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APPLICATIONS OF FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics can be applied to some simple processes such as:

- Relation between and
- 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;

- 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.
- 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

- MONO-ATOMIC GAS

This is a type of a gas whose molecule consist of a single atom

Example

Helium

Neon

- 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.

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)

- MONOATOMIC GAS

Has three degrees of freedom Mean K.E of a molecule KT 3

- DIATOMIC GAS

Has 5 – degrees of freedom

Mean K.E of a molecule = KT X 5

- 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

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

- The gas must be held in a thin-walled, highly conducting vessel, surrounded by a constant temperature bath.
- 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.

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When a quantity of heat is supplied to the gas it expands at constant pressure doing 8400J of work and heating up by 20. Calculate:

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##### Problem 82

0.15 mol of an ideal mono atomic gas is enclosed in a cylinder at a pressure of 250 KPa and a temperature of 320K. The gas is allowed to expand adiabatically and reversibly until its pressures is 100KPa