Basic Concepts of Thermodynamic

INTRODUCTION

The word 'thermodynamics' implies flow of heat. It deals with energy changes accompanying all types of physical and chemical processes.

It helps to lay down the criteria for predicting feasibility or spontaneity of a process, including a chemical reaction, under a given set of conditions. It also helps to determine the extent to which a process, including a chemical reaction, can proceed before attainment of equilibrium.

Thermodynamics is based on two generalizations called the first and second law of thermodynamics. These are based on human experience.

SOME BASIC TERMS

System

A system is defined as any specified portion of matter under study which is separated from the rest of the universe with a bounding surface. A system may consist of one or more substances.

Surroundings

The rest of the universe which might be in a position to exchange energy and matter with the system is called the surroundings.

Types of system

(i) Isolated system

A system which can exchange neither energy nor matter with its surrounding is called an isolated system.

(ii) Open system

A system which can exchange matter as well as energy with its surroundings is said to be an open system.

(iii) Closed system

A system which can exchange energy but not matter with its surroundings is called a closed system.

Macroscopic properties

The properties associated with a macroscopic system (i.e. consisting of large number of particles) are called macroscopic properties. These properties are pressure, volume, temperature, composition, density etc.

Extensive and Intensive properties

An extensive property of a system is that which depends upon the amount of the substance present in the system like mass, volume and energy.

An intensive property of a system is that which is independent of the amount of the substance present in the system like temperature, pressure, density, concentration, viscosity, surface tension, refractive index etc.

State of a system

When macroscopic properties of a system have definite values, the system is said to be in definite state. Whenever there is a change in any one of the macroscopic properties, the system is said to change into a different state. Thus, the state of a system is fixed by its macroscopic properties.

State variables

Since the state of a system changed with change in any of the macroscopic properties, these properties are called state variables or the thermodynamics parameters which depends only upon the initial and final states of the system and independent of the manner as to how the change is brought are called state functions. Some common state functions are internal energy, enthalpy, entropy, free energy, pressure, temperature, volume etc.

Thermodynamic equilibrium

A system in which the macroscopic properties do not undergo any change with time is said to be in thermodynamic equilibrium.

Thermodynamic process and their types

The operation by which a system changes form one state to another is called a process. Whenever a system changes from one state to another it is accompanied by change in energy. In case of open systems, there may be change of matter as well.

The following types of process are known

Isothermal process

A process is said to be isothermal if the temperature of the system remains constant during each stage of the process.

Adiabatic process

A process is said to be adiabatic if the heat enters or leaves the system during any step of the process.

Isobaric process

A process is said to be isobaric if the pressure of the system remains constant during each step of the process.

Illustration 1. Thermodynamics is concerned with

(A) total energy in a system (B) energy changes in a system

(C) rate of a chemical change (D) mass changes in nuclear reactions

Solution: (B)

Isochoric Process

A process is said to be isochoric if the volume of the system remains constant during each step of the process.

Reversible and Irreversible process

A process which is carried out infinitesimally slowly in such a manner that the system remains almost in a state of equilibrium at every stage or a process carried out infinitesimally slowly so that the driving force is only infinitesimally greater than the opposing force is called a reversible process.

Any process which does not take place in the above manner i.e. a process which does not take place infinitesimally slowly, is said to be an irreversible process.

In fact, all the natural processes are irreversible processes.

INTERNAL ENERGY

Every substance is associated with a definite amount of energy which depends upon its chemical nature as well as upon its temperature, pressure and volume. This energy is known as internal energy. Internal energy of the system is the energy possessed by all its constituent molecules.

Internal energy is a state property i.e. its value depends only upon the state of the substance but does not depend upon how that state is achieved. The absolute value of internal energy of a substance can not be determined. However determining the absolute values of internal energies is neither necessary nor required. It is the change in internal energy accompanying a chemical or a physical process that is of interest and this is a measurable quantity.

The first law of thermodynamics

The first law of thermodynamics states that energy can neither be created nor destroyed, although it can be transformed from one form to another. This is also known as the law of conservation of energy.

MATHEMATICAL EXPRESSION OF FIRST LAW

Let U_A be the energy of a system in its state A and U_B be the energy in its state B. Suppose the system while undergoing change from state A to state B absorbs heat q from the surroundings and also performs some work (mechanical or electrical), equal to w. The absorption of heat by the system tends to raise the energy of the system. The performance of work by the system, on the other hand, tends to lower the energy of the system because performance of work requires expenditure of energy. Hence the change of internal energy, $\Delta$U, accompanying the above process will be given by

$\Delta U=U_B-U_A=q-w$.

In general, if in a given process the quantity of heat transferred from the surrounding to the system is q and work done in the process is w, then the change in internal energy,

$\Delta$U = q + w

This is the mathematical statement of the first law of thermodynamics.

If work is done by the surroundings on the system (as during the compression of a gas), w is taken as positive so that $\Delta$U = q + w. if however work is done by the system on the surroundings (as during the expansion of a gas), w is taken as negative so that $\Delta$U = q – w.

Illustration 2. 1 mole of ideal monoatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm³ to 16 dm³.

Calculate (i) q (ii) w and (iii) $\Delta$U

Solution: (i) Since process is adiabatic $\therefore$ q = 0

(ii) As the gas expands against the constant external pressure.

W = $-P\Delta V=-1.5\left(V_2-V_1\right)$

= $-1.5\left(16-4\right)=-18atmd{m}^3$

(iii) $\Delta$U = q + w = $0+\left(-18\right)=-18atmd{m}^3$

ENTHALPY OF A SYSTEM

The quantity U + PV is known as the enthalpy of the system and is denoted by H. It represents the total energy stored in the system. Thus

H = U + PV

It may be noted that like internal energy, enthalpy is also an extensive property as well as a state function. The absolute value of enthalpy can not be determined, however the change in enthalpy can be experimentally determined.

$\Delta$H = $\Delta$U + $\Delta$(PV)

Various kinds of processes:

(i) Isothermal reversible expansion of an Ideal gas: Since internal energy of an Ideal gas is a function of temperature and it remains constant throughout the process hence

$\Delta$E = 0 and $\Delta$H = $\Delta$E + $\Delta$PV

$\because$$\Delta$E = 0and P₁V₁ = P₂V₂ at constant temperature for a given amount of the gas$\therefore$$\Delta$H= 0

Calculation of q and w:

$\because$.$\Delta$E = q + w

For an Isothermal process, w = -q

This shows that in an Isothermal expansion, the work done by the gas is equal to amount of heat absorbed.

and w = - n RT ln(V₂/V₁) = - n RT ln(P₁/P₂).

Illustration 3. 10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm Calculate the work done when the expansion is carried out (i) in single step (ii) in three steps the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.

Solution: (i) Work done = V.$\Delta$P

V = $\left(\frac{10}{4}\right)\times \frac{8.314\times 400}{100\times {10}^5}=\overrightarrow{\leftarrow }83.14\times {10}^{-5}$m³

So W = $\frac{83.14}{{10}^5}$ (100-1)x10⁵ = 8230.86 J.

(ii) In three steps

V_I = 83.14x10^-5 m³

W_I = (83.14x10^-5)x(100-60)x10⁵

= 3325.6 Jules

V _II = $\frac{2.5\times 8.314\times 400}{60\times {10}^5}=138.56\times {10}^{-5}\overrightarrow{\leftarrow }\overrightarrow{\leftarrow }{m}^3$

W_II = V. $\Delta$P

W_II = 138.56x10^-5 (60-30)x10⁵

= 4156.99 $\approx$4157 J.

V_III = $\frac{2.5\times 8.314\times 400}{30\times {10}^5}=277.13\times {10}^{-5}\overrightarrow{\leftarrow }\overrightarrow{\leftarrow }{m}^3$

W_III = 277.13x10^-5 (30-1)x10⁵

W_III = 8036.86 J.

W _total = W_I + W_II + W_III

= 3325.6+4156.909+8036.86 = 15519.45 J.

(iii) For reversible process

W = 2.303 nRT log $\frac{P_1}{P_2}$

= 2.303 x $\left(\frac{10}{4}\right)\times 8.314\times 400\times \log \left(\frac{100}{1}\right)$

W = 38294.28 Jules

(ii) Adiabatic Reversible Expansion of an Ideal gas:

$\because$q = 0

$\therefore$$\Delta$E= -w.

Total change in the internal energy is equal to external work done by the system.

$\therefore$Work done by the system = $\Delta$E= C_v$\Delta$T.

and C_p-C_v = R

On dividing all the terms by C_v.

$\frac{C_p}{C_v}-\frac{C_v}{C_v}=\frac{R}{C_v}$

$\frac{C_p}{C_v}=\gamma$

$\therefore (\gamma -1)=\frac{R}{C_v}$.

and Cv$=\frac{R}{(\gamma -1)}$

$\therefore w=\frac{R}{(\gamma -1)}{\int }_{T_1}^{T_2}dT$

$w=\frac{R}{(\gamma -1)}(T_2-T_1)$and $\Delta$H = $\Delta$E + P$\Delta$V.

Thus if T₂>T₁, w = +ve i.e. work is done on the system.

Thus if T₂<T₁, w = -ve i.e. work is done by the system.

Change of internal energy in a chemical reaction

Let us consider a chemical reaction taking place at constant temperature and at constant volume. In such a case, w = 0 and hence from the first law

$\Delta$U = q_v

Where q_v is the heat exchanged at constant volume, or heat or enthalpy of reaction at constant volume.

Change of Enthalpy in a chemical reaction

Let q_P be the heat exchanged in the chemical reaction taking place at constant pressure, Then evidently,

$\Delta$H = q_P = Heat or Enthalpy of reaction at constant pressure.

Exothermic and Endothermic reaction

Reaction that give out heat, i.e. which are accompanied by evolution of heat, are called exothermic reaction. In such reactions $\Delta$H is negative. On the other hand, reaction that intake heat, i.e. which are accompanied by absorption of heat are called endothermic reactions. In these reactions $\Delta$H is positive.

Illustration 4. Fill in the blanks with appropriate word in following:

(i) Combustion of reactions are usually ……………………..

(ii) Combustion of F₂ in oxygen is ……………………..

Solution: (ii) Exothermic

(iii) Endothermic

HEAT CAPACITY AND SPECIFIC HEAT

The heat capacity (C) of a sample of substance is the quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or Kelvin).

q = c$\Delta$t

Heat capacity is directly proportional to the amount of substance.

The specific heat capacity is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius at constant pressure.

q = s xm x $\Delta$t

where q is the heat required to raise temperature

m = mass in grams

s = specific heat of the substance

$\Delta$t = temperature difference

VARIATION OF HEAT OF REACTION WITH TEMPERATURE

The heat of reaction depends on the temperature. The relation between the two is known as Kirchoff's equation.

(i) $\frac{\Delta {\text{H}}_{\text{2}}-\Delta {\text{H}}_{\text{1}}}{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}$= $\Delta$C_P (ii) $\frac{\Delta {\text{E}}_{\text{2}}-\Delta {\text{E}}_{\text{1}}}{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}$= $\Delta$C_V

$\Delta$C_P = molar heat capacity of products – molar heat capacity of reactants (at constant pressure)

$\Delta$C_v = molar heat capacity of products – molar heat capacity of reactants (at constant volume)

Illustration 5. The standard heat of formation listed for gaseous NH₃ is -11.02 kcal/mol at

298 K. Given that at 298 K, the constant pressure heat capacities of gaseous N₂, H₂ and NH₃ are respectively 6.96, 6.89, 8.38 cal/mol. Determine $\Delta$H⁰_298K and $\Delta$H_{773 K} for the reactions,

$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right)$

Solution: $\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right)$

$\Delta H_{298K}=\sum {\left(H_f\right)}_P-\sum {\left(H_F\right)}_R=-\left(11.02-0\right)$

= -11.02 kcal mol^-1

$\frac{\Delta H_2-\Delta H_1}{T_2-T_1}=\Delta C_p$

$\frac{\Delta H_2-\left(-11.02\right)}{773-298}=\left(8.38-\frac{1}{2}\times 6.96-\frac{3}{2}\times 6.89\right)\times {10}^{-3}$

$\Delta$H₂ = -13.6 kcal mol^-1