June 2019

The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time.
    Radioactive decay is one important natural phenomenon obeying the first-order rate process. The rate of these reactions depends only on the single power of the concentration. of the reactant. The rate can be expressed as,
(dN/dt) = - k N
    Where N is the number of the atoms of the disintegrating radio-element present at any time, dt is the time over which the disintegration is measured and k is the radioactive decay rate constant.
k = - (dN/dt)/N
    Thus, the rate constant(k) is defined as the fraction decomposing in the unit time interval provided the concentration of the reactant is kept constant by adding from outside during this time interval. The negative sign shows that N decreases with time.
    Let N₀ = number of the atoms present at the time t = 0 and N = number of atom present after the t time interval. Rearranging and integrating over the limits N₀ and N and time, 0 and t.
Rate of radioactive decay half life for radioactive elements
Rate of radioactive decay half-life

Half-life period of radioactive elements

    After a certain period of time the value of (N₀/N ) becomes one half, that is, half of the radioactive elements have undergone disintegration. This period is called half-life of a radioactive element and is a characteristic property of a radioactive element.
Rate of radioactive decay half-life
Half-life and rate of radioactive decay
    If radioactivity of an element is 100%  and the half-life period of this element 4 hours. Thus after four hours, it decomposes 50% and the remaining 50%. After 8 hours it decomposes 75% and reaming 25% and the process is continued. The half-life is given by,
2.303 log(N₀/N ) = kt When t = t½, N = N₀/2
Putting in the above equation we have,
2.303 log{N₀/(N₀/2)} = k t½
or, t½ = 0.693/k
k = 0.693/t½
    The above relation shows that both the half-life and radioactive decay rate constants are independent of the amount of the radio-element present at a given time.
    ₈₄Po²¹³ has t½ = 4.2 × 10⁻⁶ sec, whereas ₈₃Bi²⁰⁹ is 3 × 10⁷ years.

Age of radioactive elements

Age of organic material

    A method of determining the age of organic martial based on the accurate determination of the ratio of carbon-14 and carbon-12. carbon 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen.
₇N¹⁴ + ₁n⁰ → ₆C¹⁴ + ₁H¹ ( Cosmic Reaction)
    The carbon-14 ultimately goes over to carbon-14 dioxide. A steady-state concentration of one ¹⁴C to ¹²C is reached in the atmospheric CO₂. This carbon dioxide is taken in or given out by plants and plant-eating animals or human beings so they all bear this ratio.
    When a plant or animals died the steady-state is disturbed since there is no fresh intake of stratospheric CO₂ the dead matter is out of equilibrium with the atmosphere.
    The ¹⁴C continues to decay so that thereafter a number of years only a fraction of it is left on the died matter. Therefore the ratio of the ¹⁴C/¹²C drops from the steady-state ratio in the living matter.
₆C¹⁴ → ₇N¹⁴ + ₋₁e⁰(t½ = 5760 years)
    By measuring this ratio and comparing it with the ratio in living plants one can estimate when the plant died.

Age of rock deposits

    Knowledge of the rate of decay of certain radioactive isotopes helps to determine the age of various rock deposits. Let us consider uranium-containing rock formed many years ago.
    The uranium started to decay giving rise to the uranium - 238 to lead -206 series. The half-lives of the intermediate members being small compared to that of uranium -238(4.5 × 10⁹ years), it is reasonable to assume that those uranium atoms that started decaying many many years ago must have been completely converted to the stable lead-206 during this extra long period.
    The uranium-238 remaining and the lead-206 formed must together account for the uranium 238 present at zero time, that is, when the rock solidified. Thus both N₀ and N are known k is known from a knowledge of the half-life of uranium -238. Therefore the age of the rock can be calculated.

Avogadro number

If 1 gm of a radioactive element contains N number of atoms.
Then, N = N₀/m
where N₀ = Avogadro number and m = Mass number
Therefore, kN = (k × N₀)/m
where k = 0.693/t½
Thus, kN = (0.693 × N₀)/(t½ × m)

∴ N₀ = (kN × t½ × m)/0.693
    If each atom of the radio-element expels one particle then kN, the rate of decay, is also the rate at which such particles are ejected.
    One gram of radium undergoes 3.7 × 10¹⁰ disintegrations per second, the mass number of Ra-226 is 226 and t½ = 1590 year so that the Avogadro Number can be calculated by the above equation,
Avogadro number
=(3.7 × 10¹⁰ sec⁻¹ × 1590 × 365 × 24 × 60 sec × 226)/0.693
= 6.0 × 10²³

Average life period of radioactive elements

    Besides half-life of a radio-element, another aspect of the life of a radio-element must also be known. It is now possible to determine the average life period of a radioactive atom present in aggregate of a large number of atoms.
    The average life period of an atom of the radio-element tells us the average span of time after which the atom will disintegrate.
    The length of time a radio-element atom can live before it disintegrates may have values from zero to infinity. This explains the gradual decay of the radio-element instead of the decay of all the atoms at the same time. The average life period may be calculated as follows-
tav = Total Life Period/Total Number of Atoms
    Let N₀ atoms of a radioactive element are present in an aggregate of large no of atoms at time zero. Now in the small-time interval t to (t+dt), dN atoms are found to disintegrate.
    Since dt is a small-time period, we can take dN as the number of atoms disintegrating at the time t. So the total lifetime of all the dN atoms is t dN.
    Again the total number of atoms N₀ is composed of many such small numbers of atoms dN₁, dN₂, dN₃, etc, each with its own life span t₁, t₂, t₃, etc.
Rate of radioactive decay half-life
Average life and the rate of radioactive decay
    The average life of Radio-element is thus the reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple alternative way.
    Since the radioactive atoms may be regarded as having an average life, then the product of the fractions of atoms disintegrating in unit time (that is k) and average life must be unity.
tav k = 1 So that, tav = 1/k
    Thus the relation between an average life and half-life is,
t½ = 0.693/k = 0.693 tav
Problem
    A piece of wood was found to have a ¹⁴C/¹²C ratio of 0.7 times that in the living plant. Calculate the approximate period when the plant died(t₁/₂ = 5760 years).
Solution
We know that radioactive decay constant,
k = 0.693/(t½) = 0.693/5760 years
= 1.20 ×10⁻⁴ yr⁻¹

N₀/N = 1.00/0.70
∴ 2.303 log(N₀/N) = kt
or, t = (2.303 log(N₀/N)/k

Putting the value, above equation,
we have, t = (2.303 × 0.155)/(1.20 × 10⁻⁴)
= 2970 years
Problem
    A sample of uranium (t₁/t₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-238 and 10.3 gm of lead-206. Calculate the age of the ore.
Solution
11.9 gm of uranium-238 = 11.9/238 = 0.05 mole of uranium
10.3 gm of lead-206 = 10.3/206 = 0.05 mole of lead -206
Thus mole of uranium -238 present in the ore at zero time = (0.05+0.05) = 0.010 mole.
And radioactive decay constant = 0.693/(4.5 × 10⁹) = 0.154 × 10⁻⁹ yr⁻¹.
Then 2.303 log(0.10/0.05) = kt
∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)
= 4.5 × 10⁹ year

What is a real gas molecule?

A real gas that does not behave as an ideal gas due to intermolecular attraction. Van der Waals modified the Ideal gas and formulate Van der Waals equation of state for real gases.

By incorporating the size effect and intermolecular attraction effect of the real gas. These above two effects are discussing under the volume correction and pressure correction of the ideal gas equation.
PiVi = RT

Size effect of real gas molecules

Real gas molecules are assumed to be a hard rigid sphere, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of one-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.

Affective volume for Avogadro number of molecules

Let us take, r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
Effective volume of a gas molecule by Van der Waals
Effective volume of a gas molecule
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.

Excluded volume = 4/3 π 𝜎³ for a pair of molecules.
Thus for a single molecule,
= 1/2 × 4/3 π 𝜎³.

The effective volume for Avogadro number of molecules of a gas,
b = N₀ × 2/3 × π 𝜎³
= N₀ × 2/3 × π (2r)³
∴ b = 4 N₀ × 4/3 × π (r)³.

Thus b measures and helps to calculate the radius or diameter of the gas molecule. The equation becomes,

Pi (V-b) = RT

Intermolecular attraction effect of real gas

The pressure of the gas developed due to the wall collision of the gas molecules. But due to intermolecular attraction, the colliding molecules will experience an inward pull.
The pressure exerted by the molecules in real gas will be less than that if there had not been an intermolecular attraction as in ideal gas Pi.

Thus, Pi 〉P
∴ Pi = P + Pa

Where Pa is the pressure correction term originating from attractive forces. Higher the intermolecular 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.

Pressure exerted by a gas molecule

Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
Pa ∝ 1/V²
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)

Van der Waals equation derivation

Using the two corrections we have Van der Waals equation for 1-mole real gas,

(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 for real gas?
Van der Waals equation
Question
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.
Also, calculate the pressure of the gas using the ideal gas equation and find the extent of deviation from ideal behavior.

Answer
P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Real gases when a=0 but b≠0 and a≠0 but b=0

We have Van der Waals equation,
(P+ an²/V²)(V - nb) = nRT
Hence, P = nRT/(V - nb) 〉Pi

Since, Pi = nRT/V only.

It means that the molecular size effect (repulsive interaction) creates higher pressure than 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, the intermolecular attraction effect reduces the pressure of 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 is called the internal pressure of a 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
Question
Write the unit dimension of Van der Waals constant a and b.

Answer
SI unit of 'a' = N m4 mol-2
CGS unit of 'a' =dyne cm4 mol-2

SI unit of 'b' = m3 mol-1
CGS unit of 'b' = cm3 mol-1

Significance of a and b

‘a’ term originates from the intermolecular 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 intermolecular attraction and more easily the gas could be liquefied.
CO₂ gas = 3.95 atm lit² mol⁻²
H₂ gas = 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.
CO₂ gas = 0.04 lit mol⁻¹
H₂ gas = 0.02 lit mol⁻¹

Boyle temperature for real gases

Mathematical condition for Boyle temperature is,
TB = [d(PV)/dP]T When P→0

Van der Waals equation for 1 mole real 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

T = TB, [d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0

Thus, we have, RTBb/(V - b)² = a/V²
TB = (a/Rb) {(V - b)/V}²
Since P → 0, V is large
Thus, (V - b)/V ≃ 1

Hence,TB = a/Rb
Question
Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1.

Answer
TB = 427 K

Question
Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.

Answer
When, a = 0, Van der Waals equation,
P (V - b) = RT
or, PV = RT + Pb

Constant temperature differentiating with respect to pressure,
[d(PV)/dP]T = b, but b ≠ 0
hence [d(PV)/dP]T ≠ 0
The gas has no Boyle temperature since at any temperature, [d(PV)/dP]T ≠ 0.
Again when a = b = 0, Van der Waals equation becomes

PV = RT
or, [d(PV)/dP]T = 0
That is at any temperature this becomes zero and so all the temperature is Boyle Temperatures.

Amagat curve from Van der Waals equation

Amagat curves for real gases
Amagat Curves
Van der Waals equation for one-mole real gas,
(P + a/V2)(V - b) = RT
∴ PV - Pb + (a/V) + (a/V2) = RT.

Neglecting the small-term (a/V2)
PV = RT + Pb - (a/V)

Using the ideal gas equation for the small-term,
a/V = aP/RT and taking Z = (PV/RT),
Z = 1 + (1/RT){b - (a/RT} P
This Shows that Z = ∫(T, P).

This equation can be used to explain the Amagat curve qualitatively at low pressure and moderate pressure region.

Van der Waals constant a is very high and small

When a is very high (a/RT) 〉b in the equation and {b - (a/RT)} = - ve
That is the slope of the Z vs P curve for carbon dioxide at a moderate pressure region is negative. That is the value of Z decreases with the increase of Pressure and it is also found in the curve of carbon dioxide.

When a is very small, a/RT〈 b,
The slope of the Z vs P curve for hydrogen becomes (+) ve and the value of Z increases with pressure.
  1. When T〈 TB
    or, T〈 a/Rb
    Thus, 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.
  2. When T = TB = a/Rb
    or, b = a/RT
    Thus, {b - (a/RT)} = 0.
    That is Z = 1 the gas shows ideal behavior. The size effect compensates for the effect due to the intermolecular attraction of the gas.
    Z runs parallel to the P axis up to the low-pressure region.
  3. When T 〉TB
    or, T 〉a/Rb
    or, b 〉a/RT hence {b - (a/RT} = (+)ve.
    That is the value of Z increases with the increase of P when T 〉TB. The size effect dominates over the effect due to the intermolecular attraction.

Hydrogen and helium

Hydrogen and helium, 0°C is greater then their TB values and Z vs P slope becomes (+)ve. At very low-pressure P→0 and high temperatures, the volume is very large.
Thus size effect and intermolecular attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys ideal behavior.

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