Priyam Study Centre

A Man would do nothing, if he waited until he could do it so well that no one would find fault with what he has done.

Showing posts with label PROPERTIES OF GASES. Show all posts
Showing posts with label PROPERTIES OF GASES. Show all posts

Nov 6, 2018

Critical Constants of a Gas

Critical Constants of a Gas:

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 tempe-rature (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. 
Critical Temperature Critical Pressure and Critical volume and Establish the Relationship of Critical constants with Van der Waals Constants
Andrew's Graphs of P Vs V
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). 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. 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.
  3. 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.
  4. At temperatures TC the horizontal portion is reduced to a mere point. At temperatures higher then TCthere 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
Critical Temperature Critical Pressure and Critical volume and Establish the Relationship of Critical constants with Van der Waals Constants
Schematic Representation of the 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:

For one mole of a gas the Van der Waals equation,
(P + a/V²)(V - b) = RT
-(b + RT/P)+(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.
Critical Temperature Critical Pressure and Critical volume and Establish the Relationship of Critical constants with Van der Waals Constants
Van der Waals Isotherms
The main characteristics of the above isotherm are as follows:
(a) At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
(b) 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 T₂ 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.
(c) 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/VC = 0
or, 2RTC/(VC -b)³ = 6a/VC
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 is
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.

a = 212.3 KPa dm⁶ mol⁻² and b = 0.0189 dm³ mol⁻¹

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/RTC = 3/8 
or, VC = (3/8) × (RTC/PC)
hence, b = (1/8)(RTC/PC)
Again, a = PC × 27b²  
= 3 × PC × (3b)² 
= 3 PCVC² 
= 3PC × (3RTC/8PC
∴ 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






Nov 5, 2018

Molar Heat Capacity Of Gases

Heat Capacity Of Gases:

Heat capacity of a substance is defined as the amount of heat required to rise of the temperature by one degree. Heat capacity per gram of substance is called specific heat and per mole called molar heat capacity.
Thus, Molar heat capacity = Molar mass × Specific heat
CP = M × cp 
CV = M × cv
Where CP and CV are the molar heat capacities at constant pressure and constant volume respectively. cp and cv are their specific heats.
The specific heat at constant pressure and constant volume are 0.125 and 0.075 Cal gm⁻¹K⁻¹ respectively. Calculate the molecular weight and atomicity of the gas. Name the gas if possible.
M = 40 and ⋎ = 1.66(mono-atomic), Ar(Argon).
For gases, there are two heat capacities at constant volume and constant pressure.
These are Represented as,
CV = (dq/dT)V = (dU/dT)V
CP = (dq/dT)P = (dU/dT)P + P(dV/dT)P
Difference in Heat Capacities of an Ideal Gas:
If the gas is assumed to be ideal, then,
PV = nRT and (dU/dT)V = (dU/dT)P
Again, P(dV/dT)P = nR
Thus for an ideal gas, PV = nRT; or, P(dV/dT)P = nR.
Again, CP = (dU/dT)P + P(dV/dT)P
or, CP = CV + {P×(nR/P)}
or, CP = CV +nR
For 1 mole ideal gas CP = Cv +R
Molar Heat capacities of Gases:
Molar heat capacities at constant volume and pressure
Molar Heat Capacity Of Gases
From the above two descriptions, it is clear that CP and CV. Since for CP, some mechanical work is required as additional energy to absorbed for definite piston from volume V₁ to V₂. Thus , CP - CV = Mechanical Work = PdV = P(V₂-V₁) = PV₂-PV₁ = R(T+1) - RT = R
Again, CP = CV+R for 1 mole ideal gas.
Now let us find the expression of CV from the point of view of Kinetic Theory.
CV = Energy required to increase transnational kinetic energy of 1 mole gas for rise of 1° temperature + energy required to increase inter molecular energy of 1 mole gas for rise temperature of 1°.
Increase of transnational K.E. = (3/2)R(T+1) - (3/2)R = (3/2) R for 1 mole gas for 1° rise in temperature.
Mono-atomic Gases:
CV = (dU/dT)v = (3/2)R and CP = (5/2)R.
Thus γ = CP/CV = 5/3 ≃ 1.667
Poly-atomic Gases:
Linear  CV = (dU/dT)v = (3/2)R + R +(3N - 5)R
Non-linear  CV = (dU/dT)v = (3/2)R + (3/2)R +(3N - 6)R
Where N is the number of particles.
Molar Heat Capacities for Diatomic Molecules:
For diatomic molecule N = 2. 
Thus CV = (3/2)R + R + R = (7/2)R and CP = (9/2)R.
∴ ⋎ = CP/CV = 9/7  ≃ 1.286
Molar Heat Capacities for Tri-atomic Molecules:
For tri-atomic molecule N = 3.
Thus CV(Linear) =  

Oct 18, 2018

Comparison Between Ideal and Real Gases

Comparison Between Ideal and Real Gases:

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 gasesA number of points can be discussed to compare the two types of gases.

Ideal Gas : 

(i) The Ideal Gas can not be liquefied since it has no inter-molecular attraction and so that molecule will not condense.
(ii) The coefficient of thermal expansion(𝛂) depends on temperature(T) and does not depends on the nature of the gas.
Show that coefficient of thermal expansion of Ideal Gas depends on the temperature of the gas.
Coefficient of thermal expansion(𝛂) is defined as 𝛂 = (1/V)(dV/dT)P.
Ideal Gas Equation for 1 mole gas is PV = RT.
Hence (dV/dT)P = R/P.
Thus 𝛂 = (1/V) × (R/P) = (R/PV) = 1/T.
This means all the gases have the same coefficient of thermal expansion.
(iii) The coefficient of compressibility(β) similarly depends on the Pressure(P) of the gas and will be same for all gases.
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. 
(iv) When P is plotted against V, at constant temperature(T) a rectangular hyperbola curve is obtained as demanded by Boyle's Low PV = Constant. 
Comparison Between Ideal and Real Gases.
P vs V Plot
The hyperbola Curve at each temperature is called one isotherm and at different tempera-ture we have different isotherms. Two isotherms will never intersect.
(v) When PV is plotted against , at constant a straight line parallel to P-axis is obtained. At different temperature there will be different parallel lines.
Comparison Between Ideal and Real Gases.
PV vs P Plot
(vi) 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. 

Real Gas :

(i) This gas could be liquefied since it has inter molecular attraction which helps to coalesce the gas molecules.
(ii) The coefficient of thermal expansion (α) is found to vary from gas to gas that is α depends on the nature of the gas.
(iii) The coefficient of compressibility (β) also is found to depend on the nature of the gas.
(iv) When P is plotted against V, a rectangular hyperbola is obtained only at high temperature (above the critical temperature).
What is the Real Gases?
P vs V Plot for Real Gas
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.
(v)  When PV is plotted against P for real gas, following plots, called Amagat Curve are obtained.

Amagat Curve:

What is the Ideal and Real Gas?
Amagat Curves for different gases
at a given Temperate(0°C)
It shows that for most gases, value of PV 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.
Comparison Between Ideal and Real Gases
Amagat Curves for a gas (CO₂)
at different temperature 
This shows that for CO₂ gas the depth of the minimum shifts towards the PV-axis with the increase of temperature (T). 
At T₃ temperature PV runs parallel to the P-axis up to certain range of P at low pressure region (P0).

Boyle Temperature(TB):

In the above Curve T₃ is called Boyle Temperature(TB) at which real gas obeys Boyle’s Law up to certain range of pressure at the low pressure region. 
The minimum coincides with the PV-axis. The mathematical condition for the calculation of Boyle Temperature(TB) is give by, [d(PV)/dP]T when P  0.
The curves obtained for Hydrogen(H₂) and Helium(He) at 0°C is above their Boyle Temperature and so with increase of P, values of PV increases from the beginning.

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, ZPV/RT where V is the molar volume.
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.
(vi) 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.
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.
From the given data, we have 
T = 273 K, Z = 0.97 and P = 100 atm.
From the definition of compressibility factor Z = PV/RT where V is the molar Volume.
Thus the molar volume of O₂ is
Vm = ZRT/P = (0.97) × (0.082 lit atm mol⁻¹ K⁻¹) × (273 K)/(100 atm) = 2.17 lit mol⁻¹
The mass of this molar volume will be equal to the molar mass of oxygen, that is 2.17 lit of oxygen equal to 32 gm.
Thus the mass of oxygen required to fill the gas cylinder 108.5 lit under the given condition
= {(32 gm)/(2.17 lit)} × (108.5 lit)
= 1600 gm = 1.6 Kg 

Sep 7, 2018

Gas Law's

Gaseous State :

If the thermal energy is much greater than the forces of attraction, then we have the matter in the gaseous state. Molecule in the gaseous state move with very large speeds and the forces of attraction among them are not sufficient to bind the molecules in one place, with the results that the molecules moved practically independent of one another. Because of this feature, gases are characterized by marked sensitivity of volume changes with the change of temperature and pressure. There exists no boundary surface and therefore gases tend to fill completely any available space i.e. they do not possess a fixed volume. 

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.
What is the Boyle’s Law?
Volume of a Given Quantity of Gas in Different Pressure, At Constant Temperature

Mathematical Representation of Boyle's Low:

V ∝ 1/P
or, 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,
P₁V₁P₂V₂

Where, V₁ and V₂ are the volume at P₁ and P₂ are the pressure respectively.

Graphical Representation :

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.

What is the Boyle's Low?
Graphical Representation of Boyle's Low

At Constant temperature a given mass of gas 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.
What is the Boyle's Low?
PV vs P Graph at Constant Temperature

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 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 and Density D₁ at Pressure P₁ and the 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/P  = D₁/D₂
P/D = Constant(K)
P = K × D
P D
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 0°C.

Mathematical Representation:

If V₀ is the volume of the gas at 0°C, then
1℃ rise of temperature the volume of the gas rise V₀/273.5 ml
∴  1°C temperature the volume of the gas (V₀+V₀/273) ml =V₀ (1 + 1/273)ml
At t°C 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
or, V T 
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 :

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.
What is the Charl's Low?
Isobars (P₁and P₂)

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 -273°C.
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 -273°C
What is the Char's Low?
Representation of Absolute Zero temperature

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

Gay – Lussac’s Law :

Pressure of a given mass of a gas at constant volume is directly proportional to its Kelvin temperature.
That is, PT at constant Volume.
P₁/T₁P₂/T₂
 What is the Gay – Lussac’s Law?
Isotherms (V₁and V₂)

Combination of Boyle’s and Charl’s Law : 

V1/P When T Constant.
From Charl's Law, 
V T 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