UNIT 04 THERMODYNAMICS - II
Syllabus
· Second Law of Thermodynamics
· Entropy
· Criteria of spontaneous and non-spontaneous process
· Internal energy, enthalpy, application of First Law of Thermodynamics.
Carriculam objectives
· To understand the concept of entropy and second law of thermodynamics through discussion , simple experiments and problem solving.
· To familiarize spontaneous and non-spontaneous process through discussion and experimentation.
· To attain the concept of free energy and criterion for spontaneity through discussion and problem solving.
· To understand the relation of free energy with emf and equilibrium constant through discussion and problem solving.
· To understand the Third Law thermodynaics through discussion, reference and problem solving.
The word ‘thermodynamics’ implies flow of heat. The subject deals with energy changes accompanying all types of physical and chemical processes. Thermodynamics may be defined as the branch of science which deals with quantitative relationship between heat and other forms of energies. When we confine our study to thermodynamics of chemical processes, it is referred to as Chemical Thermodynamics.
Thermodynamics is based upon three fundamental generalisations which have been arrived purely on the basis of human experience. These are :
· First Law which deals with equivalence of different forms of energy.
· Second Law deals with the direction of chemical change
· Third law helps to evaluate the thermodynamic parameters like entropy.
Importance of Thermodynamics
Thermodynamics is a fundamental subject of great importance in physical chemistry. Most of the important generalisations of physical chemistry can be deduced from the laws of thermodynamics.
Thermodynamics also helps to lay down the criteria for predicting the feasibility or spontaneity of a process including a chemical reaction under a given set of conditions. In other words, it helped to predict whether the given process or a chemical reaction is feasible under given conditions of temperature pressure and concentration. It also helps us to determine the extent to which a process including a chemical reaction, can proceed before attainment of equilibrium.
Limitations of Thermodynamics
(i) The laws of thermodynamics apply only to matter in bulk and not to individual molecules. In other words, the laws of thermodynamics apply to behaviour of assemblages of a vast number of molecules and not to individual molecules.
(ii) The second limitation is that thermodynamics can only predict whether a given process , including a chemical reaction, is feasible under given set of conditions. It does not tell anything about the rate at which a given process may proceed.
SOME BASIC CONCEPTS IN THERMODYNAMICS
System and surroundings
A thermodynamic system is defined as part of the physical universe which is selected for thermodynamic considerations. A system usually has a definite amount of a specific substance. It is separated from the rest of the universe called surroundings by a definite boundary. The boundary may be real or imaginary. A system is homogeneous if it is uniform throughout in all respects and is heterogeneous if it is not uniform throughout and consists of more than one phases separated from each other by sharp boundaries. A system consisting of a pure mixture of gases or a liquid or a solid etc. are examples of homogeneous systems, while a liquid and its vapour or a mixture of two immiscible liquids or two different solids etc. constitute heterogeneous systems.
System and the surroundings
The boundary (wall) between the system and surroundings differs in its ability to allow the passage of energy and matter through it. A permeable wall allows the passage of both matter and energy, a diathermal wall prevents the passage of matter but allows the flow of energy while an adiabatic wall neither allows the passage of energy nor matter.
Types of Systems
In thermodynamics there are mainly three different types of systems depending on the interactions between the system and surroundings. These are
i) Open systems
ii) Closed systems
iii) Isolated systems.
Open system
A system which can exchange both matter and energy with the surroundings is known as open system (Fig).
Open system
Due to these exchanges , matter and energy do not remain constant in open system i.e.,
dU ¹ 0 ; dm ¹ 0 .
The presence of reactants in an open beaker is an example of an open system. Here the boundary is an imaginary surface enclosing the beaker and reactants. We could have chosen only the reactants as system when walls of beakers act as boundary.
Closed system
These are systems in which exchange of energy with the surroundings is possible while the transfer of matter to and from the surroundings does not take place (Fig b) .
Closed system
Consequently, in a closed system mass remains constant and only the energy changes i.e.,
dm = 0 ; dU ¹ 0.
The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.
Isolated System
It is a system that prevents any interaction between the system and surroundings (Fig c ).
Isolated system
Both mass and energy of the system remain constant as there is no interaction of the system with the surroundings.
dm = 0 ; dU = 0
The presence of reactants in a thermos flask or any other closed isolated system.
Problem
01. Which of the following are open , close or isolated systems ?
(i) human being (open)
(ii) the earth (open)
(iii) can of tomato soup (open)
(iv) coffee in a thermoflask (isolated)
(v) ice cube tray filled with water (open)
(vi) a satellite in orbit (open)
(vii) helium filled balloon (closed)
Properties of a system
Thermodynamics deals with the behaviour of a system in bulk. The properties associated with bulk behaviour of matter are observable properties or macroscopic properties of the system. The macroscopic properties of a system are of the following two types.
1. Intensive properties
The properties which do not depend upon the quantity of matter present in the system are called intensive properties. For example, temperature, density, concentration, specific heat , viscosity, surface tension are intensive properties.
Temperature does not depend upon the quantity of matter. If the room temperature is 298 K , both small amount of water kept in a beaker and large amount of water kept in a bucket will be at 298 K. Similarly, if a solution has a concentration of 1 g/L , then the concentration of each drop of this solution would be the same, i.e., 1 g L
. Thus temperature and concentration do not depend upon the quantity of matter and hence are intensive properties. Similarly, we can understand the intensive nature of other properties mentioned above.
. Thus temperature and concentration do not depend upon the quantity of matter and hence are intensive properties. Similarly, we can understand the intensive nature of other properties mentioned above.2. Extensive properties
The properties which depend upon the quantity of matter present in the system are called extensive properties. Number of moles, mass, volume, enthalpy, entropy, free energy etc. are examples of extensive properties.
Some intensive and extensive properties are listed below.
Intensive properties | Extensive properties | ||
Temperature | Pressure | Mass | Enthalpy |
Density | Refractive index | Number of moles | Entropy |
Concentration | Surface tension | Energy | Internal energy |
Specific heat | Boiling point | Volume | Free energy |
Viscosity | Freezing point | ||
State of the system and state functions
State of a system implies the conditions of its existence. In order to understand it, let us consider H2O as a chemical system. We know that water can exist in three physical states ; ice, water and steam depending upon the conditions of temperature and pressure. For example, at 1 atmospheric pressure , H2O is :
Solid : if temperature is below 0°C
Liquid : if temperature is between 0°C - 100°C
Gas : if temperature is above 100°C.
Thus in order to define the physical state of water, we need the specification of its conditions of temperature and pressure. Consider two samples of liquid water , H2O(ℓ) at two different temperatures 25°C and 50°C respectively, both at one atmospheric pressure. Although physical state of two samples is same , yet their macroscopic properties like energy, viscosity , surface tension , refractive index etc. are different. In other words, the thermodynamic state of two samples is different. Thus, state of the system may be defined as the conditions of existence of system when its macroscopic properties have different values. When the state of a system changes, its macroscopic properties also changes and acquire new definite values. The macroscopic properties of the system which change with change in the state of the system are called state variables or State functions. The commonly used variables which are needed to define the state of a particular system are temperature, pressure , volume and chemical composition.
The values of state functions depend only upon the system but do not depend upon how that state is achieved. Some common state functions are volume, pressure, temperature , internal energy , enthalpy, entropy and free energy
THERMODYNAMIC EQUILIBRIUM
A system is said to be in a state of thermodynamic equilibrium, if none of the observable properties of the system appears to change with time. Actually, the term thermodynamic equilibrium assumes the existence of three types of equilibria in the system.
These are :
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium.
i) Thermal Equilibrium : A system is in thermal equilibrium if the temperature remains the same in all parts of the system.
ii) Mechanical equilibrium : It implies the uniformity of pressure throughout the whole of the system.
iii) Chemical equilibrium : If the concentration of the system remains constant, and uniform throughout, the system is said to be in chemical equilibrium.
THERMODYNAMIC PROCESSES
A thermodynamic process is defined as the method of operation with the help of which a change in the state of a system is effected. The various processes are :
a. Cyclic process : If a system after undergoing through a series of changes in its state comes back to its initial state, then the process is termed as a cyclic process and its path followed is known as the cyclic path.
b. Isothermal process : A process that is carried out under conditions of constant temperature is called an isothermal process. The constancy of temperature is achieved by either extracting heat from the system or supplying heat to it.
c. Adiabatic Process : A process in which there is no exchange of heat between the system and surrounding is known as an adiabatic process. The system is enclosed by adiabatic walls which do not permit heat exchanges with the surroundings. In an adiabatic process there is a change in temperature of the system.
d. Isochoric Process : A process in which volume of the system remains constant.
e. Isobaric Process : A process that is carried out under conditions of constant pressure.
f. Reversible Process : It is 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. The process is conducted in infinite number of steps and opposing force is only infinitesimally smaller than the direct force.
For example, consider a gas enclosed in a vessel fitted with a piston. The internal pressure of the gas is balanced by placing some sand on the piston. Now, if we remove a speck of the sand , the external pressure will become infinitesimally smaller than the internal pressure. Thus, the piston moves out and restore the equilibrium conditions. Now if we go on removing the specks of the sand one after another , we can ultimately achieve the finite expansion of the gas. The above process of expansion has been carried out in a number of stages and almost under equilibrium conditions and is therefore very close to the reversible process. In fact, the reversible processes are ideal processes and cannot be realised in actual practice.
g. Irreversible processes : A process that occurs rapidly or spontaneously such that it does not remain in equilibrium during transformation is called an irreversible process. Such processes do not involve a succession of equilibrium states of the system. After undergoing a change, such processes do not return themselves to their initial states, but can be reversed only with the help of external agencies. Expansion of gases against zero applied pressure, dissolution of solute in a solvent, mixing of gases, flow of liquid from higher to lower levels etc are examples of irreversible processes. All irreversible processes take finite time for their completion and are real processes in actual practice.
Problem
2. Choose the correct answer. A thermodynamic state function is a quantity :
(i) used to determine heat changes.
(ii) whose value is independent of path.
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only.
Answer 02 : (ii)
3. For the process to occur under adiabatic conditions, the correct condition is :
i) DT = 0 ii) DP = 0 iii) q = 0 iv) w = 0
Answer 3 : iii) q = 0
4. Classify the following as open, closed or isolated systems.
(i) A beaker containing boiling water.
(ii) A chemical reaction taking place in an evacuated flask.
(iii) A cup of tea placed on a table.
(iv) Hot water taken in perfectly insulated closed container.
(v) A thermos flask containing hot coffee.
Answer 04
(i) open (ii) closed (iii) open (iv) isolated (v) isolated
FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics is identical to the law of conservation of energy. It can be stated in many ways:
(i) Energy can neither be created nor destroyed although it can be converted from one form to another.
(ii) The total energy of an isolated system remains constant although it may change from one form to another.
(iii) The total energy of the universe is constant.
(iv) Whenever a quantity of some form of energy disappears , an exactly equivalent amount of some other form of energy must be produced.
(v) For a system in contact with its surroundings , the sum of the energies of the system and the surroundings at a particular moment remains constant , howsoever differently it is shared between the two.
(vi) It is impossible to construct a perpetual motion machine i.e., a machine which would produce work without consuming energy.
All the above statements mean the same and implies that we can neither produce nor destroy energy.
Justification for First Law of Thermodynamics
The first law of thermodynamics has been formulated on the basis of our day-to-day experiences and has no theoretical proof behind it. The law is universally valid and has not failed even in a single case. This strongly supports the validity of the law. Following observations are in accordance to the law and justify it.
(i) Till now , we could not construct a perpectual machine , i.e., a machine which could produce work with out consuming energy.
(ii) On supplying 286.0 kJ of electrical energy to one mole of water , it decomposes to give one mole of hydrogen and half mole of oxygen. If one mole of hydrogen is allowed to burn in half mole of oxygen , one mole of water is formed and the same amount of energy in the form of heat is liberated. This example indicates that the energy is always conserved though it may change its form.
(iii) James Joule observed that there is a definite relationship between heat energy and work energy. On the basis of several experiments, he concluded that for every 4.184 joules of work done on a system, one calorie of heat is produced.
INTERNAL ENERGY AND INTERNAL ENERGY CHANGES ( U and D U )
Since chemical changes are always accompanied by energy changes it indicates that reactants and products must be having certain amounts of energy. A fixed quantity of any substance is associated with a definite amount of energy which does not depend up on how that state is achieved. For example, CO2 can be obtained by various methods such as by heating calcium carbonate or by burning coal. However, one mole of carbon dioxide at STP is associated with a definite amount of internal energy, which does not depend upon the source from which it is obtained.
The various energies which contributes towards internal energy are :
i) Translational energy of molecules (Ut)
ii) Rotational energy of molecules (Ur
iii) Vibrational energy of molecules (Uv)
iv) Electronic energy (Ue)
v) Nuclear energy (Un)
vi) Interaction energy (Ui)
U = Ut + Ur + Uv + Ue + Un + Ui
It may be noted that the absolute value of internal energy cannot be determined because it is not possible to determine the exact values for the constituent energies such as translational, vibrational, rotational energies etc. However, we can determine changes in internal energy which occur during chemical reactions. The change in internal energy (DU) of a reaction is the difference between the internal energies of the products and reactants , i.e.,
DU = U (products) - U (reactants)
= UP - UR
where UP is the internal energy of the products , UR
If the internal energy of the of the products is less than the internal energy of the reactants, the DU would be negative. On the other hand, when the internal energy of the products is more than the internal energy of the reactants, then DU would be positive.
When a chemical reaction is carried out in such a way that the volume and temperature of the reacting system does not change, the internal energy change (DU) of the reaction is equal to the heat evolved or absorbed during the reaction. Thus,
DU = Heat evolved or absorbed in a reaction
at constant temperature and volume.
The internal energy change (DU) of the reaction can be conveniently determined by measuring the heat change occurring during the reaction at constant volume and temperature.
CONCEPT OF HEAT AND WORK
Heat and work are the two important modes by which the internal energy of the system can be changed.
Heat (q)
We can change the internal energy of a system by transfer of heat from the surroundings to the system or vice-versa without expenditure of work. This exchange of energy , which is a result of temperature difference is called heat, q .
Let us consider bringing about the same change in temperature (the same initial and final states as before) by transfer of heat through thermally conducting walls instead of adiabatic walls(fig).
A system which allows heat transfer through its boundary.
We take water at temperature TA in a container having thermally conducting walls, say made up of copper and enclose it in a huge heat reservoir at temperature TB. The heat absorbed by the system (water) , q can be measured in terms of temperature difference , TB - TA . In this case change in internal energy , DU = q , when no work is done at constant volume.
Sign conventions
According to the latest sign conventions :
The q is positive, when heat is transferred from surroundings to the system and q is negative when heat is transferred from system to surroundings.
Work
Work is said to be performed if the point of application of force undergoes displacement in the direction of force. It is equal to the force multiplied by the displacement (distance through which the force acts)
Let us first examine a change in internal energy by doing work.
We take a system containing some quantity of water in a thermos flask or in an insulated beaker. This would not allow exchange of heat between system and surroundings through its boundary and this type of system is called adiabatic. The manner in which the state of the system may be changed will be called adiabatic process. Adiabatic process is a process in which there is no transfer of heat between the system and surroundings. Here the wall separating the system and surroundings is called the adiabatic wall.
An adiabatic system which does not permit the transfer of heat through its boundary
Let us bring the change in the internal energy of the system by doing some work on it. Let us call the initial state of the system as state A and its temperature as TA. Let the internal energy of the system in state A be called UA. We can change the state of system in two different ways.
One way : We do some mechanical work, say 1 kJ , by rotating a set of small paddles and thereby churning water. Let the new state be called B and its temperature as TB. It is found that TB > TA and the change in temperature , DT = TB - TA. Let the internal energy of the system in state B be UB and the change in intermal energy , DU = UB - UA.
Second way : We do an equal amount ( i.e., 1 kJ) electrical work with the help of an immersion rod and note down the temperature change. We find that the change in temperature is same as in the earlier case , sat TB - TA .
The experiments in the above manner were done by J.P. Joule and he was able show that a given amount of work done on the system , no matter how it was done (irrespective of path) produced the same change of state , as measured by change in the temperature of the system.
So, it seems appropriate to define a quantity , the internal energy U, whose value is characteristic of the state of a system , whereby the adiabatic work , wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, DU ie.,
DU = UB - UA = wad
Therefore , internal energy , U , of the system is a state function.
The positive sign expresses that wad is positive when work is done on the system. Similarly, if the work is done by the system , wad will be negative.
Sign Convention
The latest sign conventions regarding work as per recommendations of IUPAC are :
Work done by the system is negative
Work done on the system is positive.
The positive value of w signifies that work has been done on the system by the surroundings and it has been contributed to an increase in internal energy of the system. On the other hand, negative value of w indicates that work has been done by the system ( or negative work done on the system) and it has contributed to a decrease in the internal energy of the system.
Units of work
In CGS system the work is expressed in ergs but in SI units, work is expressed in Joules. The equivalence between Joules and other units of work is :
1 J = 107 ergs = 1 Nm = 1 kg m2 s-2
The common forms of work that can come across in the study of thermodynamics are :
a) Pressure - volume work
b) Electric work
Pressure-volume work
This type of work is significant in systems which consists of gases and involve changes in volume against external pressure.
Expression for pressure-volume work
In order to derive the expression for pressure-volume (P-V) work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is P. If external pressure is Pex which is greater than P , piston is moved inward till the pressure inside becomes equal to Pex. Let this change be achieved in a single step and final volume be Vf . During this compression, suppose the piston moves a distance , ℓ and the cross-sectional area of the piston A.
Work done on an ideal gas in a cylinder when it is compressed by a constant external pressure ,
Pex (in single step) is equal to shaded area.
Then , the volume change = ℓ A = DV = ( Vf - Vi )
Therefore force on the piston = Pex A
If w is the work done on the system by the movement of piston then,
w = force x distance
= Pex A ℓ
= Pex (- DV)
= - Pex . DV
= - Pex ( Vf - Vi ) ….. (2)
The negative sign of this expression is required to obtain conventional sign for w , which will be positive. It indicates that in case of compression work is done on the system. Here ( Vf - Vi ) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive.
If the pressure is not constant at every stage of compression , but changes in number of finite steps , work done on the gas will be summed over all the steps and will be equal to -S P DV (Fig).
PV plot when pressure is not constant and changes in finite steps during compression from initial volume Vi to final volume , Vf . Work done on the gas is represented by the shaded area.
If the pressure is not constant but changes during the process such that it is always infinitesimally greater than the pressure of gas , then at each stage of compression , the volume decreases by infinitesimal amount , dV . In such a case we can calculate the work done on the gas by the relation,
…..(3)Here , Pex at each stage is equal to (Pex + dP) in the case of compression (Fig)
PV – plot when pressure is not constant and changes in infinite steps (reversible conditions) during compression from initial state , Vi to final volume , Vf . Work done on the gas is represented by the shaded area.
In an expansion process under similar conditions, the external pressure is always less than the pressure of the system ie., Pex = ( Pin - dP).
In general case we can write , Pex = ( Pin ± dP). Such processes are called reversible processes.
A process or change is said to be reversible , if a change is brought out in such a way that the process could , at any moment , be reversed by an infinitesimal change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and surrounding are always in near equilibrium with each other. Processes other than reversible process are known as irreversible processes.
Work done in isothermal reversible expansion of ideal gas
We can relate work to internal pressure of the system under reversible conditions by writing Equ (3) as follows :
Since dP x dV is very small we can write,
……. (4)Now , the pressure of the gas (Pin) which we can write P now) can be expressed in terms of its volume through gas equation. For n moles of an ideal gas ie., PV = n R T
Therefore, at constant temperature (isothermal process),
….(5)Free Expansion
Expansion of a gas in vaccum (Pex = 0) is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible (Equ 1 and 2).
We can write the equation 1 ( DU = q + w) in a number of ways depending on the type of processes.
Let us substitute w = - Pex DV (Eq 2) in Eq 1, we get :
DU = q - Pex DV
If a process is carried out at constant volume (DV = 0 ), then
DU = qv
The substript v in qv denotes that heat is supplied at constant volume.
Isothermal and Free Expansion of an Ideal Gas
For isothermal ( T = constant) expansion of an ideal gas into vaccum ; w = 0 since Pex =0. Also, Joule determined experimentally that q = 0 ; therefore DU = 0.
Equation 1 , DU = q + w can be expressed for isothermal irreversible and reversible changes as follows :
(i) For isothermal irreversible change
(ii) For isothermal irreversible change :
(iii) For adiabatic change , q = 0,
Electrical work
This type of work is significant in electrochemical cells, where redox reaction occurs and charge is carried through the external circuit. If Q is the charge flowing through the conductor and E is the potential difference across the conductor, then electric work is given by the expression :
Welectric = E Q
Electrical work is also known as non-expansion work.
Difference between heat and work
Let us compare the effect of addition of heat and work on the gaseous system.
(i) When heat is added to a gas, its molecules move faster in all the directions at random. Thus, heat can be regarded as a mode which stimulates the random motion.
(ii) When the piston compresses the gas in order to do work on it, the initial effect is to force the molecules in the direction of piston’s movement. Thus, work may be regarded as the mode that stimulates the organized motion.
In simple words, heat refers to random form of energy, whereas work refers to organized form of energy.
MATHEMATICAL FORMULATION OF FIRST LAW OF THERMODYNAMICS
Law of conservation of energy is also known as First Law of Thermodynamics. In order to derive a mathematical expression for first law of thermodynamics, let us assume that a system having internal energy U1 absorbs a certain amount of heat energy (q) .
\Its internal energy becomes = q + U1
Let 'w' amount of work be done on it, so that internal energy changes to U2.
U2 = U1 + q + w
U 2 - U1 = q + w
D U = q + w ………..(1)
i.e., 
Let the change in volume during the process be DV at constant pressure P, then the work is referred to as pressure-volume work and its expression is given as - PDV.
Thus,
w = - PDV ………..(2)
Now if there is only pressure-volume work , the expression (1) can be written as :
D U = q - PDV ………..(3)
It may be concluded that according to latest international conventions:-
Heat absorbed by the system = q is positive
Heat evolved by the system = q is negative
Work done on the system = w positive
Work done by the system = w negative.
Problems
5. Express the change in internal energy of a system when :
i) No heat is absorbed by the system from surroundings, but work(w) is done on the system. What type of wall does the system have ?
ii) No work is done on the system , but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have ?
iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be ?
Answer : 5
i) 
U = wad , wall will be adiabatic.
U = wad , wall will be adiabatic.ii) U = - q , thermally conducting walls.
U = - q , thermally conducting walls.iii) U = q - w , closed system
U = q - w , closed system6. Sate whether each of the following will increase or decrease the total energy content of the system:
(i) heat transferred to the surroundings
(ii) work done by the system
(iii) work done on the system
(iv) heat transferred from surrounding to the system
Answer : 6
i) When heat is transferred to the surroundings, the system loses a part of energy in the form of heat. Hence, the total energy content of the system decreases.
ii) Work done by the system involves loss of energy in the form of work. Hence total energy content of the system decreases.
iii) Work done on the system involves gain of energy. Hence , the total energy content of the system increases.
iv) The system gains energy resulting in an increase in total energy content of the system.
7. A system does 200 J of work and at the same time absorbs 150 J of heat. What is the internal energy change ?
Answer : 7
Work done w = - 200 J
Heat absorbed q = + 150 J
According to First Law of Thermodynamics :
D U = q + w
= + 150 J - 200 J = - 50 J
8. A system gives out 20 J of heat and also does 40 J of work.
Answer : 8
q = - 20 J
w = - 40 J
D U = q + w
= - 20 + w = - 20 + (- 40 ) = - 60 J
9. 200 J of work is done on the system and at the same time 140 J of heat is given out. What is the change in internal energy ?
Answer 8
w = + 200 J
q = - 140 J
D U = q + w
= - 140 J + 200 J = + 60 J
10. In a process 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process ?
Answer : 10
q = + 701 J
w = - 394 J
D U = q + w
= + 701 J - 394 J
= 397 J
11. A gas absorbs 125 J of heat and expands against the external pressure of 1.2 atm from a volume of 0.5 L to 1.0 L. What is the change in internal energy ?
Answer 11
w = - P V
V P = 1.2 atm
V = 1 - 0.5 = 0.5 L w = - P V
V = - (1.2 atm) (0.5 L) = - 0.6 L atm
= - (0.6 L atm) 
= - 60.78 J
= - 60.78 J12. Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vaccum until its total volume is 10 litres. How much heat is absorbed and how much work is dine in the expansion ?
Answer 12
q = - w = Pex ( 10 – 2 ) = 0 ( 8 ) = 0
No work is done ; no heat is absorbed.
13. Consider the same expansion(Problem 12) , but this time against a constant external pressure of 1 atm.
Answer 13
q = - w = Pex ( 10 – 2 ) = 1 ( 8 ) = 8 L atm =
= (8 L atm) = 810.4 J
= 810.4 J14. Consider the same expansion(Problem 12) , to a final volume of 10 litres conducted reversibly.
We have ,
q = - w = 2.303 x 10 log 
= 2.303 x 10 log 5
= 2.303 x 10 log 5 = 2.303 x 10 x 0.6990
= 16.098 L atm
= (16.098 L atm)
=1630.72 J
Significance of DU
From the First law of Thermodynamics:
D U = q - PDV
q = D U + PDV
If during the change the volume remains constant, then no work is done, i.e., PDV = 0 . Hence,
qv = D U
Thus change in internal energy D U represents the heat change taking place during the process occurring at constant volume and constant temperature.