# Introduction to thermodynamics

Thermodynamics is the branch of physics which deals with the heat and it’s relation with work and energy.First of all we are going to introduce some important terms that are the base of the thermodynamics.

# Specific Heat :

Specific heat of a substance is the quantity of heat required to raise the temperature of unit mass of substance through 1°C. Its unit is cal/ g° C or kilo cal/g°C.

Specific heat of gas:

In the case of solids and liquids the change in volume during the change of temperature is negligible; due to which there is only one specific heat for solids and liquids; but gases suffer appreciable changes of pressure and volume during change of temperature. There are in general two methods of heating the gas

(i) at constant volume and

(ii) at constant pressure.

Accordingly there are two specific heats of a gas

1. Specific heat at constant volume cv:

The specific heat at constant volume is the quantity of heat in calorie required to raise the temperature of 1 g of gas through 1°C,at constant volume.

2. Specific heat at constant pressure cp :

The specific heat at constant pressure is the quantity of heat in calorie required to raise the temperature of 1 g of gas through 1°C at constant pressure.

## Molar heat capacities :

The molar heat capacity is the product of molecular weight and specific heat i.e.
Molar heat capacity C = (Molecular weight M) (Specific heat c)

Molar heat of constant volume, $C_v = Mc_v$

Molar heat of constant pressure, $C_p = Mc_p$

## Pressure exerted by an Ideal Gas:

The pressure P exerted by a perfect gas on the walls of the containing vessel is given by

$P = \dfrac{1}{3} \dfrac{mnc^2}{V} = \dfrac{1}{3} \dfrac{M}{V} c^2 = \dfrac{1}{3} \rho c^2 = \dfrac{2}{3} \dfrac{E}{V}$     ……..(1)

where ,

m = mass of each molecule

n = number of molecules the enclosure

c = root mean square of molecules

$\rho$ = density of gas

M = mass of gas enclosed in volume V of container.

The root mean square  speed is given by

$c = v_{rms} = \sqrt{{v^2}_{mean}} \\ = \sqrt{\dfrac{{v_1}^2 + {v_2}^2 + {v_3}^2 + \dots + {v_n}^2}{N}} \\ = \sqrt{\dfrac{3RT}{m}}$        ……(2)
Kinetic Interpretation of Temperature :

Average kinetic energy per molecule

$E = \dfrac{3}{4} . kT \\ \text{as} \, E \propto T$

Obviously average translational kinetic energy per molecule is directly proportional to the absolute temperature. The temperature at which kinetic energy of gas molecules becomes zero, is called the absolute zero (0K)

### Degrees of freedom and Law of equipartition of energy :

The degrees of freedom of a dynamic system means the total number of independent coordinates required to specify the state of the system.

The law of equipartition of energy states that in equilibrium, the total kinetic energy of a dynamical system is equally distributed among its various degrees of freedom and the kinetic energy associated with each degree of freedom is $\dfrac{1}{2}$.kT, where T is absolute temperature.

Examples :

Monoatomic gas:

A monoatomic gas molecule has three degrees of freedom of translatory motion only, the translation along X, Y and Z axes.

Kinetic energy per degree of freedom

= $\dfrac{1}{2}. kT$

Kinetic energy per molecule

= $3 \times \dfrac{1}{2}. kT = \dfrac{3}{2} .kT$

Kinetic energy per mole,

= $N \dfrac{3}{2}. kT = \dfrac{3}{2} .RT$

Molar specific heat at constant volume

$C_v = \dfrac{\delta U}{\delta T} = \dfrac{3}{2}. R$

Molar specific heat at constant pressure

$C_p = C_v + R = \dfrac{5}{2}.R \\ \therefore \gamma = \dfrac{C_p}{C_v} = \dfrac{\dfrac{5}{2}R}{\dfrac{3}{2}R} = \dfrac{5}{3} = 1.67$

Diatomic gas:

Each molecule of a diatomic gas has two atoms. Such a molecule has three degrees of freedom of translation and two degrees of freedom of rotation. Thus the total number of degrees of freedom of  diatomic molecule =5.

Kinetic energy per molecule

= $5 \times \dfrac{1}{2}. kT = \dfrac{5}{2} .kT$

Kinetic energy per mole

= $N \times \dfrac{5}{2}. kT = \dfrac{5}{2} .RT$

Molar specific heat at constant volume,

$C_v = \dfrac{5}{2}.R$

Molar specific heat at constant pressure,

$C_p = \dfrac{7}{2}.R \\ \gamma = \dfrac{C_p}{C_v} = \dfrac{7}{5} = 1.4$
Triatomic or polyatomic gas :

Each molecule of a triatomic or polyatomic gas consists of 6 degrees of freedom including 3 of translational motion and 3 of rotational motion.
Kinetic energy per molecule

$6 \times \dfrac{1}{2}. kT = 3kT$

Kinetic energy per mole

= N . 3kT = 3RT

Molar specific heat at constant volume,

$C_v = 3R$

Molar specific heat at constant pressure,

$C_p = 4R \\ \gamma = \dfrac{C_p}{C_v} = \dfrac{4}{3} = 1.33$

Heat absorbed by n-moles of a gas at constant volume

$Q_v = n C_v \triangle T$

Heat absorbed of n-moles of a gas at constant pressure is

$Q_p = n C_p \triangle T$

## Zeroth Law of Thermodynamics: Thermal equilibrium.

When two bodies A and B are in thermal equilibrium with a third body (C) separately, then they (A and B) are in thermal equilibrium mutually, i.e.
If                               $T_A = T_C$

And                         $T_B = T_C$

This implies  $T_A = T_B$

# First Law of Thermodynamics:

When heat energy $\triangle Q$ is supplied to a system, it is used up in two parts,

(i) In increasing the internal energy of the system ($\triangle U$)

(ii) In doing external work against the surroundings  ($\triangle W$)

i.e

$\triangle Q = \triangle U + \triangle W$  ……………………..(1)

In this equation all quantities  $\triangle Q , \triangle U \, \text{and} \, \triangle W$ must be in the same units e.g. calorie or kilocalorie or joule.

Concept of internal energy :

Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. The energy due to molecular motion is called internal kinetic energy (Uk) and that due to molecular configuration is called internal potential energy (Up) i. e.

$\triangle U = \triangle U_k + \triangle U_p$

If there are no intermolecular forces, then

$\triangle U_p = 0 \\ \therefore \, \, \, \triangle U = \triangle U_k = mc_v \triangle T$

where m is mass of system, cv specific heat at constant volume and $\triangle T$ is change in temperature.
If Cv is molar heat capacity Cv ( = Mcv , M being molecular weight), then for n-moles of ideal gas

$\triangle U = n C_v \triangle T = \dfrac{m}{M} C_v \triangle T$
It is obvious that internal energy in the absence of intermolecular forces is simply the function of temperature and state only, it is independent of path followed. So  $\triangle U = U_f - U_i$, where $U_i - \, \text{and} \, U_f$  are internal energies in initial and final states.

Concept of work:

The work done is the flow of energy without a difference of temperature.
If P is the pressure and $\delta$V is the infinitesimal change in volume, then work done

$dW = \delta V$……………….(2)

If system undergoes a sequence of small changes in volume at different pressure, then work done

$\triangle W = \sum P \delta V = \int_{v_i}^{V^f} P d V$ …………………..(2)
where Vi and Vf are initial and final volumes.

Also work done in expansion from volume Vi to Vf

$\triangle W = \sum PdV \\ = \, \text{Area} \, ABV_f V_i A$
= Area enclosed by P-V curve and volume axis.

Work done in closed cycle is the area enclosed by the thermodynamic cycle like iAfBi.

Remark :

Work done depends on initial and final states as well as on the path followed. Work done by the system is taken as positive and that on the system is taken as negative. In other words work is positive during

expansion and negative during compression.

Mechanical equivalent of heat:

According to Joule work may be converted into heat and vice versa. The ratio of work done to heat produced is always constant.

i.e.                            $\dfrac{W}{Q} = \, \text{constant}$

J or W = JQ …………………(1)

where J is called mechanical equivalent of heat.

Its value is 4.2 joule/cal or $4.2 \times 10^3$ joule/kilo-cal. It is not a physical quantity, but simply a conversion factor. It converts unit of work into heat; and vice versa. In equation (1) W must be in joule, irrespective of nature of energy or work and Q must be calorie or kilocalorie.

## Thermodynamic state :

The thermodynamic state of a system is represented by four variables: Pressure P, Volume V, Temperature T and entropy S ,out of these only two are independent .If these two variable are known, the other two variables may be found by the relation called the equation of the state. The four variables (P, V, T, S) are also called the thermodynamic coordinates.

Equation of state:

The equation relating the thermodynamic variables is called the equation of the state.

### Ideal gas:

It is assumed that (i) ideal gas molecules have no size and (ii) ideal gas molecules do not interact with one another.
The equation of state for an ideal (or perfect) gas is

PV =RT (for 1 g-mole)                          …………………….(a)

PV = nRT (for n-moles)                        ……………………..(b)

Where

$n = \dfrac{\text{Mass in grams}}{\text{Molecular weight}} = \dfrac{m}{M}$

And PV =nkT (for n-molecules) ….(c)

Where          R = gas constant for 1 mole =8.3 J/mole K

And  $k = \dfrac{\text{Gas constant} (R)}{\text{Avogadro number }(N)} \\ = \dfrac{8.3}{6.02 \times 10^23 \\ = 1.38 \times 10^{-23} J/K}$

Equation of state for real gas (or Vander Waals Equation) :

In actual practice, the (real) gas molecules have finite size and they exert intermolecular forces. The correction in ideal gas equation due to these factors are:

(i) Correction due to finite size of molecules:

If b is volume of gas molecules and V the volume of vessel/enclosure, then the volume of gas

V’ = V – b

(ii) Correction due to intermolecular forces :

The gas molecules exert intermolecular forces on one another. Due to these forces the rate of change of momentum of molecules striking per unit area of walls of vessels decreases and so the pressure of gas decreases. This reduction in pressure

$\propto ( density of gas ) ^2 \, \\ \, \propto \, \dfrac{m^2}{V^2}$ ( m = mass of gas , V = volume of gas )

$\dfrac{km^2}{V^2} = \dfrac{a}{V^2}$

Where k and a are constants

Thus $P' = P + \dfrac{a}{V^2}$

Where P’ is pressure of ideal gas and P is pressure of real gas.

Therefore ideal gas equation P’V’ = RT for real gas takes the form

$( P + \dfrac{a}{v^2} ) ( v- b ) = RT ( for 1 mole )$

This is called Vander Waal’s equation of state.

## Isothermals:

A curve drawn on pressure volume (P – V) diagram is called an isothermal.

## Andrew curves:

Andrew plotted isothermals of real gas on P – V diagram at different temperatures; the curves obtained for CO2 are shown in figure. At temperature T < TC there is a horizontal portion which represents the change of state from gas to liquid with increase of pressure at given temperature. The portion on left of horizontal portion represents liquid state, while that on right of horizontal portion represents gaseous state.

Andrew curves

At T = TC, called critical temperature, the horizontal portion disappears and takes the form of a point, while for T the liquid state does not exist. Thus critical temperature is the temperature below which a gas can be liquefied by increase of pressure alone. No liquefaction is possible at a temperature above the critical temperature.

Critical pressure :

The pressure at critical temperature is called critical pressure.
The critical temperatures and pressures are different for different gases. They are listed in table given below :

For Vander Waal’s gas

Critical temperature $T_C = \dfrac{8 a}{27 Rb}$

Critical pressure $P_C = \dfrac{a}{27 b^2}$

Critical volume $V_C = 3b$

### Vapour and Gas:

The gaseous state of a substance above critical temperature is called the gas, while below critical temperature it is called vapour.

## Pressure-Temperature Phase Diagram:

It is well known that a substance can exist in solid, liquid and gas state out of these three phases any two may co-exist in equilibrium for a set of values of Pressure (P) and Temperature (T). If we vary the pressure and temperature and note the phase change; and plot on a diagram, then such a diagram is called the pressure- temperature phase diagram.

### Vaporization curve:

The phase diagram representing the equilibrium between liquid and vapour phases is called the vaporization curve.

For water this curve is named as steam curve. For any point above the curve the substance is in liquid state and point below the curve it is in vapour state

Fusion curve :

The phase diagram representing the equilibrium between the solid and the liquid phases is called the fusion curve.

Fusion curve(a)

These curves are of two types

(i) For substances which expand on solidification (e.g. water), the slope is negative (first fig). For

Fusion curve(b)

water it is called ice line. The melting point of such a substance decreases with rise of temperature.

(ii) For substances which contract on solidification (e.g. wax), the slope is positive ( 2nd fig). The melting point of such a substance increases with increase of pressure.

Sublimation curve

The phase diagram representing the equilibrium between solid and vapour phase is called the sublimation curve.

Sublimation Curve

Triple point :

The point at which solid, liquid and vapour phases co-exist in equilibrium is called the triple point. Triple point is a single unique point.
For water its value is 273.16 K.

Triple point

# Thermodynamic Processes :

## Isothermal Process :

This is a slow process in which temperature remains constant. A gas under isothermal process obeys Boyle’s law.
PV = constant for constant mass and temperature.
Work done in isothermal process for n moles of a gas from volume Vi to Vf is given by

$W = n RT \log _e \dfrac{V_f}{V_i} = 2.3026 n RT \log _{10} \dfrac{V_f}{V_i}$
Slope on PV diagram for isothermal process of gas

$( \dfrac{dP}{dV} )_{isothermal} = - \dfrac{P}{V}$

This is the rapid or sudden process in which heat of system remains constant i.e.
Q = constant or $\triangle$Q = 0
for an adiabatic change of perfect gas, the Poisson’s equation holds which is given by

$PV^{\gamma}$

Expansion curve

Compression curves

Or $TV^{\gamma -1}$ = constant

Or $\dfrac{T^{\gamma}}{P^{\gamma - 1}}$ = constant

Where  $\gamma$the ratio of two specific heats i.e;

$\gamma = \dfrac{C_P}{C_V}$ =

1.67 for monoatomic gas

1.4 for diatomic gas

1.33 for triatomic gas

Work done in adiabatic process moles for n moles of a gas

$W = \dfrac{1}{1 - \gamma} [ P_f V_f - P_i V_i \\ = \, \dfrac{n R}{1 - \gamma} [ T_i - T_f$

where i and f refer to initial and final states.

Slope diagram for adiabatic process of a gas

$( \dfrac{dP}{dV} )_{adiabatic} \, \, \, \,\, = \, - \dfrac{\gamma P}{V}$

Obviously

$\dfrac{( \dfrac{dP}{dV} )_{adiabatic}}{( \dfrac{dP}{dV} )_{isothermal} } \, \, \, \,\, = \, \gamma \dfrac{C_P}{C_V}$

## Isobaric process :

This is the process in which pressure remains constant
i.e.   P = constant or $\triangle$P = 0

Work done,

$W = \int_{V_i}^{V^f} P dV = P ( V_f - V_i )$
Fig (a) and (b) represent changes in volumes Vi isobaric to Vf isothermal and adiabatic changes during expansion and compression of a gas respectively.

## Isochoric Process :

This is the process in which volume remains constant
i.e. V = constant or $\triangle$ V =0.

Work done

$W = \int_{V_i}^{V^f} P dV = 0$

## Reversible Process :

A reversible process is one in which the changes in heat and work of direct process from initial to a final state are exactly retraced in opposite sense in the reverse process and the system and surroundings are left in initial positions. The reversibility is an ideal concept and cannot be realized in practice.

## Irreversible Process :

The process which is not reversible is the irreversible process. In nature the processes are irreversible.In nature the processes are irreversible.

# Heat Engine:

Heat engine is a device which converts heat into work. A heat engine, in general, consists of three parts:

(i) A source or high temperature reservoir at temperature T1.

(ii) A sink or low temperature reservoir at temperature T2.

(iii) A working substance .

In a cycle of heat engine the working substance extracts heat Q1 from source, does some work W and
rejects remaining heat Q2 to sink.

Efficiency of heat engine,

$\eta = \dfrac{\text{Work done W}}{ \text{Heat taken from source}(Q_1)} \\ = \dfrac{Q_1 - Q_2}{Q_1} \\ = 1 - \dfrac{Q_2}{Q_1}$

This is general expression for the efficiency of heat engine:

## Carnot Engine:

Carnot devised an ideal engine which is based on a reversible cycle of four operations in succession isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. If T1 and T2 are absolute temperatures of source and sink, then
Efficiency of Carnot Engine

$\eta = 1 - \dfrac{Q_2}{Q_1} = 1 - \dfrac{T_2}{T_1}$

Carnot Engine

Carnot Theorem:

No irreversible engine can have efficiency greater than Carnot reversible engine working between same hot and cold reservoirs i.e

$\eta _R > \eta _1 \\ \text{or} \, \, \, \, \, 1 -\dfrac{T_2}{T_1} > 1 - \dfrac{Q_2}{Q_1}$

In a Carnot cycle

$\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1}$

# Second Law of Thermodynamics:

First law of thermodynamics is based on conservation of energy, while second law of thermodynamics gives information about the transformation of heat energy. There are two conventional statements of second law dependent on common experience.

1. Kelvin-Planck Statement:

It is impossible for an engine working between a cyclic process to extract heat from a reservoir and convert completely into work. In other words, 100% conversion of heat into work is impossible.

2. Clausius Statement:

It is impossible for a self-acting machine, unaided by any external agency to transfer heat from a cold to hot reservoir. In other word heat cannot  flow itself from a colder to hotter body.

Refrigerator:

It is inverse of heat engine. It extracts heat (Q2) from a cold reservoir, same external work W is done on it and rejects heat (Q1) to hot reservoir.

The coefficient of performance of a refrigerator

$\beta = \dfrac{\text{Heat extracted from cold reservoir}}{\text{Work done on refrigerator}} \\ \, = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1 - Q_2} = \dfrac{1}{\dfrac{Q_1}{Q_2} - 1}$

Carnot refrigerator

For Carnot (reversible) refrigerator

$\dfrac{Q_1}{Q_2} = \dfrac{T_1}{T_2} \\ \therefore \, \, \, \beta = \dfrac{Q_2}{W} = \dfrac{1}{\dfrac{Q_1}{Q_2} - 1} = \dfrac{1}{\dfrac{T_1}{T_2} - 1} = \dfrac{T_2}{T_1 - T_2}$

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