Thermodynamics deals with the relation of heat energy to mechanical work and other forms of energy. When we rub our palms they become warm. This is due to conversion of mechanical work performed in rubbing our hands together into heat. Thermodynamics deals with macroscopic properties and does not go into the microscopic level. For example the state of gas in thermodynamics is specified by macroscopic variables such as pressure, volume, temperature, mass and composition that are felt by our sense perceptions. The distinction between mechanics and thermodynamics is worth bearing in mind. In mechanics our interest is in the motion of particles or bodies under action of forces and torques. Thermodynamics is not concerned with motion of a system as a whole. It is concerned with the internal macroscopic state of a body.
e.g.: When a bullet is fired from a gun, it is embedded into the wooden plank and stops. Then kinetic energy of the bullet gets converted into heat, which causes change of temperature of bullet and surrounding layers of the plank. Temperature is related to the energy of internal motion (disorder) of the bullet but not the whole motion of bullet.
Thermal equilibrium:
The term "equilibrium" in thermodynamics appears in different context. According to this the state of a system is in equilibrium state if the macroscopic variables that characterise the system do not change in time. For example a gas inside a closed rigid container, completely insulated from its surroundings with fixed values of mass, pressure, volume, temperature and compositions that do not change with time, is in a state of "thermal equilibrium". A state of equilibrium depends on surroundings and nature of the wall that separates the system and surroundings. There are two types of walls 1) Adiabatic wall 2) Diathermic wall.
Adiabatic wall: It is an insulating wall that does not allow flow of energy from one to another system/ surroundings.
Diathermic wall: It is a conducting wall that allows energy flow from one to another until both the systems attain equilibrium states. After that there is no change in their states.
The concept of temperature in thermodynamics is explained by "Zeroth law of thermodynamics" which states that two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other".
Let the systems A and B are separately in equilibrium with a system C. Then
TA = TC and TB = TC. This implies TA = TB. i.e., A and B are said to be in thermal equilibrium.
Heat and temperature: Heat is a form of energy. When it is added to a body it increases the internal energy of the body and is responsible for the change in its thermal condition. It flows from a hot body to a cold body, till both attain same state of hotness or coldness.
Unit: Calorie in C.G.S., Joule in SI
D.F.: [ML2T-2]
Calorie: Quantity of heat required to raise the temperature of 1 gm of water from 14.5oC to 15.5oC is called Calorie.
Temperature: It gives the thermal condition of a body which determine its ability to transfer heat to other bodies. It is a measure of degree of hotness or coldness of a body.
D.F.: [K]
Unit: oC, (K)
Joule's Law: The amount of mechanical workdone on a system is directly proportional to the amount of heat produced in that system.
If 'W' is the amount of workdone on a system, 'H' is the amount of heat produced, then according to Joule's law 
J is called Joule's constant (or) Mechanical equivalent of heat.
The amount of workdone on a system to produce unit amount of heat is called mechanical equivalent of heat.
J = 4.186 × 107 erg/ cal = 4.186 J/ cal.
Internal energy:
The concept of internal energy depends only on the state of the system.
Every system consists of a large number of molecules. Internal energy is simply the sum of kinetic energies and potential energies of these molecules. Internal KE depends on temperature and internal PE depends on volume. Internal energy of a gas
(U) = Σ PE + Σ KE (for a real gas)
(U) = Σ KE (for an ideal gas)
* Absolute internal energy of a system is not measurable. But we can measure change in internal energy of the system 
(Ui = initial value, Uf = final value)
* When temperature of the system increases, dU is +ve. When temperature of the system decreases dU is -ve. When temperature of the system remains constant, then change in internal energy 
.
* Heat and work in thermodynamics are not state variables. They are modes of transfer of energy to a system resulting in chage in its internal energy. The study of energy relations and the manner in which energy changes take place are based on two general laws i.e., the first and second law of thermodynamics. I law represents the connection between heat and mechanical work. II law depicts the manner in which this energy change takes place.
First law of Thermodynamics:
When certain amount of heat is given to a system, a part of it is used to increase the internal energy and remaining part is used in doing external work.
Where ∆Q is heat supplied to the system by the surroundings.
∆U is change in internal energy of the system.
∆W is workdone by the system on the surroundings.
This law is a particular form of law of conservation of energy.
* In case of an isolated system, there is no interaction with surroundings. No work is done by or on the system i.e., dW = 0, dQ = 0
dU = 0 
U = constant
i.e., internal energy of an isolated system is constant.
* When heat supplied is completely converted into work out changing the temperature of the system, the internal energy of the system remains constant.
Sign convention:
* If the heat energy dQ is added to the system, it is taken as positive.
* If heat is given out by the system, it is taken as negative.
* If work is done by the system is taken as +ve.
* If work is done on the system is taken as -ve.
* The increase in internal energy is taken as +ve.
* The decrease in internal energy is taken as -ve.
Applications of I law of thermodynamics:
1) When a substance melts, the change in volume (dV) is very small, the temperature of a substance remains unchanged during melting process. Internal energy changes during the melting process. From I law of thermodynamics
dQ = dU + dW
dW = P(dV) dQ = dU + PdV
* If change in volume is neglected i.e., dV = 0 then dQ = dU 
* If volume increases, then dV is +ve then dQ = dU + P(+dV)
dQ = dU + P(dV) mL = dU + (dV)P
* If volume decrease, then dV is - ve.
dQ = dU - (dV)P 
mL = dU - (dV)P 
2) In case of boiling process:
mL = (Us - Uw) + P(Vs - Vw)
Us - Uw = mL - P(Vs - Vw)
Where Us, Uw are internal energies of steam and water respectively.
Vs, Vw are volumes of steam and water respectively.
Problems:
1. When heat energy of 2000 J is supplied to a gas, the external workdone by the gas is 500 J. What is the increase in its internal energy?
Sol: dQ = 2000 J, dW = 500 J, dU = ?
dQ = dU + dW dU = dQ - dW
= 2000 - 500
= 1500 J
2. Find the external workdone by the system in K.cal, when 20 K.cal of heat is supplied to the system. So that its internal energy is 8400 J (J = 4200 J/ cal)
Sol: dQ = dU + dW
dW = 20 - 2
= 18 K.cal
Relation between specific heat capacities of gases:
I. Cp > Cv, Why?
Proof: The specific heat capacity of a gas at constant volume is represented by
∆Q = mCv ∆T (or) mCv dT
At constant volume, dV = 0 dQ = dU + P(0)
dQ = dU
... (dQ)v = dU = mCv dT
In terms of molar specific heat capacity
∆Q (or) (dQ)v = dU = nCv ∆T (or)
= µ Cv ∆T ................. (1)
Where n (or) µ are no. of moles of taken gas.
The specific heat capacity of a gas at constant pressure is represented by
dQ = mCp dT (or) mCp ∆T
At constant pressure (dQ)p = nCp dT (or) µ CpdT ............... (2)
(dQ)p = dU + dW
= dU + P(dV)
i.e., At constant pressure, heat supplied to the gas increases the internal energy and work is done during expansion.
µ Cp dT = µ Cv dT + dW
From equation (3), 
II. Cp - Cv = R
Proof: Consider 1 mole of an ideal gas contained in a conducting cylinder provided with a frictionless Piston. Temperature increases through ∆T. As heat supplied at constant volume, then
(dQ)v = nCv dT (or) µ Cv ∆T
dU = nCv dT .................... (1)
If heat supplied at constant pressure then
(dQ)p = n Cp dT (or) µ Cp ∆T ....... (2)
At constant pressure, supplied heat is used to increase the internal energy and to do the external work (dW) in moving the Piston through a small distance dx against constant pressure P, then
dW = F × dx = P(A dx) = P dV .................. (3)
From first law of thermodynamics
dQ = dU + dW
From (1), (2) and (3) equations
nCpdT = nCv dT + PdV
Put n = 1 mole
Cp dT = Cv dT + PdV
From PV = nRT, PV = RT
PdV = R dT
Cp dT = Cv dT + R dT
Thermodynamic Processess:
Quasi-Static Process: It is an infinitesimally slow process in which the system remains in thermal and mechanical equilibrium with surroundings at each and every intermediate stage.
Consider a conducting cylinder filled with a given mass of gas and fitted with a frictionless piston. A few small weights are placed on the piston.
Due to the weight the piston moves down performing work on the gas. The volume decreases and pressure increases. After attaining the thermodynamic equilibrium with surroundings, the state of gas is defined by its Pressure P, Volume V and Temperature T. If small weights are removed one by one, the piston raised slowly. The gas expands performing external work on the piston. The volume increases and attains a new value for the volume and pressure in a short time and attains a new thermodynamical equilibrium with surroundings. If weights are removed infinitesimally small, the change in pressure, volume and temperature of the gas are infinitesimally small. These small changes make the differences ∆P, ∆V, ∆T between gas and surroundings. Thus any process taking place very slowly can be considered as a quasi-static process.
Isobaric Process: Let the gas be heated at constant pressure to the same increase in its temperature dT. It is utilised to increase the internal energy (dU) and to do external work (dW) in moving the piston through a small distance dx against constant pressure P.
Then (dQ)p = dU + dW
Cp dT = dU + dW (for one mole)
nCp dT = dU + dW (for 'n' moles)
P - V graph of this process is a straight line parallel to volume axis.
"The process in which a system undergoes a change in volume and temperature at constant pressure by the exchange of heat energy with the surroundings is called isobaric process".
Isochoric Process: Let the gas heated at constant volume, its temperature increases through 'dT'. At constant volume heat supplied is utilised only to increase the internal energy of the gas.
(dQ)v = nCv (dT) = dU, dW = 0
P - V graph for this process is straight line parallel to pressure axis.
"The process in which a system undergoes a change in pressure and temperature at constant volume by the exchange of heat energy with surroundings is called Isochoric process".
Isothermal Process:
A process in which a system undergoes physical changes in such a way that the temperature remains constant by the exchange of heat energy with surroundings is known as an Isothermal process.
For an isothermal process, the cylinder and the gas should be good conductors of heat. The process should be quasi-static process. P-V graph of an isothermal process is a rectangular hyperbola.
The ideal gas equation for 'n' moles of gas is PV = nRT. At constant temperature
.
This is the equation of isothermal process. In this process, change in internal energy
dU = 0 ⇒ U = constant. Heat energy exchanged is completely converted into work form
(i.e., dQ = dW)
e.g.: When water is heated at normal atmospheric pressure, it boils at 100 °C and converted into steam.
Workdone in Isothermal Process:
Consider one mole of an ideal gas enclosed in a conducting cylinder fitted with a frictionless Piston. When the cylindar is placed on a source of heat, gas expands. If the gas expands at constant temperature (T), from initial values (P1V1) to final values (P2V2). At any intermediate stage with Pressure P and Volume changes from V to V + ∆V then ∆W = P ∆V
The above expression gives the workdone one, 1 mole of an ideal gas in isothermal process.
Adiabatic Process: In an adiabatic process, the system is insulated from surroundings and heat absorbed or released is zero (i.e., dQ = 0 (or) ∆Q = 0). In adiabatic compression, the temperature increases as work is done on the gas. In adiabatic expansion, the temperature decreases as the work is done by the gas.
So, adiabatic change is associated with change in internal energy dU and workdone dW.
From first law of thermodynamics
dU + dW = 0 (... dQ = 0)
For an adiabatic process of an ideal gas 
= constant, where 
. If an ideal gas change its state adiabatically from (P1V1) to (P2V2), then 
In adiabatic process, gas is taken in a thermally insulated cylinder fitted with a frictionless Piston.
Workdone in an adiabatic process:
Consider one mole of an ideal gas contained in a thermally insulated cylinder having a frictionless Piston. Let P, V be the pressure, volume of the gas at any instant, dx be the displacement of Piston at constant pressure. Then workdone
dW = P.(dV) .........................(1)
If the system changes from (P1V1) to (P2V2), temperature changes from T1 to T2, then workdone during the change is
If work is done by the gas in an adiabatic process (W > 0), (T2< T1). If work is done on the gas (W < 0), we get (T2 > T1). Then temperature of gas rises.
Isothermal and Adiabatic P-V graphs:
P−V graph of given mass of a gas during isothermal change is called an "isotherm". P−V graph of given mass of a gas during adiabatic change is called an "adiabatic". For isothermal process, PV = constant.
On differentiating, P dV + V dP = 0
Cyclic Process: A process in which a system undergoes series of changes and which returns the system back to its initial state is known as cyclic process. In case of cyclic process, internal energy (U) is constant.
dU = Uf − Ui = 0
But dU ∝ dT ⇒ dT = 0 ⇒ T = Constant
i.e., temperature of the system remains constant. From first law of thermodynamics,
dQ = dU + dW
dQ = dW (... dU = 0)
i.e., heat supplied is equal to workdone by the system. In cyclic process, P−V graph is a closed curve and area enclosed by the closed path represents the workdone. If the cyclic process is clockwise, workdone is positive and if the cycle is anti clockwise, workdone is negative.
Polytropic Process: For the processes like isobaric, isochoric, isothermal and adiabatic, the relation between pressure and volume can be described by an equation 
, where n is positive number. This equation is called "Polytropic law".
Specific heat of a gas during polytropic process is 
Problem: P−V diagram of a diatomic gas is a straight line passing through origin. What is molar heat capacity of the gas in the process.
Sol: From the given graph, P ∝ V
Heat engines: Heat engine is a device used to convert heat energy into useful mechanical work. It consists of three main parts. They are 1) Source 2) Sink 3) Working substance.
1) Source: It is a hot reservoir, which is at higher temperature (T1) and it can extract any quantity of heat (Q1) without any change in its temperature. So its thermal capacity is infinite.
2) Sink: It is a cold reservoir, which is at lower temperature (T2) and it can take any amont of heat (Q2) which is rejected by working substance without any change in its temperature. Its thermal capacity is infinite.
3) Working substance: In steam engines, the working substance is steam at high pressure, in diesel engine the working substance is diesel. It absorbs certain amount of heat from source, converts a part of it into work and rejects the remaining part to sink.
The working substance is taken through a cyclic process. At the end of each cycle, the working substance brought to its original condition. So, internal energy of working substance remains constant.
The efficiency of engine is the ratio between workdone (W) by the engine and the amount of heat absorbed (Q1) by the engine.
For Q2 = 0, η = 1 i.e., the engine will have 100% efficiency in converting heat into work, such engine is called an ideal engine. Such an engine with η = 1 is never possible.
* Efficiency of an ideal heat engine is independent of nature of working substance. It depends only upon the temperatures of source and sink.
* If T1 = T2, η = 0
* Practically η is less than one.
Refrigerator and Heat pumps:
A refrigerator is a heat pump which is the reverse of a heat engine.
Source: The substance kept inside the refrigerator.
Sink: The surroundings of the refrigerator (room).
Working substance: Refrigerent is the working substance. Working substance extracts heat Q2 from the sink, which is at temperature T2.
Certain amount of work (W) is done on the working substance by the compressor of the refrigerator. The quantity of heat Q1 is rejected to the hot body at temperature T1. In a refrigerator the working substance goes through the following steps.
(i) Sudden expansion of the gas from high to low pressure which cools it and converts into a vapour liquid mixture.
(ii) Absorption of heat by the cold fluid from the region to be cooled converting it into vapour.
(iii) Heating up of vapour due to external workdone on the system.
(iv) Release of heat by the vapour to the surroundings, bringing it to the initial state and completing the cycle.
Coefficient of performance defined as the ratio of quantity of heat extracted per cycle from the contents of the refregirator to the mechanical work done by the external agency
* β varies from 2 to 6.
* The coefficient of performance may be much higher than 100% and cannot be infinite.
(i.e., W ≠ 0)
Second Law of Thermodynamics:
It is a fundamental law of nature which explains that heat can flow from hot body to cold body by itself. This law gives a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of refrigerator. It says that efficiency of a heat engine can never be unity. There are several statements of this law, out of them two are most significant. They are
(i) Kelvin - Plank statement: "It is impossible by any thermodynamic process which results solely, in the removal of heat from a heat reservoir and convert it entirely into work".
It means it is not possible to get a continuous supply of work from a body by cooling it to a temperature lower than that of surroundings.
(ii) Claussius statement: "It is impossible for a self acting machine unaided by any external agency to transfer heat from a body at lower temperature to a body at higher temperature".
It means that by itself heat cannot flow from a body at lower temperature to a body at a higher temperature.
Reversible and Irreversible Processes:
A process that can be retraced back in the opposite direction in such a way that the system passes through the same states as in the direct process and finally the system and surroundings return to their original states, with no other change any where else is called a reversible process.
e.g.: The process of conversion of ice into water. Ice converts into water at 0°C by supplying heat. In the reverse process of conversion of water into ice at 0°C, heat is taken out of water at the same rate under normal pressure. (i.e., very slowly)
"A process that cannot be retraced back in opposite direction is called an irreversible process". In this process, the system does not pass through the same intermediate states as in the direct process, even if the same initial state is reached.
e.g.: In the process of rubbing of our hands, heat is produced, increasing the temperature of the hands. The mechanical work done against friction is converted into heat. In the reverse process, we cannot make the hands to move in the same way on heating them.
Carnot Engine:
Carnot engine is a reversible heat engine operating between two temperatures. The working substance in the Carnot's engine taken through a reversible cycle consisting of following steps.
(1) The cylinder containing an ideal gas is placed on the source and gas is allowed to expand slowly at constant temperature T1, absorbing heat Q1. This is ''Isothermal change" is represented by the curve AB in the indicator diagram.
(2) The cylinder is then placed on the non conducting stand and gas is allowed to expand adiabatically till the temperature falls from T1 to T2. This is adiabatic expansion, is represented by BC in the indicator diagram.
(3) The cylinder is placed on the sink and the gas is compressed at constant temperature T2 and releases the heat Q2 to the sink, is represented by the curve CD in the diagram. This is isothermal compression.
(4) Finally the cylinder is placed on the non-conducting stand and compression is continued, so that gas return to its initial values (stage), is represented by the Curve DA in the diagram.
During isothermal expansion workdone by the gas on the environment is
In the adiabatic expansion of the gas, workdone by the gas is
During isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2), heat released Q2 by the gas to the reservoir at temperature T2. Then workdone on the gas by the environment is
In adiabatic compression of the gas from (P4, V4, T2) to (P1, V1, T1) workdone on the gas is
Total workdone in Carnot's cycle is
W = area enclosed by the curve ABCD
W = W1 + W2 + (-W3) + (-W4)
If an engine consists of a no. of sources and sinks, then
Problem: The temperature entropy diagram of a reversible engine cycle is given in figure. What is its efficiency?
Sol:
