##
__Thermodynamics__

__Thermodynamics__

###
__Thermodynamics__

study of the inter-relations of
various forms of energy in a system is called thermodynamics.__Thermodynamics__

####
__Limitation of
thermodynamics__

__Limitation of thermodynamics__

i.
laws
of thermodynamics are not applicable to small particles like individual atoms
or molecule, but laws can be applied to macroscopic system or very large
system.

ii.
Thermodynamics
does not gives information about rate at which a given chemical
reaction/process may proceed and also time for this change.

###
__Thermodynamic
system__

__Thermodynamic system__

**any specified portion of the universe or matter, real or imaginary, separated from the rest of the universe, which is selected for the thermodynamic treatment is called a system.**

####
__Surroundings__

__Surroundings__

Leaving the
system the rest of the universe, which may exchange matter or energy or both
with the system is called surroundings.

####
__Types
of system__

__Types of system__

**1.**

**a system which can exchange energy as well as matter with its surroundings.**

*Open system*:-**Ex:-**water in a open beaker.

**2.**

**when a system can exchange only energy and not matter with its surroundings.**

*Closed system*:-**Ex:-**a chemical reaction taking place in a closed vessel can exchange only heat with surrounding

**3.**

**a system which can neither exchange matter nor energy with its surrounding.**

*Isolated system*:-**Ex:-**a reaction in closed vessel which can’t exchange heat or matter.

####
**Homogeneous
system:- **

A system which is uniform throughout i.e. for
chemicals it must have same composition throughout. Homogeneous system consists
of only one phase.
Ex:-
Glucose dissolved in water.

####
**Heterogeneous
system:-**

A
system which is not uniform throughout i.e. it may consists two or more phases
in equilibrium. Its phases are separated from one another by bounding surfaces.
EX:-
ice in water.

####
**Macroscopic
system:- **

A system which consist of a large no. of
atoms, particles, molecules, radicals.####
**Macroscopic
properties:- **

properties
of macroscopic system is known as macroscopic properties. EX:- pressure,
temperature, volume, composition, density, viscosity, surface tension, etc.
Change
of any macroscopic property changes the state of the system or vice-versa.

####
**State of a system:- **

It is defined by
the macroscopic properties. When the macroscopic properties a of a system have
specific or definite value it is said that the system is in definite state.###
**Thermodynamic Equilibrium:- **

If
macroscopic properties like temperature, pressure, volume composition etc. do
not change with time.
Types
of thermodynamic equilibrium

####
**1.
****Thermal equilibrium:- **

A system whose temperature do not
change along with the temperature of the surroundings.####
**2.
****Mechanical equilibrium:- **

A system which do not perform any mechanical
work. ####
**3.
****Chemical equilibrium:- **

A system whose chemical
composition does not change with time (remains same throughout).###
**Physical properties
of the system **

Physical properties
of the system are of two types-####
**1.
****Extensive property**

This property depends on quantity
or amount of matter present in the system.
Ex:- Mass, energy, no.

**of moles, enthalpy, entropy etc.**####
**2.
****Intensive property **

**This property do not depends on quantity or amount of matter present in the system.**

**Ex:**- temperature , pressure, density, viscosity, surface tension etc.

####
**State function:-**

It is the property of the
thermodynamic system whose value is definite for a particular state of the
system. When a change is brought about in this particular state of system,
change in state function also occurs. It depends only on initial and final
state of the system.
Ex:-
pressure, temperature, volume, energy are state function.

####
**Path function:- **

When a system passes from one
state A to another state B depends on the nature of the path followed, not on
initial and final state.
Ex:-
work done is path function.

###
**Thermodynamic
process**

**If a thermodynamic system changes from one state to another state the operation is known as thermodynamic process.**

####
**Types of process-**

**1.**

**in this process temperature of the system remains constant throughout the process i.e.**

*Isothermal process*:-**dT=0**

**2.**

**in this process no heat enters or leaves the system during any stage of the process i.e.**

*Adiabatic process*:-**dH=0**

**3.**

**in this process pressure of the system remains constant throughout the process i.e.**

*Isobaric process*:-**dP=0**

**4.**

**in this process volume of the system remains constant throughout the process i.e.**

*Isochoric process*:-**dV=0**

####
**Cyclic
process or cycles:-**

When a system return to its initial state after completing the process in
various stages, that is system has completed one cycle and process is known as
cyclic process.

####
**Reversible
process:**-

If a
thermodynamic process is carried out infinitesimally slowly so that at every
stage of it, the system in temperature and pressure remains in equilibrium with
surrounding, This type of process is called reversible process.

####
**Irreversible
process:**-

If a
thermodynamic process is not carried out infinitesimally slowly so that at
every stage of it, the system do not remains in equilibrium with surrounding,
This type of process is called irreversible process.####
**Endothermic
process:** -

The
process in which amount of energy is absorbed to carried out a reaction, known
as endothermic process.

####
**Exothermic
process:** -

The
process in which amount of energy is evolved to carry out a reaction, known as
endothermic process.

**Tt is a way to**

__: -__*Heat***transfer of energy from one body to another by the difference in the temperature between these bodies.**

**- Tt is a link between the system and its surroundings for the transfer of heat energy.**

__:__*Work*####
**Signs
for heat and work **

Ø
q
is positive when heat is evolved

Ø
q
is negative when heat is absorbed

Ø
w
is positive when work is done by the system

Ø
w
is negative when work is done on the system

##
**First
law of Thermodynamics**

**First law of Thermodynamics**

First
law of thermodynamics states that “

*Energy can neither be created nor destroyed but can only be transformed from one form to another*.” This law is also known as law of conservation of energy.
Let,
a system achieves a state B from A along the path 1. Let in this process heat
absorbed by the system is Q

_{1}joules and work done by the system is W_{1}joules.
Difference
in energy = Q

_{1}– W_{1}
Again, a system achieves a state B from A along the
path 2. Let in this process heat absorbed by the system is Q

_{2}joules and work done by the system is W_{2}joules.
Difference
in energy = Q

_{2}– W_{2}
Similarly
for path 3 & 4

Difference
in energy = Q

_{3}– W_{3}
Difference
in energy = Q

_{4}– W_{4}and so on…..
We
get,

Q

_{1}– W_{1 }= Q_{2}– W_{2 }= Q_{3}– W_{3 }= Q_{4}– W_{4}
From
it we come to know difference in energies for all path connecting A & B is
same so we can write as (Q-W)

Where, Q
= energy absorbed by the system

W = Energy consumed by
the system in doing work

Q - W = change or
increase in internal energy of the system when it changes from state A to B and
is independent of the path followed

If U = internal energy

Then,
U

_{A}and U_{B}shows energies at state A & B.
Now,

**D**

**U = U**[mathematical representation of

_{B}-U_{A}= Q-W**First law of thermodynamics]**

####
**First law of
thermodynamics in relation with work and heat**

Suppose,
a system be given

**q**heat. It is used in raising the internal energy of the system from initial state (U_{A}) to final state (U_{B}) and doing work W by the system on the surroundings
Then,
from first law of thermodynamics

q = (U

_{B}-U_{A}) +w
q = DU + w

It
is mathematical statement of the first law of thermodynamics.

If
changes in energy is infinitesimally small then

dU = dq - dw

dU = dq – PdV

in
a cyclic system, when system returns to its initial state i.e. U

_{B}= U_{A}or DU = 0, there will be no change in internal energy
then, q-w = DU

or, q-w = 0

or, q = w

####
**Heat changes at
constant volume: -**

When
the process is carried out at constant volume there will be neither expansion
nor contraction in volume of gas. At this condition no work is done by the
system i.e. w=0; put it in First law of Thermodynamics, we get
q = DU + w

q

_{v}= (DU)_{v}
It
means at constant volume, Heat absorbed is utilised in increasing the Internal
Energy of the system.

####
**Heat changes at
constant pressure:- **

Let
at constant pressure P, a change of state of a system is brought about from
initial state 1 to final state 2 by the absorption of **q**amount of heat.

_{p}
In
this change, volume increase from V

_{1}to V_{2}
Then, increase in volume is given by

**V**_{2}– V_{1}=**D****V ……………………………………………(1)**
Then, work done by the system in expansion is given
by

**W = P(V**_{2}– V_{1})………….(2)
Now, According to first law of thermodynamics

**q**

_{p}= (**D**

**U)**

_{p}+ W ………………………………………………………………………..(3)**q**

_{p}= (**D**

**U)**

_{p}+ P(V_{2}– V_{1}) ……………………………………………………………….(4)**q**

_{p}= (**D**

**U)**

_{p}+ P(**D**

**V) ……………………………………………………………….(5)**

if
U

_{1}& U_{2}are values of internal energies in initial and final states of system resp.
then, change in internal energy is given by

**U**_{2}– U_{1}= (**D****U)**_{p }………………………….(6)
from
eq. (4) & (6), we get

**q**

_{p}= (U_{2}– U_{1}) + P(V_{2}– V_{1})**q**

_{p}= (U_{2}+ PV_{2}) – (U_{1}+ PV_{1}) …………………………………………………….(7)
Since, P & V are definite properties of state of
system and U is also a definite property. it is clear that like internal energy
U,

**(U+ PV)****is also a definite property of state and depends only on states of system and not on path by which this state is achieved. This thermodynamic property is denoted by H, i.e. (H= U + PV) and is known as enthalpy, Total energy or Heat content at constant pressure of the system. From eq. (vii) & (H= U + PV) we get,****q**

_{p}= H_{2}– H_{1}= (**D**

**H)**

_{p}…………………………………………………………(8)**Relation between**

**D**

**H and**

**D**

**U :-**From eq. (8) & (5) we get,

**q**

_{p}= (**D**

**U)**

_{p}+ P(**D**

**V)**

**q**

_{p}= H_{2}– H_{1}= (**D**

**H)**

_{p}**(**

**D**

**H)**=

_{p}**(**

**D**

**U)**

_{p}+ P(**D**

**V) ……………………………………………………(9)**

####
**Enthalpy of
vaporisation:- **

**It is defined as change in enthalpy (DH) when liquid evaporates into vapour state or vapours condenses into liquid state.**

####
**Enthalpy of
fusion:- **

**It is defined as change in enthalpy (DH) when solid melts into liquid state or liquid freezes into solid state.**

####
**Heat Capacity:- **

The amount of heat required to raise the
temperature of known quantity of substance or system by 1°C.
If q is heat added to the system to raise the temperature from T_{1 }to T

_{2}then, heat capacity, C of the system between T

_{2 }& T

_{1}is given by

For
very small quantity of heat dq to be added to the system, it rises a small
raise in temperature by dT. Then

When,
the amount of the substance is 1 gram molecule then heat capacity is known as
the molar heat capacity.

####
**Unit of heat capacity:-
**
** **

In general

**In SI**

**,**

####
**Heat capacity at
constant volume:- **

The
amount of heat required to raise the temperature of known quantity of substance
or system by 1°C at constant volume is known as Heat capacity at constant volume (C_{V}).

From
first law of thermodynamics

dq = dU + PdV

hence,

at
constant volume dV = 0 then,

####
**Heat capacity at
constant pressure:- **

The
amount of heat required to raise the temperature of known quantity of substance
or system by 1°C at constant pressure is known
as Heat capacity at
constant pressure (C_{p}).

C

_{p}is always larger then C_{v}by an amount equal to P-V work done so,
C

_{p}= C_{v}+ external work
Since, work done in expansion = P (¶V/¶T)

Putting this value in eq. (xvi) we get

But, H = U + PV

Differentiate
this w.r.t. T at constant pressure

By
comparing this eq. with (xvii) we get,

####
**Relation Between C**_{p}
& C_{v} :-

As we know C_{p}& C

_{v}

_{p }is always greater than

_{ }C

_{v}.

Therefore,
we shall find C

_{p}-C_{v}
As
we know

Hence,

But H = U + PV

Differentiating
this eq. w.r.t. temperature at constant pressure

Combining
eq. (20) and (21) we get

Now
we have to relate 1

^{st}and 3^{rd}terms of eq. (22). For this we consider V and T as independent variables out of P, V and T, then
U = f (T, V)

Hence,

Dividing
both sides by dT, and consider pressure constant, we get

Now
this eq. is substituted in eq. (22) we get

For
an ideal gas PV = RT

Differentiated
with respect to T at constant pressure then

For
ideal gas

Hence,
with the help of eq. (27), eq. (26) will become

C

_{P}– C_{V}= R ………………………………………………..(28)####
**Joule’s Law or
Joule-Thomson Effect :-**

J.P. Joule and W. Thomson in year 1852-1862 made Joule-Thomson Law or effect.
According
to this law when a gas is made to expand adiabatically from high pressure to a
extremely low pressure cooling is produced, i.e. gas gets cooled. This
phenomenon is known as Joule-Thomson Effect or Joules law.

All
gases behaves like this except Hydrogen and Helium i.e. they get heated instead
of cooling is produced.

The
cooling effect in Joule-Thomson effect is due to decrease in kinetic energy of
the gas molecules because a part of this energy is used in overcoming the
forces of attraction existing between the molecules of the gas in expansion.
For ideal gases there is no force of attraction between the gas molecules and
therefore on expansion in vacuum through the porous plug, neither cooling nor
heating is produced. i.e., neither absorption nor evolution of heat takes place
and therefore no external work to separate the molecules and so,

**Q = 0, w = 0, and therefore**

**D**

**U = 0**

####
**Joule-Thomson
Coefficient:- **

**Enthalpy is a definite property depending upon the state of the system. Hence , dH is complete differential. Suppose that P and T are variables, then**

The
enthalpy remains constant (dH = 0) in adiabatic expansion of the real gases
therefore, in above eq. put dH = 0

Or

Or

Or

i.e.,

####
**Inversion Temperature:
-**

**As per joule-Thomson coefficient**

In
above eq. when value of 2a/RT > b, value of m

_{J.T. }is +ve.It means joule Thomson effect will be cooling the gas or substance.
When
2a/RT = b, the value of m

_{J.T.}=0 so joule Thomson effect will be nil. When 2a/RT < b. value of m_{J.T. }is -ve. It means joule Thomson effect will be warming the gas or substance.
We
know the value of a, b and R are constants so m

_{J.T }is depend on temperature only.
The
temperature at which joule Thomson coefficient changes its sign from +ve to –ve
or vice versa is known as inversion temperature. At this temperature m

_{J.T}=0.
So

Or
2a/RT

_{i}= b
Or T

_{i}= 2a/Rb
Where
T

_{i}= inversion temperature.
a and b are vander walls constants

**Work done in the expansion of ideal gases under isothermal conditions for reversible process:-**

Let
an ideal gas is enclosed in a cylinder fitted with weightless and frictionless
piston

Also
suppose that cylinder is placed in thermostat so its temperature will remain
constant throughout the process of expansion therefore the gas is in thermal
equilibrium with surroundings. if external pressure on piston be P equal to the
pressure of the gas within the cylinder in the beginning so therefore the
piston is at rest. Pressure P is lowered by infinitesimally small amount dP.
The new pressure is now (P-dP). So gas expands by infinitesimally small volume
dV. Then the volume of the gas now becomes (V+dV). so piston is pushed up until
it comes to rest (when internal and external pressure becomes equal).

In
this process gas does infinitesimally small work on the piston.

If
all the above process will be repeated again. Then infinitesimally small work done
on the piston again second time.

(We
know for isothermal expansion

**q = w**i.e. work is done by the gas is equal to heat absorbed by the gas from the surroundings)
If
the all above process are continued then

q = w = (P-dP) ´
dV

= PdV – dPdV

Neglecting
the very small term

**dPdV**we get**w = PdV.**
The
total work done by the gas In the process of expansion will be the sum of a
continuous series of PdV terms, as volume increase from V

_{1}initial state to final state V_{2}.
Thus

Since
for a reversible process the external pressure is always only infinitesimally
lower then the pressure of the gas itself, ideal gas pressure P = (RT/V) can be
substituted for P in above eq.

So

If
the gas is an ideal gas then at constant temperature P

_{1}V_{1}= P_{2}V_{2}the above equation becomes
Where
P

_{1 }and P_{2}are the external pressure in initial and final state of the gas respectively.
For
the isothermal reversible expansion of n moles of an ideal gas above equation
becomes

**q = +W = nRT ln V**

_{2}/V_{1}= nRT ln P_{1}/P_{2}
if
expansion occurs (increase in volume) V

_{2}>V_{1}then work is done by the system is +ve.
if
compression occurs (decrease in volume) V

_{1}>V_{2}then work is done on the system is -ve.
## 2 comments:

Sir ,bsc sem 2 inorganic ka notes

1}basic compound

2} electronic displacement

3} stereochemistry

4}optical isomerism

5} chemistry of aliphatic hydrocarbons

6} aromatic hydrocarbon

In sab ka notes send kijea please...

please send pdf file of thermodynamics

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