What is the radius of the Argon atom?
    1.47 × 10-8 cm
Critical Constant Values of some Commonly used Substance
Gas TC (0K) PC (atm.) VC (c.c.)
Oxygen 154.28 49.7 74.4
Hydrogen 33.2 12.8 69.7
Nitrogen 125.97 33.5 90.0
Ammonia 405.5 111.3 72.0
Carbon
dioxide
304.15 72.9 94.2
Helium 5.2 2.25 61.55
Methane 190.3 45.6 98.8

Boyle’s Law:

    At constant temperature, the volume of a definite mass of a gas is inversely proportional to its pressure.
    That is, the volume of a given quantity of gas, at a constant temperature decrees with the increasing of pressure and increases with the decreasing pressure.
    Let, a Cylinder contain 10 ml of gas, at constant temperature and 1 atm pressure. if the pressure increases to 2 atm then the volume also decrees to 5 ml.

Mathematical Representation of Boyle's Low:

V ∝ 1/P
∴ PV = K
Where, K is a constant whose value depends upon the,
a. Nature of the Gas.
b. Mass of the Gas.
For a given mass of a gas at constant temperature,
P1V1 = P2V2
    Where, V1 and V2 are the volume at P1 and P2 are the pressure respectively.

Graphical Representation of Boyle's Low:

    The relation between pressure and volume can be represented by an arm of rectangular hyperbola given below.
    As the value of the constant in equation will change with temperature, there will be a separate curve for each fixed temperature. These curves plotted at different fixed temperature are called isotherms.
Boyle's and Charl's Low with derivation and Graphical Representation.
Graphical Representation of Boyle's Low
    At Constant temperature a given mass of an Ideal Gases the product of Pressure and Volume is always same. if the product of pressure and volume represents in Y axis and Pressure represents X axis a straight line curve is obtained with parallel to X axis.
    This Graph is shows that, at constant temperature the product of pressure and volume is does not depends on its pressure.

Relation between Pressure and density of a gas:

    At constant temperature(T) a definite mass of gas has pressure P1 at volume V1 and pressure P2 at volume V2.
According to Boyle’s Law,
P₁ V1 = P2 V2
or, P1/P2 = V2/V1
Again, Let the mass of the Gas = M
Density D1 at Pressure P1
and Density D2 at Pressure P2
Thus, D1 = M/V1 and D2 = M/V2
or, V1 = M/D1 and V2 = M/D2
Again, P1/P2 = (M/D2) × (D1/M)
= D1/D2
P1/P2 = D1/D2
P/D = Constant(K)
P = K × D
PD 
    Thus, at constant temperature density of a definite mass of a gas is Proportional to its Pressure.

Relation between Volume and Temperature of a Gas:

    At constant pressure a definite mass of gas, with the increasing of temperature, volume also increases and with decreasing temperature, volume also decreases.
    That is, the volume of a given mass of gas at constant pressure is directly proportional to its Kelvin temperature.

Charl’s Law:

    At constant pressure, each degree rise in temperature of a definite mass of a gas, the volume of the gas expands 1/273.5 of its volume at 00C.

Mathematical Representation of Charl's Low:

    If V0 is the volume of the gas at 00C, then 10C rise of temperature the volume of the gas rise V0/273.5 ml.
10C temperature the volume of the gas,
(V0 + V0/273) ml
= V0 (1 + 1/273) ml.
At t0C temperature the volume of the gas,
Vt = V0 (1+ t/273) ml
= V0 (273 + t°C)/273 ml.
    It is convenient to use of the absolute temperature scale on which temperature is measured Kelvin(K). A reading on this scale is obtained by adding 273 to the Celsius scale value.
Temperature on Kelvin scale is,
T K = 273+t°C
Vt = (V0 × T)/273 = (V0/273) T
    Since V0, the volume of a gas at 0°C, has a constant value at a given pressure, the above relation expressed as,
Vt = K₂ T
VT
    Where K2 is constant whose value depends on the, Nature, mass and pressure of the gases.
    According to the above relation Charl’s Law states as,
    At constant pressure the volume of a given mass of gas is directly proportional to its Kelvin temperature.

Graphical Representation of Charl's Low:

    A typical variation of Volume of a gas with change in its kelvin temperature a straight line plot was obtained, Called isobars.
    The general term isobar, which means at constant pressure, is assigned to these plots.
Boyle's and Charl's Low with derivation and Graphical Representation.
Graphical Representation of Charl's Low.

Absolute Temperature or Absolute Zero:

    Since volume is directly proportional to its Kelvin temperature, the volume of the gas is theoretically zero at zero Kelvin or - 2730C.
    However this is indeed hypothetical because all gases liquefies and then solidity before this low temperature reached. In fact, no substance exists as a gas at the temperature near Kelvin zero, through the straight line plots can be extra plotted to zero volume.
    The temperature corresponds to zero volume is -2730C.

Relation between temperature and Density of a given Ideal Gas at constant Pressure:

From the Charl’s Law,
V1/V2 = T1/T2
Again, the mass of the gas is M.
Density D1 and D2 at the Volume V1 and V₂ respectively.
Then, V1 = M/D1 and V2 = M/D2
Thus, (M/D1 )/(M/D2 ) = T1/T2
or, D₂/D1 = T1/T2
D ∝ 1/T
    Thus at constant pressure, density of a given mass of gases is inversely proportional to its temperature.

Combination of Boyle’s and Charl’s Law:

From Charl's Law,
V ∝ 1/P When T Constant.
From Charl's Law,
VT When P Constant.
When all the variables taken into account the variation rule states as,
Then, VT/P
PV/T = K(Constant)
∴ (P1V1)/T1 = (P2V2)/T2 = Constant PV = KT
    Thus the product of the pressure and volume of a given mass of gas is proportional to its Kelvin temperature.
    At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm3, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.
From the Combination of Boyle's and Charl's Law is,
(P1V1)/T1 = (P2V2)/T2
Here, P1 = 1 atm; V1 = 2000 cm3 and T1 = 300K and P2 = 2 atm; V2= ? and T2 = 600 K.
∴ 1×2000/300 = 2×V₂/600
or, V2 = (1×2000×600)/(300×2)
= 2000 cm3

Simply one equation can be used to distinguish between ideal gas from real gas and this is,
PV = nRT
The gas which obeys this equation under all conditions of temperature and pressure is called ideal gas and the gas which does not obey this equation under all conditions of temperature and pressure is called real gases. A number of points can be discussed to compare the two types of gases.
  1. The Ideal Gas can not be liquefied since it has no inter-molecular attraction and so that molecule will not condense.
  2. The coefficient of thermal expansion(ɑ) depends on temperature(T) of the gas and does not depends on the nature of the gas.
  3. The coefficient of compressibility(β) similarly depends on the Pressure(P) of the gas and will be same for all gases.
  4. When P is plotted against V, at constant temperature(T) a rectangular hyperbola curve is obtained as demanded by Boyle's Low, 
  5. PV = Constant
    The hyperbola Curve at each temperature is called one isotherm and at different temperature we have different isotherms. Two isotherms will never intersect.
  6. When PV is plotted against P , at constant T a straight line parallel to P-axis is obtained. At different temperature there will be different parallel lines.
  7. Comparison Between Ideal and Real Gases.
    Graphical Representation of Boyle's Low
  8. When an ideal gas passes through a porous plug from higher pressure to lower pressure within insulated enclosure, there will be no change of temperature of the gas . This confirms that the ideal gas has no inter-molecular attraction.
Show that coefficient of thermal expansion of Ideal Gas depends on the temperature of the gas.
Coefficient of thermal expansion(ɑ) is defined as,
ɑ = (1/V)[dV/dT]P
Ideal Gas Equation for 1 mole gas is,
PV = RT
Hence [dV/dT]P = R/P
Thus ɑ = (1/V) × (R/P) = (R/PV) = 1/T.
This means all the gases have the same coefficient of thermal expansion.
Show that coefficient of Compressibility of Ideal Gas depends on the Pressure of the gas.
Coefficient of Compressibility(β) is defined as,
β = - (1/V)[dV/dP]T
Ideal Gas Equation for 1 mole gas is,
PV = RT
Hence [dV/dP]T = - (RT/P2)
Thus β = (1/V) × (RT/P2)
= (RT/P2V)
= (RT/PV) × (1/P)
= (1/P)
This means coefficient of Compressibility depends on the Pressure(P) of the gas.
Real Gas:
  1. This gas could be liquefied since it has inter molecular attraction which helps to coalesce the gas molecules.
  2. The coefficient of thermal expansion (ɑ) is found to vary from gas to gas that is α depends on the nature of the gas.
  3. The coefficient of compressibility (β) also is found to depend on the nature of the gas.
  4. When P is plotted against V, a rectangular hyperbola is obtained only at high temperature (above the critical temperature).
  5. But a temperature below the critical temperature(C), the gas is liquefied after certain pressure depends on temperature. The point C is the critical point where he liquid and gas can be indistinguishable.
  6. When PV is plotted against P for real gas, following plots, called Amagat Curve are obtained.
  7. When real gases pass through porous plug from higher pressure to lower pressure within insulated enclosure, there occurs a change of temperature. This is due to the fact that real gases have inter-molecular attraction and when the gas expands, the molecules have to spend energy to overcome inter-molecular attraction and so the temperature of the gas drops down.
Amagat Curve:
Comparison Between Ideal and Real Gases.
Amagat Curve
It shows that for most gases, value of Z decreases, attain minimum and then increases with the increase of P.
Only Hydrogen(H2) and Helium(He) baffle this trend and the curve rises with increase of Pressure(P) from the very beginning.
For CO2, there is a large dip in the beginning, In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low pressure region.

Boyle Temperature(TB):

At some intermediate temperature TB called Boyle temperature, the initial slope is zero.
At the Boyle temperature, the Z versus P line of an ideal gas is tangent to that of a real gas when P approaches zero. The latter rises above the ideal gas line only very slowly.
Thus, at the Boyle temperature, the real gas behaves ideally over a wide range of pressure, because the effect of size of molecules and inter-molecular forces roughly compensate each other.
Boyle Temperature(TB) is give by, 
[d(PV)/dP]T when P → 0
The Boyle temperature of Some gases are given below:
Gases TB
Hydrogen (H2) -1560C
Helium (He) -2490C
Nitrogen (N2) 590C
Methane (CH4) 2240C
Ammonia (NH3) 5870C
Thus we can see that for H2 and He, the temperature of 00C is above their respective Boyle temperature and so they have Z values greater than unity. The other gases at 00C are below their respective Boyle temperature and so hay have Z values less than unity in the low pressure region.
Compressibility Factor(Z):
An important single parameter, called Compressibility factor (Z) is used to measure the extent of deviation of the real gases from ideal behavior.
It is defined as,
Z = PV/RT
When Z=1, the gas is ideal or there is no deviation from ideal behavior.
When Z ≠ 1, the gas is non-ideal and departure of the value of Z from unity is measure of the extent of non-ideality of the gas.
When Zく1, the gas is more compressible then ideal gas and When Z 〉1, the gas is less compressible then ideal gas.
At 273 K and under pressure of 100 atm the compressibility factor of O2 is 0.97. Calculate the mass of O2 necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.
Mass of O2 = 1600 gm = 1.6 Kg 
For Solution see Question Answers Physical Chemistry Problem 8.
Questions and Answers of Physical Chemistry

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