June 2019

Radiation measurement of a radioactive sample

The rate at which a radioactive sample disintegrates can determine by counting the number of particles emitted in a given time.

Radioactivity is one important natural phenomenon obeying the first-order kinetics. The rate of these reactions depends only on the single power of the concentration of the reactant.

(dN/dt) = - k N

where N = number of the atoms of the disintegrating radio-element, dt = time over which the disintegration is measured, and k = rate constant.

k = - (dN/dt)/N

The rate constant defined as the fraction decomposing in the unit time interval provided the concentration of the reactant 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.
Radiation measurement of radio-element
Radiation measurement

How to calculate the half-life of radioactive elements?

After a certain period of time, the value of (N₀/N ) becomes one-half and half of the radioactive elements have undergone disintegration. This period is called half-life of a radioactive element.
How to calculate half-life of radio-element?
Half-life of radio-element
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.

2.303 log(N₀/N ) = kt
when t = t½, N = N₀/2.

∴ 2.303 log{N₀/(N₀/2)} = k t½
or, t½ = 0.693/k
∴ k = 0.693/t½.

This 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.

t½ for polonium - 213 = 4.2 × 10⁻⁶ sec and bismuth - 209 = 3 × 10⁷ years.

The average life period of radio-elements

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.

tav = total life/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.

The average life of radio-element is reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple way.
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
or, tav = 1/k

∴ t½ = 0.693/k = 0.693 tav

Radiocarbon dating - age of organic material

Radiocarbon dating is a method for determining the age of organic martial based on the accurate determination of the ratio of isotopes of carbon.
The radiocarbon dating method was developed by Willard Libby, the University of Chicago in 1940 and receive a Nobel prize in chemistry for his work in 1960.

Radiocarbon - 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen or cosmic reaction.

₇N¹⁴ + ₁n⁰ → ₆C¹⁴ + ₁H¹

Carbon reacts with atmospheric oxygen to form carbon dioxide. This carbon dioxide is taken by plants by photosynthesis and animals by eating plants.
When the animal or plant dies, it stops exchanging carbon with its environment since there no fresh intake of stratospheric carbon dioxide and the dead matter is out of equilibrium with the atmosphere.

The radiocarbon - 14 continues to decay so that thereafter a number of years only a fraction of carbon - 14 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.

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

Age of rock deposits by half-life

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 - 235 to lead -207 series.

The half-life of the intermediate members being small compared to that of uranium -235 (4.5 × 10⁹ years). Uranium atoms that started decaying many-many years ago must have been completely converted to the stable lead-207 during this extra-long period.

The uranium-235 remaining and the lead-207 formed must together account for the uranium 235 present at zero time when the rock solidified.
Thus both N₀ and N are known k is known from the knowledge of the half-life of uranium -235. Therefore the age of the rock can be calculated.

Problem
A sample of uranium (t₁/t₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-235 and 10.3 gm of lead-207. 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

Mole of uranium -238 present in the ore at zero tim
= (0.05+0.05)
= 0.010 mole.

∴ Radioactive decay constant = 0.693/(4.5 × 10⁹)
= 0.154 × 10⁻⁹ yr⁻¹.

2.303 log(0.10/0.05) = kt
∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)
= 4.5 × 10⁹ year

Avogadro number of radium - 226

Let 1 gm of radium - 226 contains N number of atoms.
∴ N = N₀/226
where N₀ = Avogadro number and mass number of radium = 226.

kN = (k × N₀)/m
where k = 0.693/t½
kN = (0.693 × N₀)/(t½ × m)

∴ N₀ = (kN × t½ × m)/0.693
One gram of radium - 226 undergoes 3.7 × 10¹⁰ disintegrations per second and half-life = 1590 year.
∴ Avogadro number (N₀) of radium - 226
= (kN × t½ × m)/0.693
= (3.7 × 10¹⁰ × 1590 × 365 × 24 × 60 × 226)/0.693
= 6.0 × 10²³.

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 law.
PiVi = RT

The volume measurement for 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 than the original volume. 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.

Volume for Avogadro number of molecules of real gas

Let us take, r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
Volume measurement for real gas by Van der Waals
Volume measurement for real gas
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 for a pair of molecules
= 4/3 πσ ³.
Single-molecule,
= 1/2 × 4/3 π σ³.

Volume for Avogadro number of molecules,
b = N₀ × 2/3 × π σ³
= N₀ × 2/3 × π (2r)³
∴ b = 4 N₀ × 4/3 × π (r)³.

Measurement of b helps to calculate the radius or diameter of the gas molecule. Thus the gas equation,

Pi (V-b) = RT

Pressure measurement for real gas molecules

The pressure of the gas developed due to the wall collision of the gas molecules. Due to intermolecular attraction, the colliding molecules will experience an inward pull.

The pressure exerted by the molecules in real gas will be less if the gas molecules have no intermolecular attraction as in ideal gas pressure.

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.

The 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 is a constant for the gas that measures the attractive force between the molecules.

Pi = P + (a/V²)

Van der Waals equation derivation

Using the two corrections Van der Waals equation for a one-mole real gas

(P + a/V²)(V - b) = RT.

For n moles real gases, the volume has to change because 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.
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 270°C.
Given, a = 1.4 atm liter² mol⁻² and b=0.04 lit mol⁻¹.
Also, calculate the pressure of the gas using the ideal gas law 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

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

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.

Van der Waals equation,
(P+ an2/V2)(V - nb) = nRT but b=0(no size).
P = (nRT/V - an2/V2)ㄑPi
Pi = nRT/V only.
Thus, the intermolecular attraction effect reduces the pressure of real gases.

Units of Van der Waals constant a and b

Van der Waals equation,

(P+ an²/V²)(V - nb) = nRT.
where Pa is called the internal pressure of a gas.

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

What are the SI units 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.
Carbon dioxide gas = 3.95 atm lit² mol⁻².
Hydrogen 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.

Carbon dioxide gas = 0.04 lit mol⁻¹
Hydrogen gas = 0.02 lit mol⁻¹

Boyle temperature for real gases

The mathematical definition of Boyle temperature,
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

PV = {RTV/(V - b)} - a/V
∴ 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

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

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.

Differentiating with respect to pressure at constant temperature.
[d(PV)/dP]T = b, but b ≠ 0
[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

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 by Van der Waals
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)

Ideal gas law 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.

The curve for carbon dioxide and hydrogen

When a is very high.
(a/RT) 〉b
∴ {b - (a/RT)} = - ve

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
∴ {b - (a/RT)} = + ve

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 pressure at the moderate pressure region of the curve for carbon dioxide, Z〈 1 and more compressible.
  2. When T = TB = a/Rb
    or, b = a/RT
    ∴ {b - (a/RT)} = 0.
    That is Z = 1, the gas is an ideal gas. The size effect compensates for the effect due to the intermolecular attraction of the gas.
    Z runs parallel to the pressure 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 pressure.. The size effect dominates over the effect due to the intermolecular attraction.

Hydrogen and helium gas molecules

The value of a is extremely small for hydrogen and helium gases as they are difficult to liquefy. Thus the equation of state P(V - b) = RT obtains from van der Waals equation by ignoring small-term a/V².

Hence Z is always greater than one and it increases with the increase in the pressure of the gas.

Chemistry 1

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