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.
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.
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
1. Open system:-a system which can exchange energy as well as matter with its surroundings.
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.
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.
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.
A system which consist of a large no. of atoms, particles, molecules, radicals.
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.
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.
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.
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.
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.
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.
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.
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.
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 W1joules.
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…..
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.
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= UAor 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.
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 & T1is 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:-
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
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).
Cpis 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
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)
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
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
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.
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 = wi.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.
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.
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.