Priyam Study Centre

A Page about Chemistry and Related Topics.

Feb 17, 2019

Gas Law's

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 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.
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 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






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 gases. A number of points can be discussed to compare the two types of gases.
Ideal Gas:
  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






Feb 14, 2019

Rydberg Equation

F The emission spectrum of hydrogen atom was determined experimentally and it was possible to relate the frequencies of different spectral lines by simple equation,
ν = R[1/n12 - 1/n22]
Where n1 and n2 are integers and R is a constant, called Rydberg constant after the name of the discoverer. The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.
Energy of an Electron in Bohr Orbits:
The Energy associated with the permitted orbits is given by ,
En = - 1/2(e2/r), where r = n2h2/4π2me2
Thus, En = - (2π2me4/n2h2)
Again E1 = - (2π2me4/h2)
En = E1/n2
Thus the values of E2, E3, E4, E5 etc. in terms of E1(- 13.6 eV) are given as,
E2 = -13.6/22 = - 3.4 eV
E3 = -13.6/32 = - 1.51 eV
E4 = -13.6/42 = - 0.85 eV
E5 = -13.6/52 = - 0.54 eV
As n increases the energy becomes less negative and hence the system becomes less stable.

Explanation of the Rydberg Equation:

The explanation of Rydberg equation is now simple. The most stable orbit must be the one with the lowest energy (ground state).
When n = 1 the energy of an electron is lowest and it the lowest energy state and the states with n 〉 1 are the higher energy exited states.
Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level. Conversely energy will be released in the form of light of definite frequency when the exited electron returns to its ground state.
Since energy and frequency of the emitted light are connected by Plank Relation
E = hν 
or, ν = E/h
The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the initial state 1 and the final state 2.
So that, E= E2 - E1
= - (2π2me4/n12h2) - [- (2π2me4/n22h2)]
= (2π2me4/h2)[1/n12 - 1/n22]
Then ν corresponding to the energy E is given by,
ν = E/h = (2π2me4/h3)[1/n12 - 1/n22]
ν = R[1/n12 - 1/n22]
Where R is the Rydberg Constant.
This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.
Value of the Rydberg Constant:
Using the value of 9.108 × 10-28 gm for the mass of an electron, the Bohr Theory Predicts the following value of Rydberg Constant.
R = (2π2me4/h3)
= {2 × (3.1416)2 × (9.108 × 10-28 gm) × (4.8 × 10-10 esu)4}/(6.627 × 10-27 erg sec)3 
= 3.2898 × 1015 cycles/sec
Evaluate R in terms of wave number () instead of frequency. Wave number and frequency are connected with,
λν = c and ν = c/λ = c
Thus, = ν/c
where c is the velocity of light.
Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wave number () on the other hand stands for the number of waves connected in unit length that is per centimeter (cm-1) or per metre (m-1).
Thus, R = 3.2898 × 1015 sec-1/2.9979 × 1010 cm sec-1
= 109737 cm-1
= 10973700 m-1
The experimental values of R is 109677 cm-1 (10967700 m-1) showing a remarkable agreement between experiment and Bohr's Theory.

Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom:

Putting n = 1, n = 2, n = 3, etc in Rydberg Equation we get the energies of the different stationary states for the hydrogen electron. The transition energies can be calculated from the Rydberg Equation. The experimental spectra of hydrogen atom also exhibited several series of lines .
Energy Level Diagram for the Hydrogen Spectrum
Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.
This is because any given sample of hydrogen contains almost infinite number of atoms.
Under the normal conditions the electron of the each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell). 
When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube, individual atom absorbed different amounts of energy. Some of atoms absorbed such energy to shift their electron to third energy level (n = 3) while some others may absorbed large amount of energy to shift their electron to the Fourth (n = 4), Fifth (n = 5), Sixth (n = 6) and Seventh (n = 7) energy level.
The electrons in the higher energy level are relatively unstable and hence drop back to the lower energy level with the emission of energy in the form of line spectrum containing various lines in of particular frequency and wave length.
Lyman Series:
Transition to the ground state (n = 1) from n = 2, n = 3, n = 4, etc exited states constitute the Lyman Series in the ultraviolet region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = R[1/n12 - 1/n22]
For Lyman Series, we have n1 = 1 and n2 = 2, 3, 4 ....
Thus, (i) When n2 = 2,
1/λ = 109677{1 - (1/4)}
or, 1/λ = {(109677 × 3)/4} cm-1
λ = {4/(109677 × 3)}
= 1215 × 10-8 cm
= 1215 Å
Thus, (i) When n2 = ∞,
1/λ = 109677{1 - (1/∞2)}
or, 1/λ = 109677 cm-1
λ = (1/(109677)
= 912 × 10-8 cm
= 912 Å
The wave number of the experimental spectra of hydrogen atom in Lyman Series is 82200 cm-1. Show that the transition occur to the Ground state (n =1) from n = 2.
(R = 109600 cm-1)
We know that, Wave number,
= R[1/n12 - 1/n22]
For Lyman series n = 1 and 
Wave Number () = 82200 cm-1
Thus, 82200 = 109600(1/12 - 1/n22)
or, 1/n22 = 1 - (822/1096)
= 274/1096 
=1/4
∴ n2 = 2
Thus the transition occurs to the ground state (n = 1) from n =2.
Balmer Series:
Transition to the n = 2 level from n = 3, n = 4, n = 5, etc exited states constitute the Balmer Series in the visible region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/22) - (1/n22)]
Which of the Balmer lines fall in the visible region of the spectrum wave length 4000 to 7000 Å ? (Given  R = 109737 cm⁻¹)
The Balmer Series of the hydrogen spectrum comprise the transition from n 〉2 levels to the n = 2 level.
4000 Å = 4000 × 10-8 cm = 4 × 10-5 cm and 7000 Å = 7 × 10-5 cm
Therefore wave number,
= 1/λ = (1/4) × 105 to (1/7) × 105 cm⁻¹
Transition energy of the Balmer lines,
= 1/λ = 109677[(1/22) - (1/n22)]
Taking () = (1/4) × 105 cm-1
We have,
(1/4) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/4 ⋍ 1.1{(1/4) - (1/n22)}
or, n2 ⋍ 44
So n2 = 7 ( nearest whole number).
Taking () = (1/7) × 105 cm-1
We have,
 (1/7) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/7 ⋍ 1.1 {(1/4) - (1/n22)}
or, n2 ⋍ 3
So n2 = 3 ( nearest whole number).
So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
Calculate the wave length of Hɑ and Hβ of the Balmer Series.
We know that, for the Balmer Series,
= 1/λ = 109677[(1/22) - (1/n22)]
Again for Hɑ line n2 = 3 and for Hβ line n2 = 4
Thus the wave length for Hɑ - line,
1/λ = 109677 {(1/22) - (1/32)}
= 109677 × (5/36)
λ = 36/(5 × 109677)
= 6.564 × 10-5 cm
= 6564 Å
Again the wave length for Hβ - line,
1/λ = 109677 {(1/22) - (1/42)}
= 109677 × (3/16)
λ = 16/(3 × 109677)
= 4.863 × 10-5 cm
= 4863 Å
 

 

Pschen Series:
Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Pschen Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/32) - (1/n22)]
Brackett Series:
Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Brackett Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/42) - (1/n22)]
Pfund Series:
Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/52) - (1/n22)]
Series of Lines n1 n2 Spectral Region Wavelength
Lyman Series 1 2,3,4..etc Ultraviolet ㄑ4000 Å
Balmer Series 2 3,4,5..etc Visible 4000 Å to 7000 Å 
Paschen Series 3 4,5,6..etc Near Infrared 〉7000 Å
Brackett Series 4 5,6,7..etc Far Infrared 〉7000 Å
Pfund Series 5 6,7,8..etc Far Infrared 〉7000 Å