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 T_{A} = T_{C} and T_{B} = T_{C}. This implies that T_{A }= T_{B} 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 (C_{m}) of a system is defined as the amount of heat required to raise the temperature of 1 mole of substance by 1°C.
$${\mathrm{C}}_{\mathrm{m}}\mathrm{}\mathrm{}=\frac{\mathrm{C}}{\mathrm{n}}=\frac{\mathrm{q}}{\mathrm{\Delta}\mathrm{T}}\times \frac{\mathrm{M}}{\mathrm{m}}$$
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.
$$\mathrm{c}\mathrm{}=\frac{\mathrm{C}}{\mathrm{m}}=\mathrm{}\frac{\mathrm{q}}{\mathrm{m}\mathrm{\Delta}\mathrm{T}}\mathrm{}\mathrm{}$$
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 = q_{p} – pΔV at constant pressure, where q_{p} is heat absorbed by the system and –pΔV represent expansion work done by the system.
U_{2 }– U_{1} = q_{p} – p (V_{2} – V_{1})
⇒ q_{p} = (U_{2} + pV_{2}) – (U_{1} + pV_{1})
Or ΔH = q_{p }= H_{2} – H_{1}
- 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 + Δn_{g}RT for constant T.
Here, Δn_{g} = change in the number of gas moles.
The Relationship between C_{p} and C_{V} for an ideal gas
At constant volume, the heat capacity is denoted by C_{V }
At constant pressure, the heat capacity is denoted by C_{p}.
At constant volume q_{V} = C_{V}ΔT = ΔU [since w = 0]
At constant pressure q_{p} = C_{p}ΔT = ΔH
Now
ΔH = ΔU + Δ(pV )
= ΔU + Δ(RT )
= ΔU + RΔT
On putting the values of ΔH = C_{p}ΔT and ΔU = C_{V}ΔT
C_{p}ΔT = C_{V}ΔT + RΔT
Or C_{p} = C_{V} + R
⇒ C_{p} - C_{V} = R
For monoatomic gases,
C_{V} = $\frac{3}{2}$ R
and C_{P} = $\frac{5}{2}$ R
The ratio C_{P}/C_{V} 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 (V_{f} – V_{i})
For an isothermal reversible process,
Suppose an ideal gas goes isothermally (at temperature T) from its initial state (P_{1}, V_{1}) to the final state (P_{2}, V_{2}). 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,
$$\mathrm{W}=\mathrm{}{\int}_{\mathrm{V}1}^{\mathrm{V}2}\mathrm{P}\mathrm{}\mathrm{d}\mathrm{V}=\mathrm{}{\int}_{\mathrm{V}1}^{\mathrm{V}2}\frac{\mathrm{n}\mathrm{}\mathrm{R}\mathrm{T}\mathrm{}\mathrm{d}\mathrm{V}}{\mathrm{V}}$$
$$=\mathrm{nRT}\left(\mathrm{ln}{\mathrm{V}}_{2}-\mathrm{ln}{\mathrm{V}}_{1}\right)$$
$$=\mathrm{}\mathrm{nRT}\mathrm{ln}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}$$
$$=\mathrm{2.303\; nRT}\mathrm{log}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}\mathrm{}$$
$$=\mathrm{2.303\; nRT}\mathrm{log}\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}$$
Examples of isothermal change - melting and boiling
Adiabatic Process
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.
For an adiabatic process
$$\mathrm{P}{\mathrm{V}}^{\mathrm{\gamma}}=\mathrm{K\; or\; P}=\frac{\mathrm{K}}{{\mathrm{V}}^{\mathrm{\gamma}}}$$
As also TV ^{1-γ} = constant = K’
Where, γ = $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$
Hence the work done will be
$$\mathrm{W}\mathrm{}=\mathrm{}{\int}_{\mathrm{V}1}^{\mathrm{V}2}\mathrm{P}\mathrm{}\mathrm{d}\mathrm{V}=\mathrm{}\mathrm{K}{\int}_{\mathrm{V}1}^{\mathrm{V}2}\frac{\mathrm{}\mathrm{d}\mathrm{V}}{{\mathrm{V}}^{\mathrm{\gamma}}}$$
$$=\frac{\mathrm{K}}{1-\mathrm{\gamma}}\mathrm{}\left[{{\mathrm{V}}_{2}}^{1-\mathrm{\gamma}}-\mathrm{}{{\mathrm{V}}_{1}}^{1-\mathrm{\gamma}}\right],\mathrm{}\mathrm{}$$
$$=\frac{1}{1-\mathrm{\gamma}}\mathrm{}\left[{\mathrm{P}}_{2}{\mathrm{V}}_{2}-\mathrm{}{\mathrm{P}}_{1}{\mathrm{V}}_{1}\right]$$
Isobaric Process
A process taking place in a thermodynamic system at constant pressure is called an isobaric process.
W = P (V_{2} – V_{1}) = n R (T_{2} – T_{1})
Isochoric Process
A process taking place in a thermodynamic system at constant volume is called an isochoric process.
Process equation is
$$\frac{\mathrm{p}}{\mathrm{T}}=\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$$
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 C_{V }ΔT [molar heat capacity for isochoric process is C_{V}]
Cyclic Process
When a thermodynamic system returns to initial state after passing through several states, then it is called cyclic process.
Isothermal and Adiabatic Curves
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 Q_{1} from an external reservoir at some high temperature T_{1}.
(3) In some other processes of the cycle, the working substance releases a total amount of heat Q_{2} to an external reservoir at some lower temperature T_{2}.
(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
$$\mathrm{\eta}\mathrm{}=\frac{\mathrm{W}}{{\mathrm{Q}}_{1}}$$
where Q_{1} 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 = Q_{1} – Q_{2}
Or η = 1- $\frac{{\mathrm{Q}}_{2}}{{\mathrm{Q}}_{1}}$
If Q_{2} = 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 Q_{2} from the cold reservoir at temperature T_{2}, some external work W is done on it and heat Q_{1} is released to the hot reservoir at temperature T_{1}
Performance of a refrigerator is given by
$$\mathrm{\alpha}\mathrm{}=\frac{{\mathrm{Q}}_{2}}{\mathrm{W}}$$
Since W = Q_{1} – Q_{2},
$$\mathrm{\alpha}\mathrm{}=\frac{{\mathrm{Q}}_{1}}{{\mathrm{Q}}_{1}\mathrm{}\u2013\mathrm{}{\mathrm{Q}}_{2}}$$
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 (P_{1}, V_{1}, T_{1}) to (P_{2}, V_{2}, T_{1}).
The heat absorbed by the gas from the reservoir at temperature T_{1},
$${\mathrm{Q}}_{1}\mathrm{}=\mathrm{}{\mathrm{W}}_{1\to 2}\mathrm{}=\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{1}\mathit{ln}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}$$
(b) Step 2 → 3 – Adiabatic expansion of the gas from (P_{2}, V_{2}, T_{1}) to (P_{3}, V_{3}, T_{2})
Work done by the gas,
$${\mathrm{W}}_{2\to 3}\mathrm{}=-\mathrm{}\frac{\mathrm{n}\mathrm{}\mathrm{R}({\mathrm{T}}_{1}\mathrm{}\u2013\mathrm{}{\mathrm{T}}_{2})}{1\mathrm{}\u2013\mathrm{}\mathrm{\gamma}}$$
(c) Step 3 → 4 – Isothermal compression of the gas from (P_{3}, V_{3}, T_{2}) to (P_{4}, V_{4}, T_{2}).
Heat released (Q_{2}) by the gas to the reservoir at temperature T_{2} is given by
$${\mathrm{Q}}_{2}\mathrm{}=\mathrm{}{\mathrm{W}}_{3\to 4}\mathrm{}=\mathrm{}-\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{2}\mathit{ln}\frac{{\mathrm{V}}_{4}}{{\mathrm{V}}_{3}}=\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{2}\mathit{ln}\frac{{\mathrm{V}}_{3}}{{\mathrm{V}}_{4}}$$
(d) Step 4 → 1 - Adiabatic compression of the gas from (P_{4}, V_{4}, T_{2}) to (P_{1}, V_{1}, T_{1}).
Work done on the gas,
$${\mathrm{W}}_{4\to 1}\mathrm{}=\mathrm{}\frac{\mathrm{n}\mathrm{}\mathrm{R}({\mathrm{T}}_{1}\mathrm{}\u2013\mathrm{}{\mathrm{T}}_{2})}{1\mathrm{}-\mathrm{}\mathrm{\gamma}}$$
Total work done
W = W_{1→2 } + _{ }W_{2→3} + W_{3→4} + _{ }W_{4→1}
$$=\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{1}\mathit{ln}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}-\frac{\mathrm{n}\mathrm{}\mathrm{R}({\mathrm{T}}_{1}\mathrm{}\u2013\mathrm{}{\mathrm{T}}_{2})}{1\mathrm{}-\mathrm{}\mathrm{\gamma}}-\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{2}\mathit{ln}\frac{{\mathrm{V}}_{4}}{{\mathrm{V}}_{3}}+\frac{\mathrm{n}\mathrm{}\mathrm{R}({\mathrm{T}}_{1}\mathrm{}\u2013\mathrm{}{\mathrm{T}}_{2})}{1\mathrm{}-\mathrm{}\mathrm{\gamma}}\mathrm{}$$
$$=\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{1}\mathit{ln}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}-\mathrm{}\mathrm{n}\mathrm{}\mathrm{R}{\mathrm{T}}_{2}\mathit{ln}\frac{{\mathrm{V}}_{4}}{{\mathrm{V}}_{3}}$$
So the efficiency of Carnot’ engine
$$\mathrm{\eta}\mathrm{}=\mathrm{}\frac{\mathrm{W}}{{\mathrm{Q}}_{1}}=\mathrm{}1\mathrm{}\u2013\mathrm{}\frac{{\mathrm{Q}}_{2}}{{\mathrm{Q}}_{1}}=\mathrm{}1\mathrm{}\u2013\mathrm{}\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}\mathrm{}\frac{\mathrm{ln}\left({\mathrm{V}}_{3}/{\mathrm{V}}_{4}\right)}{\mathrm{l}\mathrm{n}\left({\mathrm{V}}_{2}/{\mathrm{V}}_{1}\right)}$$
Now since step 2 → 3 and 4 →1 are both adiabatic, we have
T_{1 }V_{2 }^{γ-1} = T_{2 }V_{3 }^{γ-1 }
$$\Rightarrow {\left(\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{3}}\right)}^{\mathrm{\gamma}-1}=\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}$$
And
T_{2 }V_{4 }^{γ-1} = T_{1 }V_{1 }^{γ-1 }
$$\Rightarrow {\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{4}}\right)}^{\mathrm{\gamma}-1}=\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}$$
Therefore,
$$\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{3}}=\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{4}}$$
$$\Rightarrow \mathrm{}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}=\frac{{\mathrm{V}}_{3}}{{\mathrm{V}}_{4}}$$
Hence
$$\mathrm{\eta}\mathrm{}=\mathrm{}1\mathrm{}\u2013\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}$$
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 Q_{2} from the cold reservoir at T_{2}, doing work W on the system, and transferring heat Q_{1} to the hot reservoir. This is a reversible refrigerator.
Carnot’s Theorem
(a) Working between two given temperatures T_{1} and T_{2} 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 E_{A} and E_{B} operating between the given source at temperature T_{1} and the given sink at temperature T_{2}.
Let E_{A} be any irreversible heat engine and E_{B} be any reversible heat engine. We have to prove that efficiency of heat engine E_{B} is more than that of heat engine E_{A}.
Suppose both the heat engines receive same quantity of heat Q from the source at temperature T_{1}. Let W _{A} and W_{B} be the work output from the engines and their corresponding heat rejections be (Q – W_{A}) and (Q – W_{B}) respectively.
Assume that the efficiency of the irreversible engine is more than the reversible engine i.e. η_{A} > η_{B}.
Hence,
$$\frac{{\mathrm{W}}_{\mathrm{A}}}{\mathrm{Q}}>\frac{{\mathrm{W}}_{\mathrm{B}}}{\mathrm{Q}}\mathrm{}\mathrm{}$$
i.e. W_{A} > W_{B}
Now let us couple both the engines and E_{B} is reversed which will act as a heat pump.
It receives (Q – W_{B}) from sink and W_{A} from irreversible engine E_{A} and pumps heat Q to the source at temperature T_{1}.
The net result is that heat W_{A} – W_{B} 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.