Thermodynamics
Thermodynamics
study of the inter-relations of various forms of energy in a system is called 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
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
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
Ex:-
water in a open
beaker.
2.
Closed system:- when a system can exchange only
energy and not matter with its surroundings.
Ex:-
a chemical
reaction taking place in a closed vessel can exchange only heat with
surrounding
3.
Isolated system:- a system which can neither exchange matter nor
energy with its surrounding.
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.
Isothermal process:- in this process temperature of
the system remains constant throughout the process i.e. dT=0
2.
Adiabatic process:- in this process no heat enters or
leaves the system during any stage of the process i.e. dH=0
3.
Isobaric process:- in this process pressure of the system remains
constant throughout the process i.e. dP=0
4.
Isochoric process:- in this process volume of the
system remains constant throughout the process i.e. 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.
Heat:
- Tt is a way to transfer of energy from one body to
another by the difference in the temperature between these bodies.
Work:- Tt is a link between the system
and its surroundings for the transfer of heat energy.
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 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 Q1 joules and work done by the system is W1
joules.
Difference
in energy = Q1 – W1
Again, a system achieves a state B from A along the
path 2. Let in this process heat absorbed by the system is Q2 joules
and work done by the system is W2 joules.
Difference
in energy = Q2 – W2
Similarly
for path 3 & 4
Difference
in energy = Q3 – W3
Difference
in energy = Q4 – W4 and so on…..
We
get,
Q1 – W1 = Q2
– W2 = Q3 – W3 = Q4 – W4
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,
UA and UB shows energies at state A & B.
Now,
DU = UB-UA
= Q-W [mathematical representation of 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 (UA)
to final state (UB) and doing work W by the system on the
surroundings
Then,
from first law of thermodynamics
q = (UB-UA) +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. UB= UA
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
qv = (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 qp amount of heat.
In
this change, volume increase from V1 to V2
Then, increase in volume is given by V2
– V1 = DV ……………………………………………(1)
Then, work done by the system in expansion is given
by W = P(V2 – V1)………….(2)
Now, According to first law of thermodynamics
qp
= (DU)p
+ W ………………………………………………………………………..(3)
qp
= (DU)p
+ P(V2 – V1)
……………………………………………………………….(4)
qp
= (DU)p
+ P(DV) ……………………………………………………………….(5)
if
U1 & U2 are values of internal energies in initial
and final states of system resp.
then, change in internal energy is given by U2
– U1 = (DU)p ………………………….(6)
from
eq. (4) & (6), we get
qp = (U2 – U1) + P(V2 – V1)
qp = (U2 + PV2)
– (U1 + PV1)
…………………………………………………….(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,
qp = H2 – H1 = (DH)p …………………………………………………………(8)
Relation between DH
and DU
:- From eq. (8)
& (5) we get,
qp = (DU)p
+ P(DV)
qp = H2 – H1 = (DH)p
(DH)p = (DU)p + P(DV) ……………………………………………………(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 T1 to T2 then, heat capacity, C of the system between T2 & T1 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 (CV).
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 (Cp).
Cp
is always larger then Cv by an amount equal to P-V work done so,
Cp = Cv
+ 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 Cp & Cv :-
As we know Cp is always greater than Cv.
Therefore,
we shall find Cp-Cv
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 1st and 3rd 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
CP – CV
= 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 DU
= 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 mJ.T. is +ve.It means joule Thomson effect will be cooling the gas or
substance.
When
2a/RT = b, the value of mJ.T.=0 so joule Thomson effect will be nil.
When 2a/RT < b. value of mJ.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 mJ.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 mJ.T =0.
So
Or
2a/RTi
= b
Or Ti
= 2a/Rb
Where
Ti = 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 V1 initial
state to final state V2.
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 P1V1
= P2V2 the above equation becomes
Where
P1 and P2 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 V2/V1 = nRT ln P1/P2
if
expansion occurs (increase in volume) V2>V1 then work
is done by the system is +ve.
if
compression occurs (decrease in volume) V1>V2 then
work is done on the system is -ve.
Join us at Social Media
Hope you enjoy learning Thermodynamics B.Sc. Chemistry Notes. You can also explore these interesting science topics-
Thermochemistry ka bhi notes pdf banaiye please... It's a humble request.
ReplyDelete