Thermodynamics : First law of Thermodynamics

 

Thermodynamics

Thermodynamics (thermo means heat and dynamics means motion) is the branch of science which deals with the study of different forms of energy and the quantitative relationships between them.

The complete study of thermodynamics is based upon three generalizations celled first, second and third laws of thermodynamics. These laws have been arrived purely on the basis of human experience and there is no theoretical proof for any of these laws.

 

Basic Terms of Thermodynamics

(1) System, surroundings and Boundary : A specified part of the universe which is under observation is called the system and the remaining portion of the universe which is not a part of the system is called the surroundings.

The system and the surroundings are separated by real or imaginary boundaries. The boundary also defines the limits of the system. The system and the surroundings can interact across the boundary.

 

(2) Types of systems

(i) Isolated system : This type of system has no interaction with its surroundings. The boundary is sealed and insulated. Neither matter nor energy can be exchanged with surrounding. A substance contained in an ideal thermos flask is an example of an isolated system.

(ii) Closed system : This type of system can exchange energy in the form of heat, work or radiations but not matter with its surroundings. The boundary between system and surroundings is sealed but not insulated. For example, liquid in contact with vapour in a sealed tube forms a closed system. Another example of closed system is pressure cooker.

(iii) Open system : This type of system can exchange matter as well as  energy with its surroundings. The boundary is not sealed and not insulated. Sodium reacting with water in an open beaker is an example of open system.

(iv) Homogeneous system : A system is said to be homogeneous when it is completely uniform throughout. A homogeneous system is made of one phase only. Examples: a pure single solid, liquid or gas, mixture of gases and a true solution.

(v) Heterogeneous system : A system is said to be heterogeneous when it is not uniform throughout, i.e., it consist of two or more phases. Examples : ice in contact with water, two or more immiscible liquids, insoluble solid in contact with a liquid, a liquid in contact with vapour, etc.

(vi) Macroscopic system : A macroscopic system is one in which there are a large number of particles (may be molecules, atoms, ions etc. )

Note  All physical and chemical processes taking place in open in our daily life are open systems because these are continuously exchanging matter and energy with the surroundings.

 

(3) Macroscopic Properties of the System : Thermodynamics deals with matter in terms of bulk (large number of chemical species) behaviour. The properties of the system which arise from the bulk behaviour of matter are called macroscopic properties. The common examples of macroscopic properties are pressure, volume, temperature, surface tension, viscosity, density, refractive index, etc.

The macroscopic properties can be sub – divided into two types

(i) Intensive properties : The properties which do not depend upon the quantity of matter present in the system or size of the system are called intensive properties.  Pressure, temperature, density, specific heat, surface tension, refractive index, viscosity, melting point, boiling point, volume per mole, concentration etc. are the examples of intensive properties of the system.

(ii) Extensive properties : The properties whose magnitude depends upon the quantity of matter present in the system are called extensive properties. Total mass, volume, internal energy, enthalpy, entropy etc. are the well known examples of extensive properties. These properties are additive in nature.

Note : Any extensive property if expressed as per mole or per gram becomes an intensive property. For example, mass and volume are extensive properties, but density and specific volume, i.e. the mass per unit volume and volume per unit mass respectively are intensive properties. Similarly, heat capacity is an extensive property but specific heat is an intensive property.

 

(4) State of a system and State Variables : The state of a system means the condition of existence of the system when its macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system is also said to change. Thus, the state of the system is fixed by its macroscopic properties.

Macroscopic properties which determine the state of a system are referred to as state variables or state functions or thermodynamic parameters. The change in the state properties depends only upon the initial and final states of the system, but is independent of the manner in which the change has been brought about. In other words, the state properties do not depend upon a path followed.

Following are the state functions that are commonly used to describe the state of the system

 

(i)Pressure (P) (ii)Temperature (T) (iii) Volume   (V) (iv)Internal energy (E)
(v)Enthalpy (H) (vi) Entropy (S) (vii) Free energy (G) (viii) Number of moles (n)

 

(5) Thermodynamic equilibrium :  “A system is said to have attained a state of thermodynamic equilibrium when it shows no further tendency to change its property with time”.

The criterion for thermodynamic equilibrium requires that the following three types of equilibrium exist simultaneously in a system.

(i) Chemical Equilibrium : A system in which the composition of the system remains fixed and definite.

(ii) Mechanical Equilibrium : No chemical work is done between different parts of the system or between the system and surrounding. It can be achieved by keeping pressure constant.

(iii) Thermal Equilibrium : Temperature remains constant i.e. no flow of heat between system and surrounding.

 

(6) Thermodynamic process : When the thermodynamic system changes from one state to another, the operation is called a process. The various types of the processes are

(i) Isothermal process : The process is termed isothermal if temperature remains fixed, i.e., operation is done at constant temperature. This can be achieved by placing the system in a constant temperature bath, i.e., thermostat. For an isothermal process dT = 0, i.e., heat is exchanged with the surroundings and the system is not thermally isolated.

(ii) Adiabatic process : If a process is carried out under such condition that no exchange of heat takes place between the system and surroundings, the process is termed adiabatic. The system is thermally isolated, i.e., dQ = 0. This can be done by keeping the system in an insulated container, i.e., thermos flask. In adiabatic process, the temperature of the system varies.

(iii) Isobaric process : The process is known as isobaric in which the pressure remains constant throughout the change i.e., dP = 0.

(iv) Isochoric process : The process is termed as isochoric in which volume remains constant throughout the change, i.e., dV = 0.

(v) Cyclic process : When a system undergoes a number of different processes and finally return to its initial state, it is termed cyclic process. For a cyclic process dE = 0 and dH = 0.

(vi) Reversible process : A process which occurs infinitesimally slowly, i.e. opposing force is infinitesimally smaller than driving force and when infinitesimal increase in the opposing force can reverse the process, it is said to be reversible process.

(vii) Irreversible process : When the process occurs from initial to final state in single step in finite time and cannot be reversed, it is termed an irreversible process. Amount of entropy increases in irreversible process.

Irreversible processes are spontaneous in nature. All natural processes are irreversible in nature

 

Difference between reversible and irreversible process

Reversible process Irreversible process
It is an ideal process and takes infinite time. It is a spontaneous process and takes finite time.
The driving force is infinitesimally greater than the opposing force.          The driving force is much greater than the opposing force.
It is in equilibrium at all stages. Equilibrium exists in the initial and final stages only.
Obtained work is maximum.               Obtained work is not maximum
It is difficult to realise in practice. It can be performed in practice.

 

Internal energy, Heat and work.

(1) Internal energy (E) : “Every system having some quantity of matter is associated with a definite amount of energy. This energy is known as internal energy.” The exact value of this energy is not known as it includes all types of energies of molecules constituting the given mass of matter such as translational vibrational, rotational, the kinetic and potential energy of the nuclei and electrons within the individual molecules and the manner in which the molecules are linked together, etc.

            E = Etranslational + Erotational + Evibrational + Ebonding + Eelectronic + …..     

(i) Characteristics of internal energy

(a) Internal energy of a system is an extensive property.

(b) Internal energy is a state property.

(c) The change in the internal energy does not depend on the path by which the final state is reached.

(d) There is no change in internal energy in a cyclic process.

(e) The internal energy of an ideal gas is a function of temperature only.

(f) Internal energy of a system depends upon the quantity of substance, its chemical nature, temperature, pressure and volume.

(g) The units of E are ergs (in CGS) or joules (in SI)

                             1 Joule = 107 ergs.

(ii) Change in internal energy (ΔE) : It is neither possible nor necessary to calculate the absolute value of internal energy of a system. In thermodynamics, one is concerned only with energy change which occurs when the system moves from one state to another. Let ΔE be the difference of energy of the initial state (Ein) and the final state (Ef), then, ΔE = Ef – Ein ; ΔE is positive if Ef > Ein and negative if Ef > Ein.

 

(2) Heat (q) and Work (w) : The energy of a system may increase or decrease in several ways but two common ways are heat and work.

Heat is a form of energy. It flows from one system to another because of the difference in temperature. Heat flows from higher temperature to lower temperature. Therefore, it is regarded as energy on the move.

Work is said to be performed if the point of application of force is displaced in the direction of the force. It is equal to the force multiplied by the displacement (distance through which the force acts).

There are three main types of work which we generally come across. These are, Gravitational work, electrical work and mechanical work. Gravitational work is said to be done when a body is raised through a certain height (h) against the gravitational field (g).

Electrical work is important in systems where reaction takes place between ions whereas mechanical work is performed when a system changes its volume in the presence of external pressure. Mechanical work is important specially in systems that contain gases. This is also known as pressure – volume work.

Note : Heat is a random form of energy while work is an organised form of energy.

(i) Units of Heat and Work : The heat changes are measured in calories (cal), Kilo calories (kcal), joules (J) or kilo joules (kJ). These are related as, 1 cal = 4.184 J; 1 kcal = 4.184 kJ

The S.I. unit of heat is joule (J) or kilojoule. The Joule (J) is equal to Newton – metre (1 J= 1 Nm).

Work is measured in terms of ergs or joules. The S.I. unit of work is Joule.

                                      1 Joule =  ergs = 0.2390 cal.

                                      1 cal   > 1 joule > 1 erg

(ii) Sign conventions for heat and work : It is very important to  understand the sign conventions for q and w. The signs of w or q are related to the internal energy change. When w or q is positive, it means that energy has been supplied to the system as work or as heat. The internal energy of the system in such a case increases.

When w or q is negative, it means that energy has left the system as work or as heat. The internal energy of the system in such a case decreases.

Thus, to summarise,

Heat absorbed by the system = q positive ; Heat evolved by the system = q negative

 

Work done on the system = w positive ; Work done by the system = w negative.        

 Zeroth law of thermodynamics

This law was formulated after the first and second laws of thermodynamics have been enunciated. This law forms the basis of concept of temperature. This law can be stated as follows,

If a system A is in thermal equilibrium with a system C and if B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other whatever the composition of the system.”

              

First law of Thermodynamics

First law of thermodynamics was proposed by Helmholtz and Robert Mayer. This law is also known as law of conservation of energy. It states that,

“Energy can neither be created nor destroyed although it can be converted from one form into another.”

(1) Justification for the law : The first law of thermodynamics has no theoretical proof. This law is based on human experience and has not yet been violated. The following observations justify the validity of this law

(i) The total energy of an isolated system remains constant although it can undergo a change from one form to another.

(ii) It is not possible to construct a perpetual machine which can do work without the expenditure of energy, If the law were not true, it would have been possible to construct such a machine.          

(iii) James Joule (1850) conducted a large number of experiments regarding the conversion of work into heat energy. He concluded that for every 4.183 joule of work done on the system, one calorie of heat is produced. He also pointed out that the same amount of work done always produces same amount of heat irrespective of how the work is done.   

(iv) Energy is conserved in chemical reactions also. For example, the electrical energy equivalent to 286.4 kJ mol-1 of energy is consumed when one mole of water decomposes into gaseous hydrogen and oxygen. On the other hand, the same amount of energy in the form of heat is liberated when one mole of liquid water is produced from gases hydrogen and oxygen.

                   H2O(l) + 286.4 kJ → H2(g) + ½ O2(g) ;

                   H2(g) + ½ O2(g) → H2O(l) + 286.4 kJ

These examples justify that energy is always conserved though it may change its form.

 

(2) Mathematical expression for the law : The internal energy of a system can be changed in two ways

(i) By allowing heat to flow into the system or out of the system.

(ii) By doing work on the system or by the system.

Let us consider a system whose internal energy is E1. Now, if the system absorbs q amount of heat, then the internal energy of the system increases and becomes E1 + q.

If work (w) is done on the system, then its internal energy further increases and becomes E2. Thus,

          E2 = E1 + q +w or E2 – E1 = q +w or ΔE = q + w

i.e.

(Change in internal energy) = (Heat added to the system) + (Work done on the system)

If a system does work (w) on the surroundings, its internal energy decreases. In this case, work is taken as negative (–w). Now, q is the amount of heat added to the system and w is the work done by the system, then change in internal energy becomes, ΔE = q + (–w) = q – w

i.e.

(Change in internal energy) = (Heat added to the system) – (Work done by the system)

The relationship between internal energy, work and heat is a mathematical statement of first law of thermodynamics.

 

(3) Some useful conclusions drawn from the law : ΔE = q + w

(i) When a system undergoes a change ΔE = 0, i.e., there is no increase or decrease in the internal energy of the system, the first law of thermodynamics reduces to

                   0 = q +w  or  q = –w

          (heat absorbed from surroundings = work done by the system)

                               or w = – q

          (heat given to surroundings = work done on the system)

(ii) If no work is done, w = 0 and the first law reduces to

                                ΔE – q

i.e. increase in internal energy of the system is equal to the heat absorbed by the system or decrease in internal energy of the system is equal to heat lost by the system.

(iii) If there is no exchange of heat between the system and surroundings, q = 0, the first law reduces to

                                      ΔE = w

It shows if work is done on the system, its internal energy will increase or if work is done by the system its internal energy will decrease. This occurs in an adiabatic process.

(iv) In case of gaseous system, if a gas expands against the constant external pressure P, let the volume change be ΔV. The mechanical work done by the gas is equal to –P × ΔV.

          Substituting this value in ΔE = q + w; ΔE = q – PΔV ;

          When, ΔV = 0,   ΔE = q or qv

          The symbol qv indicates the heat change at constant volume.  

 

(4) Limitations of the law : The first law of thermodynamics states that when one form of energy disappears, an equivalent amount of another form of energy is produced. But it is silent about the extent to which such conversion can take place. The first law of thermodynamics has some other limitations also.

(i) It puts no restriction on the direction of flow of heat. But, heat flows spontaneously from a higher to a lower temperature.

(ii) It does not tell whether a specified change can occur spontaneously or not.

(iii) It does not tell that heat energy cannot be completely converted into an equivalent amount of work without producing some changes elsewhere.

 

Enthalpy and Enthalpy change

Heat content of a system at constant pressure is called enthalpy denoted by ‘H’.

          From first law of thermodynamics, q = E + PV               ……….(i)

          Heat change at constant pressure can be given as

                             Δq = ΔE + PΔV                                                  ……….(ii)

          At constant pressure heat can be replaced at enthalpy.

                             ΔH = ΔE = PΔV                                                 …..…..(iii)

Constant pressures are common in chemistry as most of the reactions are carried out in open vessels.

          At constant volume, ΔV = 0; thus equation (ii) can be written as,

                             ΔqV = ΔE

ΔH = Heat change or heat of reaction (in chemical process) at constant pressure

ΔE = Heat change or heat of reaction at constant volume.

(1) In case of solids and liquids participating in a reaction,

                             ΔH = ΔE(PΔV ≈ 0)

(2) Difference between ΔH and ΔE is significant when gases are involved in chemical reaction.

                             ΔH = ΔE + PΔV

                             ΔH = ΔE + ΔnRT                           

                             PΔV = ΔnRT

Here, Δn = Number of gaseous moles of products – Number of gaseous moles of reactants.

Using the above relation we can interrelate heats of reaction at constant pressure and at constant volume.

 

Specific and Molar heat capacity

(1) Specific heat (or specific heat capacity) of a substance is the quantity of heat (in calories, joules, kcal, or kilo joules) required to raise the temperature of 1g of that substance through 1oC. It can be measured at constant pressure (Cp) and at constant volume (CV).

(2) Molar heat capacity of a substance is the quantity of heat required to raise the temperature of 1 mole of the substance by 1oC.

       ∴ Molar heat capacity = Specific heat capacity × Molecular weight, i.e.,

                   CV = cv × M and CP = cp × M

(3) Since gases on heating show considerable tendency towards expansion if heated under constant pressure conditions, an additional energy has to be supplied for raising its temperature by  relative to that required under constant volume conditions, i.e.,

  CP > CV or CP = CV + work done on expansion, PΔV(=R)

Where, CP = molar heat capacity at constant pressure; CV = molar heat capacity at constant volume.

Note :  CP and CV for solids and liquids are practically equal. However, they differ considerable in case of gas because appreciable change in volume takes place with temperature.

 

(4) Some useful relations of Cp  and Cv

(i) CP – CV = R = 2 caloried = 8.314 J

(ii) C=  \frac { 3 }{ 2 } R (for monoatomic gas) and C=  \frac { 3 }{ 2 } + x (for di and polyatomic gas), where x varies from gas to gas.

(iii)  \frac { { C }_{ p } }{ { C }_{ V } } =\gamma (Ratio of molar capacities)

(iv) For monoatomic gas CV = 3 calories whereas, CP = CV + R = 5 calories

(v) For monoatomic gas, (γ) =  \frac { { C }_{ p } }{ { C }_{ V } } =\frac { \frac { 5 }{ 2 } R }{ \frac { 3 }{ 2 } R } = 1.66

(vi) For diatomic gas (γ) =  \frac { { C }_{ p } }{ { C }_{ V } } =\frac { \frac { 7 }{ 2 } R }{ \frac { 5 }{ 2 } R } = 1.40

(vii) For triatomic gas (γ) =  \frac { { C }_{ p } }{ { C }_{ V } } =\frac { 8R }{ 6R } = 1.33

 

Values of Molar heat capacities of some gases,

Gas Cp Cv Cp– Cv Cp/Cv= γ Atomicity
He 5 3.01 1.99 1.661 1
N2 6.95 4.96 1.99 1.4 2
O2 6.82 4.83 1.99 1.4 2
CO2 8.75 6.71 2.04 1.30 3
H2S 8.62 6.53 2.09 1.32 3

 

Expansion of an Ideal gas

(1) Isothermal Expansion : For an isothermal expansion, temperature remains fixed i.e. ΔE = 0 (ΔE of ideal gases depends only on temperature)   

According to first law of thermodynamics,

                                      ΔE = q + w

                                      ∴ q = – w

This shows that in isothermal expansion, the work is done by the system at the expense of heat absorbed.

Since for isothermal process, ΔE and ΔT  \left( \frac { \pi }{ 2 } -\theta \right) are zero respectively, hence, ΔH = 0

(i) Work done in reversible isothermal expansion : Consider an ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The cylinder is not insulated. The external pressure, Pext is equal to pressure of the gas, Pgas. Let it be P.

                                      Pext = Pgas = P

If the external pressure is decreased by an infinitesimal amount dP, the gas will expand by an infinitesimal volume, dV. As a result of expansion, the pressure of the gas within the cylinder falls to Pgas – dP, i.e., it becomes again equal to the external pressure and, thus, the piston comes to rest. Such a process is repeated for a number of times, i.e., in each step the gas expands by a volume dV.

Since the system is in thermal equilibrium with the surroundings, the infinitesimally small cooling produced due to expansion is balanced by the absorption of heat from the surroundings and the temperature remains constant throughout the expansion.

The work done by the gas in each step of expansion can be given as,

                   dw = –(Pext – dP)dV = –Pext.dV = –PdV

dP.dV, the product of two infinitesimal quantities, is neglected.

The total amount of work done by the isothermal reversible expansion of the ideal gas from volume V1 to volume V2 is, given as,

                   w = -nRTloge   \frac { V_{ 2 } }{ V_{ 1 } }      or   w = -2.303nRTlog10   \frac { V_{ 2 } }{ V_{ 1 } }

At constant temperature, according to Boyle’s law,

                   P1V1 = P2V2        or      \frac { V_{ 2 } }{ V_{ 1 } } =  \frac { P_{ 1 } }{ P_{ 2 } }      So,    w = -2.303nRTlog10   \frac { P_{ 1 } }{ P_{ 2 } }             

Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with positive sign.

 { w }_{ compression }=2.303nRTlog\frac { V_{ 1 } }{ V_{ 2 } } =2.303nRTlog\frac { P_{ 2 } }{ P_{ 1 } }       

(ii) Work done in irreversible isothermal expansion : Two types of irreversible isothermal expansions are observed, i.e., (a) Free expansion and (b) Intermediate expansion. In free expansion, the external pressure is zero, i.e., work done is zero when gas expands in vacuum. In intermediate expansion, the external pressure is less than gas pressure. So, the work done when volume changes from V1 to V2 is given by

 { w }=-\int _{ { V }_{ 1 } }^{ { V }_{ 2 } }{ { P }_{ ext }\quad \times \quad dV\quad =\quad - } { P }_{ ext }({ V }_{ 2 }-{ V }_{ 1 })

Since Pext is less than the pressure of the gas, the work done during intermediate expansion is numerically less than the work done during reversible isothermal expansion in which Pext is almost equal to Pgas.

Note The work done by the system always depends upon the external pressure. The higher the value of Pext, the more work is done by the gas. As Pext cannot be more than Pgas, otherwise compression will occur, thus the largest value of Pext can be equal to Pgas. Under this condition when expansion occurs, the maximum work is done by the gas on the surroundings.

(2) Adiabatic Expansion : In adiabatic expansion, no heat is allowed to enter or leave the system, hence, q = 0.

According to first law of thermodynamics,

                             ΔE = q + w

                             ΔE = w

work is done by the gas during expansion at the expense of internal energy. In expansion, ΔE decreases while in compression ΔE increases.

The molar specific heat capacity at constant volume of an ideal gas is given by

              Cv \left( \frac { dE }{ dT } \right) _{ V }  or dE = CV.dT  and for finite change ΔE = CV.ΔT  So,   w = ΔE = CVΔT 

The value of ΔT depends upon the process whether it is reversible or irreversible.

(i) Reversible adiabatic expansion : The following relationships are followed by an ideal gas under reversible adiabatic expansion.

PVγ = Constant  

where, P = External pressure, V = Volume

 \gamma =\frac { { C }_{ P } }{ { C }_{ V } }

where, CP = molar specific heat capacity at constant pressure, CV = molar specific heat capacity at constant volume.

 \left( \frac { { T }_{ 1 } }{ T_{ 2 } } \right) ^{ \gamma }\left( \frac { P_{ 1 } }{ P_{ 2 } } \right) ^{ \gamma -1 }\left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) ^{ 1-\gamma }

knowing γ, P1, P2 and initial temperature T1, the final temperature T2 can be evaluated.

(ii) Irreversible adiabatic expansion : In free expansion, the external pressure is zero, i.e, work done is zero. Accordingly,  which is equal to w is also zero. If ΔE is zero, ΔT should be zero. Thus, in free expansion (adiabatically), ΔT = 0, ΔE = 0, w = 0 and ΔH = 0.

In intermediate expansion, the volume changes from V1 to V2 against external pressure, Pext.

 { w\ }=\ -{ P }_{ ext }({ V }_{ 2 }-{ V }_{ 1 })\ =\ -{ P }_{ ext }\ \left( \frac { { RT }_{ 2 } }{ P_{ 2 } } -\frac { { RT }_{ 1 } }{ P_{ 1 } } \right) \ =\ -{ P }_{ ext }\ \left( \frac { T_{ 2 }P_{ 1 }-T_{ 1 }P_{ 2 } }{ P_{ 1 }P_{ 2 } } \right) \times R

Or   { w }=\quad C_{ V }({ T }_{ 2 }-{ T }_{ 1 })\ =\ -{ RP }_{ ext } \left( \frac { T_{ 2 }P_{ 1 }-T_{ 1 }P_{ 2 } }{ P_{ 1 }P_{ 2 } } \right)

 

Example­1 :       The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25o C is

(a) 2.303 × 298 × 0.0821 log 2               (b)–298 × 107 × 8.31 × 2.303 log 2

(c) 2.303 × 298 × 0.0821 log 0.5             (d) 2.303 × 298 × 2 log 2

Solution (b) :   w=-2.303nRTlog\frac { V_{ 2 } }{ V_{ 1 } } =-2.303\times 1\times 8.31\times 1{ 0 }^{ 7 }\times 298\quad log\frac { 20 }{ 10 }   

= 298 × 107 × 8.31 × 2.303 log 2

 

Example ­2 : An ideal gas expands from 10–3 m3 to 10–2 m at 300 K against a constant pressure of 105 Nm–2. The work done is   

(a) –103 kJ           (b) 102 kJ            (c) 0.9 kJ             (d) – 900 kJ

Solution (c) : w = –P(V2 – V1) = – 105 (10–2 – 10–3) = – 105 × 10–2 (1–0.1) = – 0.9 × 103 J = – 0.9 kJ   

 

Example ­3 :      At 27°C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The values of ΔE and q are (R = 2)

(a) 0, –965.84 cal                                     (b) – 965.84 cal, + 965.84 cal      

(c) + 865.58 cal, –865.58 cal                   (d) –865.58 cal, –865.58 cal         

Solution (a) : w = – 2.303 RTlog   \frac { P_{ 1 } }{ P_{ 2 } } = 2.303 × 1 × 2 × 300log   \frac { 2 }{ 10 } = 965.84

                      At constant temperature ΔE = 0

                    ΔE = q + w; q = – w = – 965.84 Kcal

 

Example ­4 :      The work done by a system is 8 joule, when 40 joule heat is supplied to it, what is the increase in internal energy of system

(a) 25 J                (b) 30 J                (c) 32 J                (d) 28 J     

Solution (c) :     q = 40J

                   W = – 8J (work done by the system)

                   ΔE = q + w = 40 – 8 = 32J

 

Example ­5 :      One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27oC. If the work done during the process is 3 kJ, the final temperature will be equal to (Cv = 20 JK-1)

(a) 150 K             (b) 100 K             (c) 26.85 K           (d) 295 K

Solution (a) :    Work done during adiabatic expansion = CV(T2 – T1)

                    – 300 = 20(T2 – 300) or T2 = 150 K

 

Joule – Thomson effect

The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure, is known as Joule-Thomson effect.

During this process, enthalpy remains constant. It is, therefore, called isoenthalpic process.

The actual change in temperature on expansion of the gas is expressed in terms of Joule–Thomson coefficient (m), defined as

  \mu =\left( \frac { \vartheta T }{ \vartheta T } \right) _{ H }

i.e. it is the temperature change in degrees produced by a drop of one atmospheric pressure when the gas expands under conditions of constant enthalpy.

(1) For cooling, μ = + ve (because dT and dP both will be – ve)

(2) For heating, μ = – ve (because dT = +ve, dP = – ve

(3) μ = 0 means dT = 0 for any value of dP, i.e., there is neither cooling nor heating.

The temperature at which a real gas shows no cooling or heating effect on adiabatic expansion (i.e. μ = 0), is called Inversion temperature. Below this temperature it shows cooling effect while above this temperature, it shows heating effect.

Most of the gases have inversion temperature above room temperature. H2 and  have low inversion temperatures (being – 80°C and –240°C respectively). That is why they show heating effect at room temperature.