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APPLICATIONS OF FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics can be applied to some simple processes such as:
  1. Relation between and
  2. Boiling process iii) Isobaric process (no change in pressure) iv) Isochoric process (no change in volume)
v) Isothermal process (no temp change) vi)Adiabatic process (no heat learning or enters) RELATION BETWEEN (Mayer’s equation)
Consider n – moles of an ideal gas
Suppose the gas is heated at constant volume so that its temperature increases by The amount of heat Q supplied is given by
From the 1st law of thermodynamics
Q
The gas is heated at constant volume and hence
Equation (1) = Equation (2)
Suppose now that n – moles of the same gas are heated at constant pressure so that its temperature increases by the same amount T
The heat Q‘ – supplied is given by
From
From ideal gas equation
Substitute equation (3), equation (4) and equation (6) in equation (5)
This equation shows that is greater than
Moreover;
  1. When the gas is heated at constant volume, no work is done ( V 0) and all the heat goes into raising the internal energy and thus the temperature of the gas.
  2. When the gas is heated at constant pressure, it expands and does work so that only a part of heat is used up in increasing the internal energy and hence temperature of the gas.
Therefore, in a constant- pressure process, more heat is needed to achieve a given temperature change than that of constant volume process.
NOTE:
If are specific heat capacities of the gas at constant pressure volume respectively then:
Where r = gas constant for a unit mass
It is given by
TYPES OF GASES
  1. MONO-ATOMIC GAS
This is a type of a gas whose molecule consist of a single atom
Example
Helium
Neon
  1. DIATOMIC GAS
This is a type of gas whose, molecule consist of two atoms
Examples
Hydrogen (H2)
Oxygen (O2)
Chloride (Cl2)
POLYATOMIC GAS
This is a type of gas whose molecules consists of more than two atoms.
Examples
Carbon dioxide (CO2)
Ammonia (NH3)
Hydrogen peroxide (H2O2)
Steam (water gas) (H2O)
Etc.
DEGREES OF FREEDOM Degrees of freedom of a gas molecule are the number of independent ways the molecule can possess energy.
It refers to the axes x, y and z to which the gas molecule can move freely.
DEGREES OF FREEDOM FOR DIFFERENT GASES
MONOATOMIC GAS
The atom can move freely along x, y and Z- axes.
Hence a monatomic gas has three degrees of freedom
DIATOMCI GAS
The molecule has three degree of freedom in translation and two degree of freedom in rotation. Thus, a diatomic gas has a total 5 – degrees of freedom
POLYATOMIC GAS
The molecule has 3-degree of freedom in translation and 3-degrees of freedom in rotation.
Hence, a polyatomic gas has a total of 6-degrees of freedom.
EQUIPARTITION PRINCIPLE
The mean kinetic energy of a molecule is given by
= KT ……………………………………… (1)
Where all symbols carry their usual meaning.
Since gas molecules are in random motion
………………………………….. (2)
If is the resultant mean square speed of a gas molecule then: –
= + + ………………………….. (3)
Substitute equation (2) in equation (3)
= = ………………………….. (4)
Substitute equation (4) in equation (1)
= = = KT
When are the main square speed of gas molecules along x, y and z-axes respectively, and the
axes of the degrees of freedom.
Equation (5) above express the principal of equitation of energy which says: The mean energy of the molecules of a gas is equally divided among their available degrees of freedom, the average for each degree of freedom being KT
Where K Boltzmann Constant
T Thermodynamic temperature
DEDUCTIONS FROM EQUATION (5)
  1. MONOATOMIC GAS
Has three degrees of freedom Mean K.E of a molecule KT 3
  1. DIATOMIC GAS
Has 5 – degrees of freedom
Mean K.E of a molecule = KT X 5
  1. POLYATOMIC GAS
Has 6 – degrees of freedom
Mean K.E of a molecule = KT 6
RATIO OF MOLAR HEAT CAPACITIES OF A GAS
The ratio is called gamma
This ratio has different values for different gases
FOR MONOATOMIC GAS
This means the internal energy U is given by:
U = KT
Where K = = Boltzmann
constant
U = T
If Cv is the molar heat capacity of a gas at constant volume then
U =
For 1 mole of a gas n = 1
U = …………………………………… (2)
Equation (1) equation (2)
………………….(3)
According to Mayer‘s equation
– = R
= R +
Substitute equation (3) in this equation:
Now = = = 1.67
...
FOR DIATOMIC GAS
The internal energy U is given by:
U = KT
Where K
U = T
The internal energy of the gas can also be given by
Equation (1) equation (2)
RT =
According to Mayer‘s equation
– = R
= R +
Substitute equation (3) in this equation
Now, = = = 1.40
FOR POLYATOMIC GAS
The internal energy (U) of a gas is given
U 3KT
Where K
U 3 T
The internal energy U of the gas can also be given by:
Equation (1) = Equation (2)
3RT
=
Cv = 3R ………………………………(3)
According to Mayer‘s equation
– = R
= R +
Substitute equation (3) in this equation
+ 3R
Now, = = = 1.33
APPLICATION OF =
It is used to solve adiabatic problems
BOILING PROCESS
Suppose a liquid of mass m vaporizes at constant pressure P
Let be volume in the liquid state.
Let be volume in the vapor state
Here expansion takes place at constant pressure and hence the work done by the system is
W =
)
Where = Increase in internal energy of the system
MELTING PROCESS
When a solid changes into liquid state (melting), its internal energy increases.
This can be calculated from, the 1st law of thermodynamics.
Let m = mass of the solid
L = specific latent heat of fusion
Heat (Q) absorbed during the melting process is
Q = …………………………….. (1)
Since during melting process, the change in volume ( = 0
From the 1st law of thermodynamics
Q = + W
Q = +
= + P x 0
... ……………………………(2)
During melting process, internal energy increase by .
Since temperature remains constant during melting the kinetic energy remains the same.
Therefore, the increase in potential energy.
ISOBARIC PROCESS
This is the process which occurs at constant pressure.
According to 1st law of thermodynamics
Q = +
In this case, the heat Q added increases the does internal energy of the gas as well as the gas does external work.
ISOCHORIC PROCESS
This is the process which occurs at constant volume (i.e = 0)
In such a process, external work done W is zero
W P 0 0
According to 1st law of thermodynamics:
Q = + W
We conclude that if heat is added to a system at constant volume, all the heat goes into increasing the internal energy of the system.
ISOTHERMAL CHANGE /PROCESS
Definition
An isothermal change is that change which takes place at constant temperature.
In such a process heat is, if necessary, supplied or removed from the system at just the right rate to maintain constant Temperature.
Conditions for isothermal change
  1. The gas must be held in a thin-walled, highly conducting vessel, surrounded by a constant temperature bath.
  2. The expansion or contraction must take place slowly. So that the heat can pass in or out to maintain the temperature of the gas at every instant during expansion or contraction.
Isothermal change represented graphically
When the temperature is constant the pressure of a gas varies with volume and a graph which
shows this variation is a curve known as isothermal curve
Where
This graph is also called PV – curve or PV – Indicator diagram
When a gas expands, or is compressed, at constant temperature, its pressure and volume vary along the appropriate isothermal, and the gas is said to undergo an isothermal compression
expansion
Isothermal reversible change
When the gas is compressed isothermally from , T to , T then a graph which show this variation is:
If the gas is allowed to expand isothermally so that the state of the gas is brought back from
(, T) through exactly the same intermediate stage then the gas is said to undergo isothermal reversible change:
Definition
An isothermal reversible change is that change which goes to and from through exactly the same intermediate stages at constant temperature.
Isothermal reversible change equation
Since the temperature is constant, and is isothermal change obeys Boyle‘s law.
PV = constant.
... P1V1 = P2V2 Isothermal reversible change equation
Work done during isothermal change
Consider a gas pressure (P) expanded isothermally from volume , to volume .
The gas does some work during the expansion given by:
= …………………….. (1)
Where all symbols carry their usual meaning.
According to ideal gas equation
PV =
P = …………………………. (2)
Substitute equation (2) in equation (1)
dW = dV ……………………………….(3)
When n, R and T are constants
The total work done W is obtained by integrating equation (3) above from volume , to volume .
=
W =
W
W =
The work done W can either be positive or negative depending on whether the gas undergoes isothermal expansion or compression.
If it is isothermal expansion then the work done is positive and the work is said to be done by the gas.
Problem 65
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= 2400 J/kg K




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