Thermodynamic State: It is the state in which a thermodynamic system exists. Thermodynamic states are characterized by a set of macroscopic properties, which determine the internal properties of a system in that state and the interaction of the system with external bodies. These properties include temperature, pressure, volume, electric polarization, and magnetization.

Thermal Equilibrium: When the temperature throughout a system is uniform, the system is in thermal equilibrium.

Zeroth law of thermodynamics: The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with each other.If two systems A and B are separately in thermal equilibrium with a third system C, then the three systems are in thermal equilibrium with each other. Zeroth law of thermodynamics states that two systems which are individually in thermal equilibrium with a third one, are also in thermal equilibrium with each other. This Zeroth law was stated by Flower much later than both first and second laws of thermodynamics. This law helps us to define temperature in a more rigorous manner.

Internal Energy - U is the most common symbol used for internal energy.Internal energy is defined as the energy associated with the random, disordered motion of molecules.

Concept of Heat • Heat may be defined as energy in transit.

• Word heat is used only if there is a transfer of energy from one thermodynamic system to the another.

• When two systems at different temperatures are kept in contact with each other then after some time temperatures of both the systems become equal and this phenomenon can be described by saying that energy has flown from one system to another.

• This flow of energy from one system to another on account of temperature difference is called heat transfer.

• Flow of heat is a non-mechanical mode of energy transfer.

• Heat flow depends not only on initial and find states but also on path it's.

P-V Indicator Digram • Only two thermodynamic variables are sufficient to describe a system because third vaiable can be calculated from equation of state of the system.• P-V Indicator Digram is just a graph between pressure and volume of a system undergoing an operation.• When a system undergoes an expansion from state A (P1 V1) to a state B (P2V2) its indicator diagram is shown as follows.

In case of compression system at state A(P1 V1) goes to a state B(P2V2) its indicator digram is as follows.

In isothermal process temperature of the system remains constant throughout the process. For an iso-thermal process equation connecting P, V and T gives PV = constant i.e., pressure of given mass of gas varies inversly with its volume this is nothing but the Boyle's law. In isothermal process there is no change in temperature, since internal energy for an ideal gas depends only on temperature hence in iso thermal process there is no change in internal energy. Thus, ΔU=0 therefore, ΔQ =ΔW Thus during isothermal process Heat added (or substacted) from the system = wok done by (or on) the system

Process in which no heat enters or leaves a system is called an adiabatic process (Temperature doesn’t remain constant) For every adiabatic process Q=0 Prevention of heat flow can be accomplished by surrounding system with a thick layer of heat insulating material like cork, asbestos etc. Flow of heat requires finite time so if a process is performed very quickly then process will be practically adiabatic. On applying first law to adiabatic process we get      ΔU=U2 - U1= +ΔW               (adiabatic expansion) Here, internal energy of systems decreases resulting a drop in temperature. But, the work done is positive. On applying first law to adiabatic compression we get      ΔU=U2 - U1= - ΔW               (adiabatic compression) Here, internal energy of systems increases resulting an increase in temperature. Bur, the work done is negative.

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A process taking place at constant pressure is called isobaric process. we see that work done in isobaric process is W = P(V2 - V1) = nR (T2-T1) where pressure is kept constant. Here in this process the amount of heat given to the system is partly used in increasing temperature and partly used in doing work.

In an isochoric process volume of the system remain uncharged throughout i.e.

ΔV = O. When volume does not change no work is

done ; ΔW = 0 and therefore from first law U2 - U1 = ΔU =ΔQ All the heat given to the system has been used to increase the intenal energy of the system.

Workdone in an isothermal expansion

Consider one mole of an ideal gas enclosed in a cylinder with perfectly conducting walls and fitted with a perfectly frictionless and conducting piston. Let P1, V1 and T be the initial pressure, volume and temperature of the gas. Let the gas expand to a volume V2 when pressurereduces to P2, at constant temperature T. At any instant during expansion let the pressure of the gas be P. If A is the area of cross section of the piston, then force F = P × A.Let us assume that the pressure of the gas remains constant during an infinitesimally small outward displacement dx of the piston.

Work donedW = Fdx = PAdx = PdVTotal work done by the gas in expansion from initial volume V1 to final volume V2 is

Work done in an adiabatic expansion

Consider one mole of an ideal gas enclosed in a cylinder with perfectly non conducting walls and fitted with a perfectly frictionless, non conducting piston. Let P1, V1 and T1 be the initial pressure, volume and temperature of the gas. If A is the area of cross section of the piston, then force exerted by the gas on the piston is F = P × A, where P is pressure of the gas at any instant during expansion. If we assume that pressure of the gas remains constant during an infinitesimally small outward displacement dx of the piston,then work done dW = F × dx = P × A dx

Adiabatic relations of system for perfect gas

Consider 1 gram of the working substance (ideal gas) perfectly insulated from the surroundings. Let the external work done by the gas be δW.Applying the first law of thermodynamics

δH = dU + δW

But δH = 0

and δW = P.dV

Therefore,

Where P is the pressure of the gas and dV is the change in Volume.

0= dU + P dV

As the external work is done by the gas at the cost of its internal energy, there is fall in temperature by dT.

Let r = Cp - Cv CvP dV + Cv V dP+ Cp P dV− Cv P dV = 0

Cv V dP+ Cp P dV = 0

Dividing by Cv PV, 𝑑𝑃𝑃 + 𝐶𝑃𝐶𝑉 𝑑𝑉𝑉 = 0

Substitute 𝐶𝑃𝐶𝑉 = 𝛾

𝑑𝑃𝑃 + 𝛾 𝑑𝑉𝑉 = 0

Integrating,

log P + 𝛾 log V = Constant

log P + log 𝑉𝛾 = Constant

log P𝑉𝛾 = Constant

or

P𝑉𝛾 = Constant …………………..(4)

This is the equation connecting pressure and volume during an adiabatic process.

Taking PV = r T

Or P = 𝑟𝑇𝑉

Substitute in eqn (4)

𝑟𝑇𝑉 𝑉𝛾= Constant

rT 𝑉𝛾-1 = Constant

or

T 𝑉𝛾-1 = Constant …………..(5)

Taking PV = r T

Or V = 𝑟𝑇𝑃

Substitute in eqn (4)

P ( 𝑟𝑇𝑃 )𝛾= Constant

𝑟𝛾𝑇𝛾𝑃𝛾−1 = Constant

Or

𝑃𝛾−1𝑇𝛾 = Constant …………………(6)

Thus during adiabatic process,

(i) P𝑉𝛾 = Constant (ii) T 𝑉𝛾-1 = Constant

(iii) 𝑃𝛾−1𝑇𝛾 = Constant

Reversible processA thermodynamic process is said to be reversible when (i) the various stages of an operation to which it is subjected can be reversed in the opposite direction and in the reverse order and (ii) in every part of the process, the amount of energy transferred in the form of heat or work is the same in magnitude in either direction. At every stage of the process there is no loss of energy due to friction, inelasticity, resistance, viscosity etc. The heat losses to the surroundings by conduction, convection or radiation are also zero.Condition for reversible process(i) The process must be infinitely slow.(ii) The system should remain in thermal equilibrium (i.e) system and surrounding should remain at the same temperature.Examples(a) Let a gas be compressed isothermally so that heat generated is conducted away to the surrounding. When it is allowed to expand in the same small equal steps, the temperature falls but the system takes up the heat from the surrounding and maintains its temperature.(b) Electrolysis can be regarded as reversible process, provided there is no internal resistance.

Irreversible processAn irreversible process is one which cannot be reversed back. Examples : diffusion of gases and liquids, passage of electric current through a wire, and heat energy lost due to friction. As an irreversible process is generally a very rapid one, temperature adjustments are not possible. Most of the chemical reactions are irreversible.

Efficiency of Carnot’s cycle is independent of the working substance, but depends upon the temperatures of heat

source and sink. Efficiency of Carnot’s cycle will be 100% if T1 = ∞ or T2 = 0 K. As neither the temperature of heat source can be made infinite, nor the temperature of the sink can be made 0 K, the inference

is that the Carnot heat engine working on the reversible cycle cannot have 100% efficiency.

Relation between Cp and Cv (Meyer’s relation)

Let us consider one mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A. Let P, V and T be the pressure, volume and absolute temperature of gas respectively (Fig.). A quantity of heat dQ is supplied to the gas. To keep the volume of the gas constant, a small weight is placed over the piston. The pressure and the temperature of the gas increase to P + dP and T + dT respectively. This heat energy dQ is used to increase the internal energy dU of the gas. But the gas does not do any work (dW = 0).

∴ dQ = dU = 1 × Cv × dT ... (1)

The additional weight is now removed from the piston. The piston now moves upwards through a distance dx, such that the pressure of the enclosed gas is equal to the atmospheric pressure P. The temperature of the gas decreases due to the expansion of the gas. Now a quantity of heat dQ’ is supplied to the gas till its temperature becomes T + dT. This heat energy is not only used to increase the internal energy dU of the gas but also to do external work dW in moving the piston upwards.