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Ideal Gas

Real GasIdeal GasA gas which obeys the ideal gas equation, PV =nRT under all conditions of temperature and pressure is called an ideal gas. There is no gas which obeys the ideal gas equation under all conditions of temperature and pressure. The gases are found to obey the gas laws if the pressure is low or the temperature is high.Such gases are known as real gases. It is found that gases which are soluble in water or are easily liquefiable show larger deviation than gases like H2, 02, N2 etc. Deviation from ideal behaviour According to Boyle's law , PV = constant at constant temperature.Hence , at constant temperature ,plot of PV vs P should be a straight line parallel to x-axis. For gases like H2 and He ,PV increases continuously with increase of pressure whereas for gases like CO , CH4, PV first decreases with increase of pressure and reaches a minimum value and then increases continuously with increase of pressure. If we plot experimental value of pressure versus volume at constant temperature and theoretically calculated values from Boyle's Law, the two curves do not coincide.

Cause of Deviation form ideal behaviour The real gases obey ideal gas equation only if the pressure is low or the temperature is high.If the pressure is high or temperature is low, the real gases show marked deviation from ideal behaviour. At high pressure or low temperature, the following two assumptions of Kinetic theory of gases are faulty: 1)The volume occupies by the gas molecules is negligible as compared to the total volume of the gas. 2)The force of attraction or repulsion between the gas molecules are negligible. If the pressure is high or the temperature is low, the gas molecules come close together.Hence under these conditions: 1)The forces of attraction or repulsion between the molecules may not be negligible. 2)The volume occupied by the gas may be small that the volume occupied by the molecules may not be negligible. Van der Waals EquationJ.D. van der waal in 1873 , modified the ideal gas equation by applying correction so as to take into account1) The volume of the gas molecules2) The forces of attraction between the gas moleculesVan der waal equation for 1 mole of the gas,

(

P+

)

(v-b)=RTVan der waal equation for n mole of the gas,

(

)

(V-nb)=nRT Derivation: ven der Waals EquationThe equation of state (i.e., ideal gas equation) for perfect gas can be stated as: PV=RT (for 1 mole of gas) .... (i) The basic assumption while deriving, the perfect gas equation PV=RT are:(i) There exist no intermolecular interaction.(ii) The size of individual gas molecules can be neglected with respect to size of its container.But from different observation and experiments the gas equation for real gas is deviated from equation (i). The deviation is small at low pressure and high temperature. But, the deviation is largeat high pressure and low temperature. From J - T expansion, it has been observed that, there exist intermolecular interaction in real gases. Similarly large deviation of real gases from peffect gas equation PV=RT; leads Vander Waal's to correct it. Therefore, it is corrected by considering intermolecular interaction and finite size of gas molecules.(a) Correction due to intermolecular interaction For the gas molecules, which are well inside of the container, the net cohesive (attractive) force becomes zero. But for the gas molecules which are striking to the walls of the container, the observed pressure is less than actual pressure. This decrease in pressure depends on the number of gas molecules; striking to the wall of container and the number of gas molecules attracting from back sides. Both of these factors depend upon the number of molecules per unit volume, which means they depend on the density of gas. ∴ Decrease in pressure (p)∝𝜌

∝

, Where ' a ' is constant.So the real pressure is: (P+p)=

(

P+

)

.(ii) Correction in volume As the molecules have finite size, they occupy a space Hence the actual space for their movement is less than observed volume V of the container. Therefore the volume V in the equation of can be written as: (V-b) Where b is constant for unit mass of the gas (as volume occupies gas molecules). Thus the Vander Waal's equation for a real gas becomes:

(

P+

)

(V-b)=RT:

for

mole of

gasThis equation is called Van der Waal's equation. The constants a and b are called Van der Waal's constant and they are characteristies values for a particular gas. Critical Constants The temperature below which a gas can be liquefied by the application of pressure alone is called critical temperature and is denoted by T

The corresponding value of pressure and volume are called critical pressure

(

)

and volume

(

)

respectively. From Vander Waal's equation of state, we have:

(

P+

)

(V-b)=RTWhere, a and b are Vander Waal's constants.or, P=

V-b

...... (i) Differentiating with respect to V, we get:or,

(V-b)

..............(ii)Again, differentiating,

2RT

(V-b)

.....................

(

iii

)

At critical point, (i.e., at point of inflexion):

and

with

T=T

,P=P

and

V=V

}

...............

(

)

So, from equation (ii), (iii) and (iv), we get:

(

-b

)

..........

(

)

and

2RT

(

-b

)

..........(vi)Dividing equation (v) and (vi), we get: V

=3b ..................

(

vii

)

Then, from equation (v), using (vii), we get:T

27Rb

...........

(

viii

)

Using the value of V

and T

equation (i) becomes at critical point a P

-b

(using value of T

and

)

On solving, we get:P

27b

............

(

)

Thus, the equation (vii), (viii) and (ix) gives the expression of critical values of thermodynamies respectively.Q. Write the physical significance of terms ' a ', ' b ' and

in Vander Waal's equation.Solution:The Vander Waal's equation for real gas can be stated as.

(

P+

)

(V-b)=RT (

for

mole real gas)

Here, the term ' a ' is related to the correction of pressure and this term determines the strength of intermolecular attraction between the gas molecules. Its unit is litre

atm. mole

-2

. Similarly the term b signifies to the correction of volume and this term determines the size of gas molecules as well as the molecules. Its unit is litre mol " or free space available for gas m

~mol

-1

The values of a and b are different for different gases and are non-significant (i.e., zero) for perfect gas. Thus, a and b are the Vander Waal's constant for correction of and volume respectively.Further, the term

signifies that as the pressure (or decrease in pressure) of the system of real gas molecules. Q. What is critical temperature? Draw the Andrew curve for CO

at different temperature and discuss the result obtained.Solution:

Andrew's experimental isothermals of CO

gas at different temperature are as shown in figure.At temperature 13.1

∘

C, the curve AB represents gascous state of CO

and from A and B the gas obeys Boyle's law. From B to C gas changes to liquid phase and at point C whole gas has been liquefied. The curve CD represents incompressible liquid state of CO

gas. At temperature 21.5

∘

C, the nature of curve is similar but the horizontal portion FG of the curve gets decreased then that of 13.1

∘

C. At temerature 31.1

∘

C horizontal portion converges to the single points O indicating that gas cannot be liquefied whatever large pressure be applied to decrease volume. So, the point ' O ' is called the point of inflection. Above the temperature 31.1

∘

C, the horizontal portion of compressibility decreases and at temperature 48.1

∘

C curve becomes smooth and similar to prefect gas obeying Boyle's law. Thus, above figure, it is clear that CO

gas liquefies only at a temperature below 31.1

∘

C. So, this temperature is called critical temperature which is different for different gases. Hence, we can conclude that; the temperature below which a gas can be liquefied by the application of pressure alone is called the critical temperature. Above this temperature, the gas cannot be liquefied however high pressure may be applied. The corresponding value of pressure and (Pc) and volume

(

)

. temperature are called critical pressure respectively.Joules -Thomson ExpansionDefine enthalpy. Show that enthalpy remains constant in Joule-Thomson experiment. Enthalpy is defined as total heat content in the system. Mathematically, enthalpy: H=U+PV

When a gas is expanded from high constant pressute to low constant pressure under adiabatic condition, its temperature gets changed. Such effect is called Joule Thomson effect. Consider a cylinder CC' with perfectly non conducting wall having porous plug G and two movable pistons 1 and 2 respectively as shown in fig above. Let 1 mole of gas having volume V

filled between piston 1 and porous plug. When the gas is compressed with piston 1 , then the gas flows through porous plug through orifices and becomes throttled. Let P

be the initial constant high pressure and volume before expansion. Similarly let P

be the final constant low pressure and volume of gas after expansion, and x

and x

be the distances through which pistons A and B move respectively. Now, work done by the piston A on the gas, W

= Force × Displacement =P

×A×x

….(i)Where, A is the cross section area of cylinder \& work done by the gas on the piston B is W

×A×x

............(ii)∴ Net work done by the gas (W)=P

-P

… (iii) Since, the gas is expanded under adiabatic condition. So, this work is done at the expenses of its internal energy. Let U

and U

be the total internal energy (i.e., K.E+P.E) of gas before and after expansion respectively through the porous plug.The change in internal energy dU = U1- U2 ............................ (iv)According to first law of thermodynamics: dQ=dU+dW. Then, we have: U

-U

-P

(:dQ=0 for adiabatic process) or, U

............(v) or, H

or, H= Constant.Here, the quantity H is the total heat function or enthalpy of the gas and is defined as H=U+PV. Therefore, enthalpy remains conserved during Joule-Thomson expansion. This expansion is also called iso-enthalphic effect. Note: Special case from equation (iii)(i) If P

. Results in cooling(i) If P

: Results in heating(ii) If P

: No change (i.e. the gas is perfect and obeys Boyle's law).Short Questions # Write Merits and Demerits of Van Der Waals Equation of State.Merits• It can predict the behaviour of gas much better and accurately than the ideal gas equation.• It is also applicable to fluids in spite of gases.• The arrangement is made in a manner of cubic equation in volume. The cubic equation can give three volumes which can be used for calculating the volume at and below the critical temperatures. Demerits• It can only get accurate answers for real gases which are above the critical temperature.• Below critical temperature results also get accepted.• In the transition phase of gas, the equation is a failure.# What are the results of Porous Plug experiment? Joule and Thomson performed the experiment for different gaseous system on measuring the temperature values of gaseous system before and after the expansion by thermometers T

, then they get decrease in temperature. This decrease in temperature, during the porous plug experiment is called Joule Thomson's effect. So, they concluded that there exist intermolecular forces of attractions in between gas molecules.# What is throttling process? In Joule Thomson experiment, the porous plug" (such as compressed form of cotton or wool etc.) provides a large number of very narrow parallel paths for the gas molecules while crossing these orifices under at constant high pressure to at constant low pressure. As a result gas molecules are pulled apart to each other. This phenomenon is called throttling process.# Show that for Vander Waal's gas: R=

. Hence, calculate critical coefficient. Or, Starting from Vander Waal's equation, derive an equation which does not involve Vander Waal's constants a and b. Since, V

=3b, Critical pressure

(

)

27b

and critical temperature

(

)

27Rb

∴

27b

.3b.

27Rb

R∴ R=

This equation is independent of Vander Waal's constants a and b.Critical coefficient: The quantity

is called critical coefficient of a gas. Its calculated value is

and is same for all gases.# Write limitations of Vander Waal's equation? Vander Waal's equation of state includes qualitatively all the features of actual gases. So, it is a convenient the behavior of real gases. But, it has certain limitations as below:(i) It fails over very wide pressure and temperature ranges to describe features of gases.(ii) It breaks down when applied to the difference between the various gases:(iii) The values of Vander Waal's constants a and b that varies with vary in temperature and also they are different for different gases.(iv) The Vander Waal's equation is not obeyed exactly by any gas in the neighborhood of its critical point.# Derive relation between Boyle temperature of inversion and critical temperature. We know that, temperature of inversion T

…. (i)

Boyle's temperature

…

(ii)

and critical temperature compositeblock dir="ltr" class="math-container-symbol role-mathmode-area inline" style="font-size: 17px;">

(

)

27Rb

… (iii) From (i) and (iii), T

⋅

or, T

.... (iv)Also from (i) and (ii), T

=2T

…... (v) From (iv) and (v), we get: T

=2T

⋅T

or,

=2T

= 6.75T

.............(vi)or, T

Equation (vi) gives relation between T

and T

. # Calculate the critical tempcrature and Boyle temperatare of CO

, for which the Vander Waal's constant are given to be: a=0.0072 and b=0.002. The unit of pressure is atmosphere \& the unit of volume is that of a gm. mol. of the gas at NTP. From Vander Waal's equation, we have:

(

P+

)

(V-b)=RT

At N.T.P., P=1 and V=1 (given)∴

[

0.0072

]

(1-0.02)=R×273This gives, R=

1.0072×0.998

273

Also, the critical temperature is given by;

TC	=8a 27Rb=8×0.0072 27×(1.0072×0.998 273)×0.002
	=289.6K=(289.6-273)∘C=16.6∘C.

& the Boyle temperature T

in terms of critical temperature is given by,

TB	=6.75 2TC=6.25 2×289.6K
	=977.4K=(977.4 -273)∘C=664.4∘C

# Calculate the critical temperature of a gas for which the Vander Waal's constants are: a=0.00874,b=0.0023 and the gas constant is given by: 273R=1.00646.Given a=0.00874,b=0.0023 and R=

1.00646

273

The critical temperature is given by, T

27Rb

…..(i)Substituting these values in equation (i), we get:T

8×0.00874

27×

(

1.00646

273

)

×0.0023

=305.5

∘

# T

and P

of CO

are 31

∘

C and 73 atm respectively. Assuming that CO

gas obeys the Van der Walls equation, find (1) critical density and (2) diameter of CO

molecule.Solution,

Critical Temperature

(

)

=31

∘

C=(31+273)K=204K

Pressure

(

)

=73atm=73 atm

Molar Mass of CO

(

)

=44

(i) Critical density

(

𝜌

)

= ? (ii) Diameter of Molecule

(

)

=? We have,i) For a gas which obeys van der Waal's equation,R =

∴ critical volume

(

)

3 ⨯ 82.07 ⨯304

8 ⨯ 73

= 128.16 cm

We know,

Critical density

(

𝜌

)

molar mass

critical volume

128.16

= 0.3433 gram/cm

ii) Diameterwe know,

= 3bor, b=

= 42.72 cm

But, b= 4nv where n is number of molecules per cm3 and v is the volume per molecule. Avogadro's number is = 6.02 ⨯ 10

mol

-1

and gram molecular volume = 22400 cm

So, number of molecules per unit volume

(

)

6.02⨯ 10

22400

= 2.6875 ⨯10

Hence, b= 4nvor, 42.72 = 4⨯2.6875⨯10

vor, v=

42.72

4⨯2.6875 ⨯10

= 3.97 ⨯10

-19

Now volume of a molecule

(

)

𝜋r

(

r= radius of molecule

)

or, r =

(

4𝜋

)

1/3

(

3 ⨯ 3.97 ⨯10

-19

4⨯ 3.142

)

1/3

= 4.56 ⨯10

-7

cm∴ diameter of molecule = 2⨯r = 2 ⨯ 4.56 ⨯10

-7

= 9.12 ⨯10

-7

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