HEAT-3
Problem 46
The gas hydrogen at 
has a density of 0.09g/litre. Find the gas constant for a unit mass of hydrogen
has a density of 0.09g/litre. Find the gas constant for a unit mass of hydrogen Problem 47
Calculate the density of air at 1000C and 200K pa given its density at 00C and 101 k pa is
1.29
Problem 48
Calculate the density of hydrogen gas at 200C and 101Kpa given the molar mass of hydrogen molecule is 2 10-3 kg assume the molar gas constant R
10-3 kg assume the molar gas constant R 
Problem 49
A faculty barometer tube has some air at the top above the mercury. When the length of air column is 260mm the reading of the mercury level is 740mm when the length of air column is decreased to 200mm by depressing the barometer tube further into the mercury , the reading of the mercury above the outside level becomes 747mm calculate the atmospheric pressure. (Density of mercury is 13600 kgm-3)
KINETIC THEORY OF AN IDEAL GAS
This explains the behaviour of gases by considering the motion of their molecules.
Assumptions
- Gas molecules are in random motion colliding in one another and with the walls of the container.
- The molecules are like perfectly elastic spheres.
- The attraction between the molecules is negligible.
- The volume of the molecules is negligible compared with the volume occupied by the gas.
- The duration of a collision is negligible compared with the time between the collisions.
- A sample of gas consist a large number of identical molecules for statistics to be applied.
PRESSURE EXTENDED BY A GAS
The pressure of a gas is due to the molecules bombarding the walls of its container.
When the molecule collide with the walls they transfer momentum to it.
The total change in moment per second is the force exerted on the wall by the gas.
The pressure on the wall is the force divided by the area of the wall.
EXPRESSION OF THE PRESSURE EXERTED BY AN IDEAL GAS
Consider a cube of side L containing N molecules of gas each mass m
Let u, v and w be component velocities of a gas molecule along ox, oy and oz axes respectively
If C is the resultant velocity of a gas molecule
Then,
Extension of Pythagoras theorem
Consider the force exerted on the face x of the cube due to the component U
Forward momentum
Rebounding momentum =
Where sign shows that the two momentum are in opposite directions
sign shows that the two momentum are in opposite directions Momentum change on impact = mu – (-mu)
= 2mu
The time taken for the molecule to move across the cube to the opposite face and back to X = Force F
on the wall is
Force Fon the wall is
The pressure exerted on face by one molecule is
This is the pressure exerted on face by one molecule
by one molecule For N- molecule let them have velocities towards face
towards face The pressure P exerted on face x by N molecule is
Let be mean square speed of gas molecules along OX direction
be mean square speed of gas molecules along OX direction 
Substitute equation [5] in equation [ 4 ]
………..6 For random motion the mean square of the components velocities are the same
From equation 1
- Substitute equation [7] in equation [8]
Substitute equation [9] in equation [6]
Where L3 volume of the containers
volume of the containers 
volume of the gas 
We may write:-
NOTE
-The equation indicates that the quantities on the left hand side of the equation [P and V] are macroscopic quantities. While the quantities on the right hand side of the equation are microscopic quantities
EXPRESSION FOR PRESSURE IN ALTERNATE FORMS
- The pressure equation says
Where
The equation (12) above Shows the expression of Pressure in terms of density () of the gas.
) of the gas. From equation (11) above
Since
From equation (11) above:-
OUTCOME/RESULTS OF THE KINETIC THEORY OF AN IDEAL GAS
OUTCOME 1
INTRODUCTION OF TEMPERATURE TO THE PRESSURE EQUATION
From the pressure equation
From ideal gas equation for 1 mole
Equation (1) above can be written as
Where If
If 
This equation shows that the translational means K.E of a gas molecule is proportional to its absolute temperature.
It is the gas constant per molecule
From equation (3) above
The SI unit of K is J/K
Numerical value of K
From the definition of K
The value of boltzman’s constant is
DEDUCTIONS FROM THE EQUATION (3)
Equation 3 above says
Deduction 1
Since and K are constant, we have
and K are constant, we have 
The translational mean K.E of a gas molecule is proportional to absolute temperature [T].
Deduction 2
Since ½ . and K are constant we have
and K are constant we have Therefore
Since are constant we have
are constant we have 
The mean square speed of a gas molecule is proportional to absolute temperature. 
Where K constant proportionality
constant proportionality If are the mean square speeds of a gas molecule at temperature T1 and T2 respectively then we have
are the mean square speeds of a gas molecule at temperature T1 and T2 respectively then we have 
Dividing equation (i) by equation (ii)
ROOT MEAN SQUARE [ R.M.S] SPEED OF A GAS MOLECULE
The root mean square [] speed of a gas molecule is the square root of the mean square speed of a gas molecule
] speed of a gas molecule is the square root of the mean square speed of a gas molecule R.M.S speed =
From deduction 3,
The root mean square speed of a gas molecule is proportional to the square root of absolute temperature 
Where
K constant of proportionality
constant of proportionality At two different temperatures T1 and T2 we have
Where
= R.M.S speed of a gas molecule at a temperature T1 
Speed of the gas molecules at a temperature T2 DIFFERENT EXPRESSIONS OF R.M.S SPEED OF GAS MOLECULES
The R.M.S speed of gas molecule can be given in several alternate forms
From the pressure equation terms of density
- From the pressure equation:
If
Where
- From equation (1)
If N is the total number of molecules making up the gas, then
Nm total mass of gas (MT)
total mass of gas (MT) V total volume of the gas
total volume of the gas 
- From the equation (2) above
Ideal gas equation for 1 mole requires:
PV RT
RT 
- From
OUTCOME 2
DERIVATION OF GAS LAWS
(1) BOYLE’S LAW
Boyle’s law requires
V or PV constant
or PV constant Condition of the law
Temperature must be kept constant
Derivation
According to kinetic theory of gases, the pressure exerted by a gas is given by
But Nm
From kinetic theory of an ideal gas
Where
For a given mass of a gas at constant temperature
- CHARLE’S LAW
Charles’ law requires
Condition of the law
Pressure must be kept constant
Derivation
According to kinetic theory of gases
But, Nm
According to kinetic theory of an ideal gas
V T (Charles law)
T (Charles law) - PRESSURE LAW
Pressure law requires
Condition of the law
Volume must be kept constant
Derivation
According to kinetic theory of gases:
But Nm
According to kinetic theory of gases
For a given mass of a gas at constant volume
IDEAL GAS EQUATION
Ideal gas equation requires
PV RT (for 1 mole of a gas)
RT (for 1 mole of a gas) Derivation
According to kinetic theory of gases:
P =
PV =
But Nm
According to | kinetic | theory | of | gases:  |
For a given | mass | of a | gas |  |
PV = RT (Ideal Gas equation)
OUTCOME 3
PROOF OF AVOGADRO’S HYPOTHESIS
Avogadro’s hypothesis
Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules
PROOF
Consider two gases (gas 1 and gas 2 say)
Let T be same temperature of gases
Let P be same pressure of gases
Let V be same volume of gases
The pressure equations for the gases are
GAS 1
P =
Therefore
PV = ………………1
………………1 GAS 2
Similarly PV = ……………………2
……………………2 Where,
N1m1 and N2m2 are number of molecules mass of each molecule and mean square speed of molecule of gas 1 and gas 2 respectively
and N2m2
are number of molecules mass of each molecule and mean square speed of molecule of gas 1 and gas 2 respectively Equation (1) equation (2)
equation (2) 
According to kinetic theory of ideal gas
Since the absolute temperature T is the same for both gases we have
...N1 = N2 (Avogadro’s hypothesis)
OUT COME 4
PROOF OF DALTON’S LAW OF PARTIAL PRESSURE
Definition
Partial pressure that could be exerted by a gas if it were present alone and occupies the same volume as the mixture of gases
DALTON’S LAW OF PARTIAL PRESSURE
The law states that
“The total pressure exerted by a mixture of gases is always equal to the sum of the partial pressure of the constituent of gases”
P P1 +P2
P1 +P2 Where;
P total pressure of the mixture
total pressure of the mixture P1 and P2 are partial pressures of the constituent gases
PROOF
Consider two gases nitrogen (N2) and hydrogen (H2) in a container of volume V at the surrounding temperature T
Let
N1, m1, , and N2, m2 be number of molecules mass of each molecule and the mean square speed of the molecules of N2 and H2 gases respectively
, and N2, m2 be number of molecules mass of each molecule and the mean square speed of the molecules of N2 and H2 gases respectively If P1 and P2 are partial pressure of N2 and H2 gases respectively then the pressure equation for N2 and H2 gases are
For N2
P1 = N
N 
For H2
P2 =
Adding equation (1) and equation (2) above
According to kinetic theory of an ideal gas
Since the absolute temperature T is the same, we have
Equation (3) above becomes
Compare this equation with the pressure equation of an ideal gas
Where N total number of molecules
total number of molecules Therefore;
N N1 + N2
N1 + N2 P total pressure of the mixture
total pressure of the mixture Therefore
P P1 + P2
P1 + P2 Dalton’s law of partial pressure
CALCULATION FORMULA OF DALTON’S LAW
Consider two gas contai
ners of volumes V1 and V2 containing nitrogen (N2) and hydrogen
ners of volumes V1 and V2 containing nitrogen (N2) and hydrogen
(H2) gases respectively at pressure P1 and P2
Let the two containers to be connected by a narrow tube of negligible volume with a stopper’s
Before the stopper S is opened
Ideal gas equation gives
For N2
P1 V1 = n1 RT
…………………………………..(1)Where n1 number of moles of N2 in the container of volume V1
number of moles of N2 in the container of volume V1 For H2
P2 V2 = n2 RT
………………………………..(2) Where n2 = number of moles of H2 in the container of volume
When the stopper ―S‖ is opened
The two gases are mixed up and finally the system will have a common pressure
P = total pressure of the mixture
Volume of mixture = V1 + V2
Total number of moles of the mixture = n1 + n2
Ideal gas equation for the mixture of the two gases gives:
Substitute equation (1) and equation (2) in equation (3)
NOTE
OUTCOME 5
MEAN FREE PATH
Symbol, λ
Definition
The mean free path a gas molecule is the average distance covered by a gas molecule during successive collisions.
It is given by:
Condition
For the collision to happen the distance between centres of the molecule must be equal to the diameter of a molecule.
Assumption
We assume that during collision, one molecule move and others are stationary target
Assumption molecule C moves while A and B are stationary.
The molecule with which C just make collision are those whose centers lie inside the volume of the cylinder of radius. d = 2r = diameter of a molecule. Where r = radius of a molecule
Volume of the cylinder =
Let n be the number of the molecules per unit volume in the cylinder.
Total number of molecules in collision =
Where = total distance covered by a gas molecule
From the definition of mean free path
This equation is true for a stationary target
From kinetic theory of an ideal gas all molecules move colliding one another and with the walls of the container, hence equation (1) above is statistically modified to:
This equation is true for moving target
ALTERNATIVE EXPRESSION OF MEAN FREE PATH
From equation (2) above :
From ideal gas equation for 1 mole
PV = RT
But, R = KN
PV = KTN
Where K = Boltzmann‘s constant
N = Avogadro‘s number of molecule
P = KTN / V
Thus equation (2) above can be written as:
……………………….(3) This is an alternative expression of mean free path.
FACTORS DETERMINING THE MEAN FREE PATH OF A GAS MOLECULE
(1) Absolute temperature (T) From equation (3) above:
First dark ring, n = 1
Second dark ring, n =2
Since a phase change of radians occurs when light is reflected at the glass plate H, the central spot (for which t = 0) is dark.
At any position where there is a bright ring:
difference = 2t = (n + 1/2) λ
Where n = 0, 1, 2, 3, ………………..
First bright ring, n = 0
Second bright ring, n = 1
Third bright ring, n = 2
COLLISION FREQUENCY (f)
This is the number of collisions made by a gas molecule in unit time.
It is given by:
Since n
Where K Boltzmann‘s constant
Boltzmann‘s constant P pressure of the gas
pressure of the gas means speed of a molecule d diameter of a molecule T absolute temperature
absolute temperature Equation 7 shows that
i) At constant temperature
ii) At constant pressure REAL GASES
Definition
A real gas is that gas which does not have the properties assigned to an ideal gas
A real gas satisfies the equation
equation THE VAN DER WAAL’S EQUATION
modified the ideal gas equation to take account that two of the Kinetic theory of an ideal gas may not be valid - The volume of the molecules may not be negligible in relation to the volume V occupied by the gas.
- The attractive forces between the molecules may not be negligible.
Problem 50
. Find the gas constant for unit mass of hydrogen Problem 51
The gas hydrogen has a density of 0.09g/liter at. Find the mean square speed and hence root mean square speed of hydrogen at 42oC
. Find the mean square speed and hence root mean square speed of hydrogen at 42oC Problem 52
- List down any four assumptions of the kinetic theory of an ideal gas.
- Determine the absolute temperature of a gas in which the average molecules of mass
8 x 10-26kg are moving with speed of 500ms-1. Given that universal gas constant R 8.31 jmol-1K-1 and number 6.023 x 1023
8.31 jmol-1K-1 and number 6.023 x 1023 Problem 53
Helium gas occupies a volume of 0.04m3 at a pressure of 2 105pa and temperature of 300K
105pa and temperature of 300K Calculate
- The mass of helium
- Thespeed of its molecules
- Thespeed at 432K when the gas is heated at constant pressure to this temperature.
Problem 54
- Derive the gas equation obeyed by a system consisting of N molecules each of mass m and mean square speed. Hence obtain the kinetic energy per molecule in terms of absolute temperature.
- A vessel of volume 6 x 10-3m3 contains nitrogen at a pressure of 2 102 pa and a temperature of 27oC. What is
- The number of nitrogen molecules in the vessel, and
- Theirspeed
Given that
R 8.3 Jmol-1K-1
8.3 Jmol-1K-1 MN 28 gmol-1
28 gmol-1 NA 6.0 x 1023
6.0 x 1023 Problem 55
- Write down the equation of a state of an ideal gas, defining all symbols used.
- How does the average translation kinetic energy of a molecule of an ideal gas change if
i) The pressure is doubled while the volume is kept constant? ii) The volume is double while the pressure is kept constant?
- Calculate the value of the root mean square of the molecules of helium at
Problem 60
- i) What is meant by the mean free path of a molecule
ii) Show that the mean free path of a molecule of an ideal gas at pressure p and temperature T is given by
Where KB is the Boltzmann’s constant and d is the molecule diameter
- i) Derive an expression for the average kinetic energy of one molecule of a gas assuming the formula for the pressure of an ideal gas.
- A cylinder of volume 2 10-3 m3 contains a gas at pressure of 1.5 106Nm-2 and temperature 300K. Calculate
i) The number of moles of the gas ii) The number of molecules the gas contains iii) The mass of the gas if its molar mass is 320 10-3 Kg.
10-3 Kg. iv)The mass of one molecule of the gas
Problem 61
- (i) Write down theequation and define each term in its usual meaning
- State the assumptions upon which the equation you have written in (a)
- above is derived from the ideal gas equation
- i) On the basis of the kinetic theory of gases, shows that two different gases at the same temperature, will have the same average value of the kinetic energy of the molecules.
- Define means free path, λ of the molecules of a gas and state hoe it is affected by temperature.
- If the mean free path of molecules of air at 0oC and 1.0 atmospheric pressure is 2
10-7m, what will be mean free path be at 1.0 atmospheric pressure and 27oC INTERNAL ENERGY OF GAS(Symbol, U)
The internal energy (U) of a gas is the total mean kinetic energy of all molecules making up the gas.
U total mean K.E of all molecules
total mean K.E of all molecules But mean K.E of a molecule =
= Where all symbols carry their usual meaning
U means K.E of a molecule N
means K.E of a molecule N U
Where N total number of molecules making up the gas
total number of molecules making up the gas For 1 mole of a gas N N Avogadro‘s number of molecules
Equation I above becomes
U
Since K Boltzmann‘s constant, we have
Boltzmann‘s constant, we have U T
T U
This equation is true for 1 mole of a gas
For n moles of a gas, the equation becomes
U
U 
At constant temperature the change in internal energy U of the gas is zero.
U of the gas is zero. THERMODYNAMICS
This deal with the study of the laws that govern the conversion of energy from one form to another, the direction in which heat will flow and the availability of energy o do work.
It is based on the facts in isolated system everywhere in the universe there is a measurable quantity of the energy known as internal energy of the system.
THE FIRST LAW OF THERMODYNAMICS
The law states “The heat energy Q supplied to a system is equal to the sum of the increase in internal energy (u) of the system and external work done (w)”.
u) of the system and external work done (w)”. Q U + W
U + W 1) Q: comes from outside at constant pressure
U: depends on the temperature of the gas since W: is given by:
W P (V2 – V1)
P (V2 – V1) V change in volume
V1 initial volume
initial volume V2 final volume
final volume Thus, the first law of the thermodynamic can be written as:
Q
The first of thermodynamics represents the principle of conversation of energy
MOLAR HEAT CAPACITY OF A GAS (C)
Definition
The molar heat capacity of a gas is the amount of heat required to raise the temperature of one mole of a gas through one Kelvin.
UNIT OF C
By definition
Hence, the SI unit used is jmol-1 K-1
PRINCIPLE MOLAR HEAT CAPACITIES OF A GAS
There are two principle molar heat capacities of a gas
i) Molar heat capacity of a gas at constant pressure ii) Molar heat capacity of a gas at constant volume
- MOLAR HEAT CAPACITY OF A GAS AT CONSTANT PRESSURE
Symbol, Cp
This is the amount of heat required to raise the temperature of one mole of a gas through one Kelvin when the gas is at constant pressure
Where Q heat supplied to the gas at constant pressure. - MOLAR HEAT CAPACITY OF A GAAS AT CONSTANT VOLUME
Symbol,
This is the amount of heat required to raise the temperature of one mole of a gas through one Kelvin when the gas is at constant volume.
Where Q heat supplied to the gas at constant volume
heat supplied to the gas at constant volume 
