CBSE NOTES CLASS 11 CHAPTER 12

THERMODYNAMICS

Thermodynamics

The branch of physics that deals with the study of transformation of heat energy into other forms of energy and vice versa is called thermodynamics.

A thermodynamic system is said to be in thermal equilibrium when macroscopic variables (like pressure, volume, temperature, mass, composition etc) that characterize the system do not change with time.

Thermodynamic System

An assembly of an extremely large number of particles whose state can be expressed in terms of pressure, volume and temperature, is called thermodynamic system.

Thermodynamic system is classified into the following three systems

• Open System - Exchange of both energy and matter with surrounding.

• Closed System - Exchange of energy, but no exchange of matter with surroundings.

• Isolated System - Exchange of neither energy nor matter with the surrounding.

The boundary between the system and surrounding is called a wall. Wall can be,

• Adiabatic Wall – an insulating wall (can be movable) that does not allow flow of energy (heat) from system to surrounding and vice versa or from one system to another.

• Diathermic Wall – a conducting wall that allows energy flow (heat) from system to surroundings and vice versa or from one system to another.

Thermodynamic Parameters or Coordinates or Variables

The state of thermodynamic system can be described by specifying pressure, volume, temperature, internal energy and number of moles, etc. These are called thermodynamic parameters or coordinates or variables.

Zeroth Law of Thermodynamics

‘Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’.

If systems A and B are separately in equilibrium with system C, then TA = TC and TB = TC. This implies that TA = TB i.e. the systems A and B are also in thermal equilibrium with each other.

Internal Energy (U)

The total energy possessed by any system due to molecular motion and molecular configuration, is called its internal energy and is represented by U. It is the total of all types of energies.

Internal energy of a thermodynamic system depends on temperature. It is a state function.

Work Done

 Isobaric process Reversible isothermal process

Work done by a thermodynamic system is given by

W = p × ΔV

Where p = pressure and ΔV = change in volume.

Work done by a thermodynamic system is equal to the area enclosed between the p-V curve and the volume axis.

Work done by a thermodynamic system depends not only upon the initial and final states of the system but also depend upon the path followed in the process.

Work done by the Thermodynamic System is taken as

• Positive if volume increases.

• Negative if volume decreases.

First Law of Thermodynamics

If

ΔQ = Heat supplied to the system by the surroundings

ΔW = Work done by the system on the surroundings

ΔU = Change in internal energy of the system

then

ΔQ = ΔU + ΔW

First law of thermodynamics is a re-statement of the principle of conservation of energy.

Types of Thermodynamic Properties

Intensive Properties

Properties of the system which depend only on the nature of matter but not on the quantity of matter are called Intensive properties, e.g., pressure, temperature, specific heat, etc

Extensive Properties

Properties of the system which are dependent on the quantity of matter are called extensive properties, e.g., internal energy, volume, enthalpy, etc.

Heat Capacity of a System

Heat capacity (C) of a system is defined as the amount of heat required to raise the temperature of a system by 1°C.

q = C × ΔT or C =q/ΔT

• Heat capacity is an extensive property.

Molar Heat Capacity

Molar heat capacity (Cm) of a system is defined as the amount of heat required to raise the temperature of 1 mole of substance by 1°C.

where n = number of moles, m = given mass, M = molar mass

• Molar heat capacity is an intensive property.

Specific Heat Capacity

Specific heat capacity (c) of a system is defined as the amount of heat required to raise the temperature of 1gram of substance by 1°C.

where

m = mass of substance,

ΔT = change in temperature.

Enthalpy (H)

It is the sum of internal energy and pV-energy of the system. It is a state function and extensive property. Mathematically,

H = U + pV

Like U, absolute value of H also cannot be known

ΔH is determined experimentally.

Now, ΔU = qppΔV at constant pressure, where qp is heat absorbed by the system and –pΔV represent expansion work done by the system.

U2 – U1 = qp – p (V2 – V1)

qp = (U2 + pV2) – (U1 + pV1)

Or ΔH = qp = H2 – H1

• For exothermic reaction (the reaction in which heat is evolved), ΔH = -ve

• For endothermic reaction (the reaction in which heat is absorbed), ΔH = +ve.

Relationship between ΔH and ΔU

ΔH = ΔU + Δ(pV)

or ΔH = ΔU + ΔngRT for constant T.

Here, Δng = change in the number of gas moles.

The Relationship between Cp and CV for an ideal gas

At constant volume, the heat capacity is denoted by CV

At constant pressure, the heat capacity is denoted by Cp.

At constant volume qV = CVΔT = ΔU [since w = 0]

At constant pressure qp = CpΔT = ΔH

Now

ΔH = ΔU + Δ(pV )

= ΔU + Δ(RT )

= ΔU + RΔT

On putting the values of ΔH = CpΔT and ΔU = CVΔT

CpΔT = CVΔT + RΔT

Or Cp = CV + R

⇒ Cp - CV = R

For monoatomic gases,

CV = $\frac{3}{2}$ R

and CP = $\frac{5}{2}$ R

The ratio CP/CV is denoted by γ

Thermodynamic Processes

A thermodynamic process is said to take place when some changes’ occur in the state of a thermodynamic system i.e., the thermodynamic parameters of the system change with time.

Isothermal Process

A process taking place in a thermodynamic system at constant temperature is called an isothermal process.

Isothermal processes are very slow processes.

These processes follow Boyle’s law and ideal gas equation, according to which

pV = nRT

Work done by the system against a constant pressure P is

ΔW = P ΔV

Where ΔV is the change in volume of the gas.

ΔQ = ΔU + P ΔV

In isothermal process, change in internal energy is zero (ΔU = 0), therefore,

ΔQ = ΔW = P ΔV

For isothermal irreversible process,

ΔW = P (Vf – Vi)

For an isothermal reversible process,

Suppose an ideal gas goes isothermally (at temperature T) from its initial state (P1, V1) to the final state (P2, V2). At any intermediate stage with pressure P and volume change from V to V + ΔV (ΔV small)

ΔW = P ΔV

Taking (ΔV → 0) and summing the quantity ΔW over the entire process,

$=\mathrm{nRT}\left(\mathrm{ln}{\mathrm{V}}_{2}-\mathrm{ln}{\mathrm{V}}_{1}\right)$

$=\mathrm{2.303 nRT}\mathrm{log}\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}$

Examples of isothermal change - melting and boiling

A process, in which there is no exchange of heat between the system and its surroundings, is called adiabatic.

When a system expands adiabatically, work done is positive and hence internal energy decreases, i.e., the system cools down and vice-versa.

Examples - sudden compression or expansion of a gas in a container with perfectly non-conducting wall, sudden bursting of the tube of a bicycle and propagation of sound waves in air and other gases.

$\mathrm{P}{\mathrm{V}}^{\mathrm{\gamma }}=\mathrm{K or P}=\frac{\mathrm{K}}{{\mathrm{V}}^{\mathrm{\gamma }}}$

As also TV 1-γ = constant = K’

Where, γ = $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$

Hence the work done will be

Isobaric Process

A process taking place in a thermodynamic system at constant pressure is called an isobaric process.

W = P (V2V1) = n R (T2T1)

Isochoric Process

A process taking place in a thermodynamic system at constant volume is called an isochoric process.

Process equation is

p-V curve is a straight line parallel to pressure axis.

Volume is constant so work done is zero, i.e., ΔW = 0,

Therefore, ΔQ = ΔU

Or, ΔQ = n CV ΔT [molar heat capacity for isochoric process is CV]

Cyclic Process

When a thermodynamic system returns to initial state after passing through several states, then it is called cyclic process.

The graph drawn between the pressure p and the volume V of a given mass of a gas for an isothermal process is called isothermal curve and for an adiabatic process it is called adiabatic curve.

The slope of the adiabatic curve (p vs ln V) = γ × the slope of the isothermal curve.

Quasi-Static Process

A quasi-static process is a thermodynamic process that happens slowly enough for the system to remain in internal equilibrium.

Any reversible process is a quasi-static process, but the reverse is not necessarily true.

An example of a quasi-static process that is not reversible is a compression against a system with a piston subject to friction — although the system is always in thermal equilibrium, the friction ensures the generation of dissipative heat, which directly goes against the definition of reversibility.

Heat Engines

Heat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work.

(1) It consists of a working substance – the system. For example, a mixture of fuel vapour and air in a gasoline or diesel engine or steam in a steam engine are the working substances.

(2) The working substance goes through a cycle consisting of several processes. In some of these processes, it absorbs a total amount of heat Q1 from an external reservoir at some high temperature T1.

(3) In some other processes of the cycle, the working substance releases a total amount of heat Q2 to an external reservoir at some lower temperature T2.

(4) The work done (W) by the system in a cycle is transferred to the environment via some arrangement (e.g. the working substance may be in a cylinder with a moving piston that transfers mechanical energy to the wheels of a vehicle via a shaft).

(5) The cycle is repeated again and again to get useful work for some purpose.

The efficiency (η) of a heat engine is defined by

where Q1 is the heat input i.e., the heat absorbed by the system in one complete cycle.

According to the First Law of Thermodynamics, over one complete cycle,

W = Q1Q2

Or η = 1- $\frac{{\mathrm{Q}}_{2}}{{\mathrm{Q}}_{1}}$

If Q2 = 0, then η = 1.

Practically an ideal engine with η = 1 is never possible.

Types of Heat Engines

External Combustion Engine

In this engine fuel is burnt in a chamber outside the main body of the engine. e.g., steam engine. In practical life thermal efficiency of a steam engine varies from 12% to 16%.

Internal Combustion Engine

In this engine, fuel is burnt inside the main body of the engine. e.g., petrol and diesel engines. In practical life thermal efficiency of a petrol engine is 26% and a diesel engine is 40%.

Refrigerator or Heat Pump

A refrigerator is the reverse of a heat engine. Here the working substance extracts heat Q2 from the cold reservoir at temperature T2, some external work W is done on it and heat Q1 is released to the hot reservoir at temperature T1

Performance of a refrigerator is given by

Since W = Q1 – Q2,

Hence α can be greater than 1.

A refrigerator cannot work without some external work being done on it. Therefore, α cannot be infinite.

Second Law of Thermodynamics

The second law of thermodynamics gives a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of a refrigerator.

It says that efficiency of a heat engine can never be unity (or 100%). This implies that heat released to the cold reservoir can never be made zero.

Kelvin-Planck statement

No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.

Clausius statement

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object without any external work.

The two statements are completely equivalent

Carnot’s Cycle

A reversible heat engine operating between two temperatures is called a Carnot engine. Carnot’s engine uses ideal gas as the working substance.

A Carnot’s cycle (p-V diagram of working of Carnot’s engine) contains the following four processes

(a) Step 1 → 2 - Isothermal expansion of the gas from (P1, V1, T1) to (P2, V2, T1).

The heat absorbed by the gas from the reservoir at temperature T1,

(b) Step 2 → 3 – Adiabatic expansion of the gas from (P2, V2, T1) to (P3, V3, T2)

Work done by the gas,

(c) Step 3 → 4 – 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 is given by

(d) Step 4 → 1 - Adiabatic compression of the gas from (P4, V4, T2) to (P1, V1, T1).

Work done on the gas,

Total work done

W = W1→2 + W2→3 + W3→4 + W4→1

So the efficiency of Carnot’ engine

Now since step 2 → 3 and 4 →1 are both adiabatic, we have

T1 V2 γ-1 = T2 V3 γ-1

$⇒{\left(\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{3}}\right)}^{\mathrm{\gamma }-1}=\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}$

And

T2 V4 γ-1 = T1 V1 γ-1

$⇒{\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{4}}\right)}^{\mathrm{\gamma }-1}=\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}$

Therefore,

$\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{3}}=\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{4}}$

Hence

Carnot engine is the only reversible engine possible that works between two reservoirs at different temperatures.

Each step of the Carnot cycle can be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on the system, and transferring heat Q1 to the hot reservoir. This is a reversible refrigerator.

Carnot’s Theorem

(a) Working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, no engine can have efficiency more than that of the Carnot engine and

(b) The efficiency of the Carnot engine is independent of the nature of the working substance.

Proof of Carnot’s Theorem

Suppose there are two engines EA and EB operating between the given source at temperature T1 and the given sink at temperature T2.

Let EA be any irreversible heat engine and EB be any reversible heat engine. We have to prove that efficiency of heat engine EB is more than that of heat engine EA.

Suppose both the heat engines receive same quantity of heat Q from the source at temperature T1. Let W A and WB be the work output from the engines and their corresponding heat rejections be (Q – WA) and (Q – WB) respectively.

Assume that the efficiency of the irreversible engine is more than the reversible engine i.e. ηA > ηB.

Hence,

i.e. WA > WB

Now let us couple both the engines and EB is reversed which will act as a heat pump.

It receives (Q – WB) from sink and WA from irreversible engine EA and pumps heat Q to the source at temperature T1.

The net result is that heat WA – WB is taken from sink and equal amount of work is produced, without doing any change anywhere else.

This violates second law of thermodynamics.

Hence the assumption we made that irreversible engine is having higher efficiency than the reversible engine is wrong.

Hence it is concluded that reversible engine working between same temperature limits is more efficient than irreversible engine thereby proving Carnot’s theorem.

A similar argument can be constructed to show that a reversible engine with one particular substance cannot be more efficient than the one using another substance.