Chapter 21 Thermodynamics – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 21 Thermodynamics – Physics free study material by TEACHING CARE online tuition and coaching classes
Introduction.
 Thermodynamics : It is a branch of science which deals with exchange of heat energy between bodies and conversion of the heat energy into mechanical energy and viceversa.
 Thermodynamic system : A collection of an extremely large number of atoms or molecules confined with in certain boundaries such that it has a certain value of pressure, volume and temperature is called a thermodynamic Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings.
Example : Gas enclosed in a cylinder fitted with a piston forms the thermodynamic system but the atmospheric air around the cylinder, movable piston, burner etc. are all the surroundings.
Thermodynamic system may be of three types
 Open system : It exchange both energy and matter with the
 Closed system : It exchange only energy (not matter) with the
 Isolated system : It exchange neither energy nor matter with the
 Thermodynamic variables and equation of state : A thermodynamic system can be described by specifying its pressure, volume, temperature, internal energy and the number of moles. These parameters are called thermodynamic variables. The relation between the thermodynamic variables (P, V, T) of the system is called equation of
For m moles of an ideal gas, equation of state is PV = mRT and for 1 mole of an it ideal gas is PV = RT
æ am 2 ö
æ a ö
For m moles of a real gas, equation of state is ç P + ÷ (V – m b) = m RT and for 1 mole of a real gas it is ç P + ÷(V – b) = RT
V 2 V 2
è ø è ø
 Thermodynamic equilibrium : When the thermodynamic variables attain a steady value e. they are independent of time, the system is said to be in the state of thermodynamic equilibrium. For a system to be in thermodynamic equilibrium, the following conditions must be fulfilled.
 Mechanical equilibrium : There is no unbalanced force between the system and its
 Thermal equilibrium : There is a uniform temperature in all parts of the system and is same as that of
 Chemical equilibrium : There is a uniform chemical composition through out the system and the
 Thermodynamic process : The process of change of state of a system involves change of thermodynamic variables such as pressure P, volume V and temperature T of the system. The process is known as thermodynamic Some important processes are
(i) Isothermal process (ii) Adiabatic process (iii) Isobaric process (iv) Isochoric (isovolumic process)
 Cyclic and noncyclic process (vi) Reversible and irreversible process
Zeroth Law of Thermodynamics.
If systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.
 The zeroth law leads to the concept of temperature. All bodies in thermal equilibrium must have a common property which has the same value for all of This property is called the temperature.
 The zeroth law came to light long after the first and seconds laws of thermodynamics had been discovered and numbered. Because the concept of temperature is fundamental to those two laws, the law that establishes temperature as a valid concept should have the lowest Hence it is called zeroth law.
Quantities Involved in First Law of Thermodynamics.
 Heat (DQ) : It is the energy that is transferred between a system and its environment because of the temperature difference between them. Heat always flow from a body at higher temperature to a body at lower temperature till their temperatures become
 Heat is a form of energy so it is a scalar quantity with dimension [ML^{2}T ^{2} ] .
 Unit : Joule (S.I.), Calorie (practical unit) and 1 calorie = 2 Joule
 Heat is a path dependent quantity g. Heat required to change the temperature of a given gas at a constant pressure is different from that required to change the temperature of same gas through same amount at constant volume.
 For solids and liquids : DQ = mL [For change in state] and DQ = mcDT [For change in temperature] For gases when heat is absorbed and temperature changes :
(DQ)V
= mCV DT
[For constant volume] and
(DQ)P
= mCP DT
[For constant pressure]
 Work (DW) : Work can be defined as the energy that is transferred from one body to the other owing to a force that acts between them
If P be the pressure of the gas in the cylinder, then force exerted by the gas on the piston of the cylinder F = PA
In a small displacement of piston through dx, work done by the gas
dW = F.dx = PA dx = P dV
\ Total amount of work done DW =
ò dW =
Vf P dV = P(V


Vi
 Vi )
 Like heat, work is also a path dependent, scalar physical quantity with dimension [ML^{2}T ^{2} ]
 From DW = PDV = P(Vf – Vi )
DW = positive if Vf
DW = negative if Vf
 Vi e. system expands against some external force.
< Vi i.e. system contracts because of some external force exerted by the surrounding.
 In PV diagram or indicator diagram, the area under PV curve represents work
W = area under PV diagram
It is positive if volume increases (for expansion)
It is negative if volume decreases (for compression)
 In a cyclic process work done is equal to the area under the It is positive if the cycle is clockwise.
It is negative if the cycle is anticlockwise.

 W = Vf P dV
Vi
From this equation it seems as if work done can be calculated only when PV equation is known and limits Vi
and Vf
are known to us. But it is not so. We can calculate work done if we know the limits of temperature.
For example, the temperature of n moles of an ideal gas is increased from T_{0}
to 2T0
through a process
P = a
T
and we are interested in finding the work done by the gas. Then
PV = nRT (ideal gas equation) …..(i)
and
P = a
T
…..(ii)
Dividing (i) by (ii), we get
V = nRT ^{2}
a
or dV = 2nRT dT
a



\ W = òVf P dV = ò2T0 æ a ö æ 2nRT ö dT = 2nRT
Vi T0
è T ø è a ø
So we have found the work done without putting the limits of volume.
 If mass less piston is attached to a spring of force constant K and a mass m is placed over the If the
external pressure is
P_{0} and due to expansion of gas the piston moves up through a distance x then
Total work done by the gas W = W1 + W2 + W3
where W_{1} = Work done against external pressure (P_{0} )
W_{2} = Work done against spring force (Kx)
W_{3} = Work done against gravitational force (mg)
\ W = P V + 1 Kx ^{2} + mgx
0 2
 If the gas expands in such a way that other side of the piston is vacuum then work done by the gas will be zero
As W = PDV = 0 [Here P = 0]
 Internal energy (U) : Internal energy of a system is the energy possessed by the system due to molecular motion and molecular
The energy due to molecular motion is called internal kinetic energy U_{K} and that due to molecular configuration is called internal potential energy U_{P}.
i.e. Total internal energy U = UK + UP
 For an ideal gas, as there is no molecular attraction U p = 0
i.e. internal energy of an ideal gas is totally kinetic and is given by U = UK
and change in internal energy DU = 3 mR DT
2
 In case of gases whatever be the process
= 3 mRT
2
DU = m f RDT
= mC
DT = m R DT = m R(Tf – Ti ) = mRTf – mRTi
= (Pf Vf – Pi Vi )
2 ^{V} (g – 1)
g – 1
g – 1
g – 1
 Change in internal energy does not depends on the path of the So it is called a point function i.e.
it depends only on the initial and final states of the system, i.e. DU = Uf – Ui
 Change in internal energy in a cyclic process is always zero as for cyclic process Uf= Ui
So DU = Uf – Ui = 0
Problem 1. A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
[Orissa JEE 2002]
 20 J
 – 20 J
 400 J
(d) – 374 J
Solution : (b) Work done by the system = Area of shaded portion on PV diagram
= (300 – 100)10 ^{–}^{6} ´ (200 – 10) ´ 10^{3} = 20 J
and direction of process is anticlockwise so work done will be negative i.e. DW = – 20 J.
Problem 2. An ideal gas is taken around ABCA as shown in the above PV diagram. The work done during a cycle is
[KCET (Engg./Med.) 2001]
 2PV
 PV
 1/2PV
 Zero
Solution : (a) Work done = Area enclosed by triangle ABC = 1 AC ´ BC =
2
1 ´ (3V – V) ´ (3P – P) = 2 PV
2
Problem 3. The PV diagram shows seven curved paths (connected by vertical paths) that can be followed by a gas. Which two of them should be parts of a closed cycle if the net work done by the gas is to be at its maximum value [AMU (Engg.) 2000]
 ac
 cg
 af
 cd
Solution : (c) Area enclosed between a and f is maximum. So work done in closed cycles follows a and f is maximum.
Problem 4. If Cv = 4.96 cal / mole K , then increase in internal energy when temperature of 2 moles of this gas is
increased from 340 K to 342 K [RPET 1997]
(a) 27.80 cal (b) 19.84 cal (c) 13.90 cal (d) 9.92 cal
Solution : (b) Increase in internal energy
DU = m.Cv .DT = 2 ´ 4.96 ´ (342 – 340) = 2 ´ 4.96 ´ 2 = 19.84 cal
Problem 5. An ideal gas of mass m in a state A goes to another state B via three different processes as shown in figure. If
Q1, Q2 and Q3
denote the heat absorbed by the gas along the three paths, then [MP PET 1992]
 Q1 < Q2 < Q3
 Q1 < Q2 = Q3
 Q1 = Q2 > Q3
Q1 > Q2
 Q3
Solution : (a) Area enclosed by curve 1 < Area enclosed by curve 2 < Area enclosed by curve 3
\ Q1 < Q2 < Q3
[As DU is same for all curves]
Problem 6. The relation between the internal energy U and adiabatic constant g is
(a)
U = PV
g – 1
(b)
U = PV ^{g}
g – 1
(c)
U = PV
g
(d)
g

U PV
Solution : (a) Change in internal energy = DU = m cv DT Þ U 2 – U1 = m cv (T2 – T1 )
Let initially T1 = 0 so U1 = 0 and finally T2 = T and U 2 = U
U = m cv T = m T ´ cv = PV ´ R = PV
[As PV = m RT
\ mT = PV and cv = R ]
Joule’s Law.
R g – 1 g – 1
R g – 1
Whenever heat is converted into mechanical work or mechanical work is converted into heat, then the ratio of work done to heat produced always remains constant.
i.e. W µ Q or
W = J Q
This is Joule’s law and J is called mechanical equivalent of heat.
 From W = JQ if Q = 1 then J = W. Hence the amount of work done necessary to produce unit amount of heat is defined as the mechanical equivalent of
 J is neither a constant, nor a physical quantity rather it is a conversion factor which used to convert Joule
or erg into calorie or kilo calories viceversa.
 Value of J = 2
J
calorie
= 4.2 ´ 10^{7}
erg calorie
= 4.2 ´ 10^{3}
J .
kilocalorie
 When water in a stream falls from height h, then its potential energy is converted into heat and temperature of water rises
From W = JQ
mgh = J ms Dt [where m = Mass, s = Specific heat of water]
\Rise in temperature Dt = gh °C
Js
 The kinetic energy of a bullet fired from a gun gets converted into heat on striking the By this heat the temperature of bullet increases by Dt.
From W = JQ
1 mv^{2} = J ms Dt
2
[where m = Mass, v = Velocity of the bullet, s = Specific heat of the bullet]
\ Rise in temperature Dt =
v 2 °C
2Js
If the temperature of bullet rises upto the melting point of the bullet and bullet melts then. From W = JQ
1 mv^{2} = J[ms Dt + mL] 2
[where L = Latent heat of bullet]
\ Rise in temperature
Dt = ìïé v 2

îë
– Lù

û
süï°C

ïþ
 If iceblock falls down through some height and melts partially then its potential energy gets converted into heat of
From W = JQ
mgh = J m‘ L
So m‘ = mgh kg
JL
[where m = mass of ice block, m^{1} = mass which melts]
If iceblock completely melts down then mgh = J mL
\ Height required for complete melting h = JL meter
g
Problem 7. Water falls from a height of 210 m. Assuming whole of energy due to fall is converted into heat the rise in temperature of water would be (J = 4.3 Joule/cal) [Pb. PMT 2002]
(a) 42°C (b) 49°C (c) 0.49°C (d) 4.9°C
Solution : (c) Loss in potential energy of water = Increment in thermal energy of water
Þ mgh = J ´ msDt
Þ 9.8 ´ 210 = 4.3 ´ 1000Dt \
Dt = 0.49°C
Problem 8. A block of mass 100 gm slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s
to 5 m/s, the thermal energy developed in the process is [UPSEAT 2002]
(a) 3.75 J (b) 37.5 J (c) 0.375 J (d) 0.75 J
Solution : (a) Thermal energy developed = Loss in kinetic energy = 1 m(v^{2} – v^{2}) 1 0.1´(10^{2} – 5^{2}) = 3.75 J
2 2 1 = 2 ´
Problem 9. The weight of a person is 60 kg. If he gets 10^{5} calories heat through food and the efficiency of his body is 28%, then upto how much height he can climb (approximately) [AFMC 1997]
(a) 100 m (b) 200 m (c) 400 m (d) 1000 m
Solution : (b) Increment in potential energy of mass = 28% of heat gained
Þ mgh = J æ 28 ´ 10^{5} ö Þ 60 ´ 9.8 ´ h = 4.2 ´ æ 28 ´ 10^{5} ö
Þ h = 200 m




ç 100 ÷ ç 100 ÷
Problem 10. Hailstone at 0°C falls from a height of 1 km on an insulating surface converting whole of its kinetic energy into
heat. What part of it will melt (g = 10 m / s ^{2} )
[MP PMT 1994]
(a)
1 (b)
33
1 (c)
8
1 ´ 10^{–}^{4}
33
 All of it will melt
Solution : (a) Energy required for melting = Loss in potential energy Þ J ´ m¢L = mgh
Þ 4.18 ´ (m‘ ´ 80 ´ 10 ^{–}^{3} ) = m ´ 10 ´ 1000 Þ
m‘ = 1
[As L = 80 ´ 10^{–3} calorie/kg]
m 33
Problem 11. A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be
m, specific heat S, initial temperature 25°C, melting point 475°C and the latent heat L. Then
(a)
(c)
mL = ms (475 – 25) + mv 2
2J
ms (475 – 25) + mL = mv2
J
(b)
(d)
ms (475 – 25) + mL = mv2
2J
ms (475 – 25) – mL = mv2
2J
Solution : (b) K.E. of bullet = Heat required to raise the temperature of bullet from 25°C to 475°C + heat required to melt the bullet
Þ 1 mv ^{2} = J[ms(475 – 25) + mL] Þ ms(475 – 25) + mL = mv 2 .
2 2J
Problem 12. A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking (M.P. of lead = 327°C, specific heat of lead = 0.03 cal/gm°C, latent heat of fusion of lead = 6 cal/gm and J = 4.2 J/cal) [IIT–JEE 1981]
 410 m/sec (b) 1230 m/sec (c) 5 m/sec (d) None of these
Solution : (a) Using expression obtained in problem (11) we get 75%æ 1 mv ^{2} ö = J [ms(327 – 27 + mL]
ç
è
Substituting s = 0.3 ´ 10^{3} cal/kg ^{o} C and L = 6 ´ 10^{3} cal/kg
÷

ø
we get v = 410 m / s
Problem 13. A drilling machine of power P watts is used to drill a hole in Cu block of mass m kg. If the specific heat of Cu is
5 J kg ^{–}^{1}°C^{–}^{1}
(will be in °C)
and 40% of power lost due to heating of machine the rise in temperature of the block in T sec
(a)
0.6 PT ms
0.6 P msT
0.4 PT ms
 P msT
Solution : (a) As we know Power (P) = Work (W)
Time (T)
\ W = P ´ T
As 40% energy is lost due to heating of machine so only 60% energy will increase the temperature of the block
\ 60% of W = m ´ s ´ Dt Þ Dt = 0.6 W = 0.6PT .
First Law of Thermodynamics.
ms ms
It is a statement of conservation of energy in thermodynamical process.
According to it heat given to a system (DQ) is equal to the sum of increase in its internal energy (DU) and the work done (DW) by the system against the surroundings.
DQ = DU + DW
 It makes no distinction between work and heat as according to it the internal energy (and hence temperature) of a system may be increased either by adding heat to it or doing work on it or
 DQ and DW are the path functions but DU is the point
 In the above equation all three quantities DQ, DU and DW must be expressed either in Joule or in calorie.
 Just as zeroth law of thermodynamics introduces the concept of temperature, the first law introduces the concept of internal
 Sign conventions
DQ  Positive  When heat is supplied to a system 
Negative  When heat is drawn from the system  
DW  Positive  When work done by the gas (expansion) 
Negative  When work done on the gas (compression)  
DU  Positive  When temperature increases, internal energy increases 
Negative  When temperature decreases, internal energy decreases 
 When a thermos bottle is vigorously shaken :
No heat is transferred to the coffee DQ = 0 [As thermos flask is insulated from the surrounding] Work is done on the coffee against viscous force DW = (–)
Internal energy of the coffee increases DU = (+)
and temperature of the coffee also increases DT = (+)
 Limitation : First law of thermodynamics does not indicate the direction of heat It does not tell anything about the conditions, under which heat can be transformed into work and also it does not indicate as to why the whole of heat energy cannot be converted into mechanical work continuously.
Problem 14. If 150 J of heat is added to a system and the work done by the system is 110 J, then change in internal energy will be [AMU (Engg.) 1999; BHU 2000]
(a) 260 J (b) 150 J (c) 110 J (d) 40 J
Solution : (d)
DQ = DU + DW
Þ 150 = DU + 110
Þ DU = 40 J
Problem 15. 110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is [CBSE PMT 1993; AFMC 1999; JIPMER 2000]
(a) 150 J (b) 70 J (c) 110 J (d) 40 J
Solution : (b)
DQ = DU + DW
Þ 110 = 40 + DW
Þ DW = 70 J
Problem 16. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas, is [IITJEE 1990; RPET 2000]
 25
3 (c)
5
3 (d) 5
7 7
Solution : (d)
DU = C_{v} DT =
R /(g – 1) = 1 = 1 = 5
DQ CP DT
g R /(g – 1) g
7 / 5 7
Problem 17. An electric fan is switched on in a closed room. The air in the room is [MP PET 1996]
 Cooled
 Heated
 Maintains its temperature
 Heated or cooled depending on the atmospheric pressure
Solution : (b) When an electric fan is switched on in a closed room conventional current of air flows. Hence due to viscous force mechanical energy is converted into heat and some heat is also produced due to thermal effect of electric current in motor of fan.
Problem 18. A gas is compressed at a constant pressure of
50 N / m^{2}
from a volume of
10 m^{3}
to a volume of
4 m^{3} .
Energy of 100 J is then added to the gas by heating. Its internal energy is [MNR 1994]
 Increased by 400 J (b) Increased by 200 J (c) Increased by 100 J (d) Decreased by 200 J
Solution : (a)
DQ = DU + DW
Þ DQ = DU + DdV Þ 100 = DU + 50(4 – 10) Þ 100 = DU – 300
\ DU = 400 J
Problem 19. A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in
the figure are : PA = 3 ´ 10^{4} Pa, PB = 8 ´ 10^{4} Pa
and
VA = 2 ´ 10 ^{–}^{3} m^{3} , VD = 5 ´ 10 ^{–}^{3} m^{3}
In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be [CBSE PMT 1992]
 560 J
 800 J
 600 J
 640 J
Solution : (a) By adjoining graph WAB = 0
and WBC = 8 ´ 10^{4} [5 – 2] ´ 10 ^{–}^{3} = 240 J
\ WAC = WAB + WBC = 0 + 240 = 240 J
Now, DQAC = DQAB + DQBC = 600 + 200 = 800 J
From first law of thermodynamics DQAC = DUAC + DWAC
Þ 800 = DUAC + 240 Þ DUAC = 560 J.
Problem 20. If R = universal gas constant, the amount of heat needed to raise the temperature of 2 mole of an ideal monoatomic gas from 273 K to 373 K when no work is done [MP PET 1990]
(a) 100 R (b) 150 R (c) 300 R (d) 500 R
æ R ö R 5
Solution : (c) DQ = DU = m Cv DT = m ç g – 1 ÷ DT = 2 ´ 5
[373 – 273] = 300 R [As for monoatomic gas g = ]
3
è ø – 1
3
Isothermal Process.
When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as isothermal changes.
In this process, P and V change but T = constant i.e. change in temperature DT = 0
(1) Essential condition for isothermal process
 The walls of the container must be perfectly conducting to allow free exchange of heat between the gas and its
 The process of compression or expansion should be so slow so as to provide time for the exchange of heat. Since these two conditions are not fully realised in practice, therefore, no process is perfectly
 Equation of state : From ideal gas equation PV = mRT
If temperature remains constant then PV= constant i.e. in all isothermal process Boyle’s law is obeyed. Hence equation of state is PV = constant.
(3) Example of isothermal process
 Melting process [Ice melts at constant temperature 0°C]
 Boiling process [water boils at constant temperature 100°C].
(4) Indicator diagram
 Curves obtained on PV graph are called isotherms and they are hyperbolic in
 Slope of isothermal curve : By differentiating PV = We get
P dV + V dP = 0
Þ P dV = –V dP
Þ dP dV
= – P
V
\ tanq = dP dV
= – P
V
 Area between the isotherm and volume axis represents the work done in isothermal If volume increases DW = + Area under curve and if volume decreases DW = – Area under curve
 Specific heat : Specific heat of gas during isothermal change is
As C =
Q
mDT
= Q = ¥
m ´ 0
[As DT = 0]
 Isothermal elasticity : For isothermal process PV = constant
Differentiating both sides
PdV + VdP = 0 Þ
P dV = –V dP
Þ P = dP = Stress = Eq
\ Eq = P
i.e. isothermal elasticity is equal to pressure
 dV / V
Strain
At N.T.P. isothermal elasticity of gas = Atmospheric pressure = 1.01 ´ 10^{5} N / m^{2}
(7) Work done in isothermal process
W = òVf P dV = òVf mRT dV
[As PV = mRT]
Vi Vi V
æ V_{f} ö
æ V_{f} ö
W = mRT log e ç V ÷ = 2.303 mRT log10 ç V ÷
è i ø è i ø
æ Pi ö æ Pi ö
or W = mRT log e ç ÷ = 2.303mRT log10 ç ÷
ç Pf ÷ ç Pf ÷
è ø è ø
(8) FLTD in isothermal process
DQ = DU + DW
\ DU = 0
but
DU µ DT
[As DT = 0]
\ DQ = DW i.e. heat supplied in an isothermal change is used to do work against external surrounding.
or if the work is done on the system than equal amount of heat energy will be liberated by the system.
Problem 21. One mole of O2
gas having a volume equal to 22.4 litres at 0°C and 1 atmospheric pressure in compressed
isothermally so that its volume reduces to 11.2 litres. The work done in this process is [MP PET 1993; BVP 2003]
(a) 1672.5 J (b) 1728 J (c) – 1728 J (d) – 1572.5 J
Solution : (d) Work done in an adiabatic process
æ V_{f} ö
æ 11.2 ö
W = m RT log _{e} ç V ÷ = 1 ´ 8.3 ´ 273 ´ log _{e} ç 22.4 ÷ = 8.3 ´ 273 ´ ( 0.69) » 1572 J
è i ø è ø
Problem 22. An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy [CBSE PMT 1993; JIPMER 2001, 2002]
(a) Will be same in both A and B (b) Will be zero in both the gases
(c) Of B will be more than that of A (d) Of A will be more than that of B
Solution : (c) In real gases an additional work is also done in expansion due to intermolecular attraction.
Problem 23. Which of the following graphs correctly represents the variation of b = (dV / dP) / V
with P for an ideal gas at
constant temperature [IITJEE (Screening) 2002]
 b
 b
b
 b
P P P P
Solution : (a) For an isothermal process PV = constant Þ PdV + VdP = 0 Þ – 1 æ dV ö = 1
ç ÷
So,
b = 1
P
\ graph will be rectangular hyperbola.
V è dP ø P
Problem 24. Consider the following statements
Assertion (A): The internal energy of an ideal gas does not change during an isothermal process
Reason (R) : The decrease in volume of a gas is compensated by a corresponding increase in pressure when its temperature is held constant.
Of these statements [SCRA 1994]
 Both A and R are true and R is a correct explanation of A
 Both A and R are true but R is not a correct explanation of A
 A is true but R is false
 Both A and R are false
 A is false but R is true
Solution : (b) As DU µ DT so for isothermal process internal energy will not change and also P µ 1
V
(Boyle’s law)
Problem 25. How much energy is absorbed by 10 kg molecule of an ideal gas if it expands from an initial pressure of 8 atm
to 4 atm at a constant temperature of 27°C [Roorkee 1992]
(a)
1.728 ´ 10^{7} J (b)
17.28 ´ 10^{7} J (c)
1.728 ´ 10^{9} J (d)
17.28 ´ 10^{9} J
Solution : (a) Work done in an isothermal process
æ Pi ö 3
æ 8 ö _{4} _{7}
W = m RT log e ç ÷ = (10 ´ 10
) ´ 8.3 ´ 300 ´ log _{e} ç ÷ = 10 ´ 8.3 ´ 300 ´ 0.693 = 1.728 ´ 10 J
ç Pf ÷
è 4 ø
è ø
Problem 26. 5 moles of an ideal gas undergoes an isothermal process at 500K in which its volume is doubled. The work done by the gas system is
(a) 3500 J (b) 14400 J (c) 17800 J (d) 5200 J
æ V_{f} ö
æ 2V ö
Solution : (b)
DW = m RT log _{e} ç V ÷ = 5 ´ 8.3 ´ 500 ´ log _{e} ç V ÷ = 5 ´ 8.3 ´ 500 ´ 0.69 » 14400 J.
è i ø è ø
Problem 27. Work done by a system under isothermal change from a volume V_{1}
to V2
for a gas which obeys Vander
æ
Waal’s equation (V – bn)ç P +
an^{2} ö

= nRT

ç ÷
è ø
æ V2 – nb ö _{2}æ V1 – V2 ö
æ V2 – ab ö _{2}æ V1 – V2 ö
(a)
nRT loge ç V – nb ÷ + an ç V V ÷
(b)
nRT log10 ç V – ab ÷ + an ç V V ÷
è 1 ø
è 1 2 ø
è 1 ø
è 1 2 ø
æ V2 – na ö
_{2}æ V1 – V2 ö
æ V1 – nb ö
_{2}æ V1V2 ö
(c)
nRT loge ç V
– na ÷ + bn ç V V ÷
(d)
nRT loge ç V
÷ + an ç ÷
 nb V – V
è 1 ø
è 1 2
nRT
ø
an^{2}
è 2 ø
è 1 2 ø
Solution : (a) By Vander Waal’s equation P = V – nb – V 2
Work done, W =
V2 V2 dV



PdV nRT
 an2 V2 dV
òV òV
V – nb

V
òV V 2

_{2} é 1 ù ^{V}2
V2 – nb
_{2} æ V1 – V2 ö

Adiabatic Process.
= nRT [log e (V – nb )] 2 + an
êë V úû V
= nRT log e
V1 – nb
+ an ç
è
÷
V1 V2 ø
When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between it and the surroundings, the process is known as adiabatic process.
In this process P, V and T changes but DQ = 0.
(1) Essential conditions for adiabatic process
 There should not be any exchange of heat between the system and its All walls of the container and the piston must be perfectly insulating.
 The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its
Since, these two conditions are not fully realised in practice, so no process is perfectly adiabatic.
(2) Example of some adiabatic process
 Sudden compression or expansion of a gas in a container with perfectly nonconducting
 Sudden bursting of the tube of bicycle
 Propagation of sound waves in air and other
 Expansion of steam in the cylinder of steam
 FLTD in adiabatic process : DQ = DU + DW
but for adiabatic process DQ = 0 \ DU + DW = 0
If DW = positive then DU = negative so temperature decreases i.e. adiabatic expansion produce cooling. If DW = negative then DU = positive so temperature increases i.e. adiabatic compression produce heating.
 Equation of state : As in case of adiabatic change first law of thermodynamics reduces to,
DU + DW = 0 , i.e., dU + dW = 0 ….. (i)
But as for an ideal gas dU = mCV dT
and dW = PdV
Equation (i) becomes
mCvdT + P dV = 0
….. (ii)
But for a gas as PV = mRT,
P dV + V dP = mRdT
….. (iii)
So eliminating dT between equation (ii) and (iii)
mC (P dV + V dP) + P dV = 0
^{v} mR
or (P dV + V dP) + P dV = 0
éas C = R ù
(g – 1)

or gP dV + V dP = 0
i.e.,
ê v
ë
g dV + dP = 0
(g – 1)ú
V P
Which on integration gives
g log e V + log e P = C ,
i.e.,
log(PV ^{g} ) = C
or PV ^{g}
= constant
….. (iv)
Equation (iv) is called equation of state for adiabatic change and can also be rewritten as
TV ^{g} ^{1} = constant [as P = (mRT/V)] ….. (iv)
and
T g = constant
éas V = mRT ù
….. (vi)
P g 1
(5) Indicator diagram
ëê P úû
 Curve obtained on PV graph are called adiabatic
 Slope of adiabatic curve : From PV ^{g} = constant
By differentiating, we get
dP V ^{g}
+ PgV ^{g} ^{1} dV = 0
dP = –g
dV
PV g 1
V g
= –g æ P ö

V
è ø
\ Slope of adiabatic curve
tanf = –g æ P ö

V
è ø
But we also know that slope of isothermal curve tanf = – P
V
So,
Slope of adiabatic curve = – g (P / V) = g
= Cp > 1
Slope of isothermal curve
 (P / V) Cv
 Specific heat : Specific heat of a gas during adiabatic change is zero
As C = Q = 0 = 0
[As Q = 0]
mDT mDT
 Adiabatic elasticity : For adiabatic process
PV ^{g}
= constant
Differentiating both sides
d PV ^{g}
+ PgV ^{g} ^{1}dV = 0
gP = dP = Stress = Ef
 dV / V Strain
Ef = gP
i.e. adiabatic elasticity is g times that of pressure but we know isothermal elasticity Eq = P
So Ef
Eq
= Adiabatic elasticity Isothermal elasticity
= gP = g
P
i.e. the ratio of two elasticities of gases is equal to the ratio of two specific heats.
(8) Work done in adiabatic process
W = Vf P dV =
Vf K dV
éAs æ P = K öù


òV òV V g
ê ç g ÷ú

ë è øû
or = 1 é K – K ù
éAs
V ^{–}^{g} dV =
V –g +1 ù
[1 – g ] ê V ^{g} ^{1}
V g 1 ú
ê ò (g + 1)ú
ëê _{f}
i úû ë û
or = [Pf Vf – Pi Vi ]
[As K = PV ^{g} = P V ^{g}
= P V ^{g} ]
(1 – g )
or = mR [T – T ] (1 – g ) ^{f} ^{i}
[As Pf Vf = mRTf
f f i i
and Pi Vi = mRTi ]
So = [Pi Vi – Pf Vf ] = mR(Ti – Tf )
(g – 1) (g – 1)
 Free expansion : Free expansion is adiabatic process in which no work is performed on or by the system. Consider two vessels placed in a system which is enclosed with thermal insulation (asbestoscovered). One vessel contains a gas and the other is evacuated. The two vessels are connected by a stopcock. When suddenly the stopcock is opened, the gas rushes into the evacuated vessel and expands The process is adiabatic as the vessels are placed in thermal insulating system (dQ = 0) moreover, the walls of the vessel are rigid and hence no external work is performed (dW = 0).
Now according to the first law of thermodynamics dU = 0
If Ui
and U f
be the initial and final internal energies of the gas
then
Uf – Ui = 0
[As Uf
= Ui ]
Thus the final and initial energies are equal in free expansion.
(10) Special cases of adiabatic process
g g
PV ^{g}
= constant
\ P µ 1 ,
V g
PT 1g
= constant
\ P µ T ^{g} ^{1}
and
TV ^{g} ^{1} = constant
\ T µ
1
V g 1
(11) Comparison between isothermal and adiabatic process
 Compression : If a gas is compressed isothermally and adiabatically from volume V_{i} to V_{f} then from the slope of the graph it is clear that graph 1 represents adiabatic process where as graph 2 represent isothermal
Work done  Wadiabatic > Wisothermal 
Final pressure  Padiabatic > Pisothermal 
Final temperature  Tadiabatic > Tisothermal 
 Expansion : If a gas expands isothermally and adiabatically from volume V_{i} to V_{f} then from the slope of the graph it is clear that graph 1 represent isothermal process, graph 2 represent
adiabatic process.
Work done  Wisothermal > Wadiabatic 
Final pressure  Pisothermal > Padiabatic 
Final temperature  Tisothermal > Tadiabatic 
Problem 28. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp / Cv for the gas is [AIEEE 2003]
(a)
3 (b)
2
4 (c) 2 (d) 5
3 3
g
Solution : (a) Given P µ T ^{3} . But for adiabatic process P µ T ^{g} ^{–}^{1} . So, g
= 3 Þ g = 3
Þ Cp = 3
g – 1 2 C_{v} 2
Problem 29. An ideal gas at 27°C is compressed adiabatically to 8
27
of its original volume. If
g = 5 , then the rise in
3
temperature is [CPMT 1984; CBSE PMT 1999; KCET 2003; UPSEAT 2003]
(a) 450 K (b) 375 K (c) 225 K (d) 405 K
Solution : (b) For an adiabatic process TV ^{g} ^{–}^{1} = constant
é ùg 1
é V ùg 1
5 1
2 / 3
\ T1 = ê V2 ú
Þ T = T ê 1 ú
= 300é 27 ù 3
= 300é 27 ù
= 675 K
Þ DT = 675 – 300 = 375 K
T2 ë V1 û
^{2} ^{1} ë V2 û
êë 8 úû
êë 8 úû
Problem 30. If g = 2.5 and volume is equal to 1/8 times to the initial volume then pressure P¢ is equal to (initial pressure = P)
[RPET 2003]
(a)
P¢ = P
(b)
P¢ = 2P
P é V
(c)
ùg
P¢ = P ´(2)^{15} ^{/} ^{2}
P¢
(d)
P¢ = 7P
Solution : (c) For an adiabatic process PV ^{g}
= constant Þ ^{ } ^{2} = ê 1 ú Þ
= 85 / 2
Þ P ¢ = P ´ (2)^{15} ^{/} ^{2}
P1 ë V2 û P
Problem 31. In adiabatic expansion of a gas [MP PET 2003]
 Its pressure increases (b) Its temperature falls
(c) Its density decreases (d) Its thermal energy increases
Solution : (b)
DQ = DU + DW , In an adiabatic process DQ = 0 \ DU = DW
In expansion DW = positive \ DU = negative. Hence internal energy i.e. temperature decreases.
Problem 32. PV plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively to [IITJEE (Screening) 2001]
 He and O_{2}
 O_{2} and He
 He and Ar
 O_{2} and N_{2}
Solution : (b) Slope of adiabatic curve µ g µ
1 . So g is inversely proportional to atomicity of the gas.
Atomicity of the gas
[As g = 1.66 for monoatomic gas, g = 1.4 for diatomic gas and g = 1.33 for triatomic nonlinear gas.]
From the graph it is clear that slope of the curve 1 is less so this should be adiabatic curve for diatomic gas (i.e. O_{2}). Similarly slope of the curve 2 is more so it should be adiabatic curve for monoatomic gas (i.e. He).
Problem 33. A monoatomic ideal gas, initially at temperature T_{1} , is enclosed in a cylinder fitted with a frictionless piston.
The gas is allowed to expand adiabatically to a temperature T_{2}
by releasing the piston suddenly. If
L1 and
L2 are the lengths of the gas column before and after expansion respectively, then T1 / T2 is given by
[IITJEE (Screening) 2000]
æ L ö2 / 3 L
L æ L
ö2 / 3
(a)
ç 1 ÷
è L2 ø
 ^{1}
L2
T é V
(c)
ùg 1
2
L1
é L A ù5 / 31
(d)
é L ù2 / 3
ç 2 ÷
è L1 ø
Solution : (d) For an adiabatic process T1V ^{g} ^{–}^{1} = T2V ^{g} ^{–}^{1} Þ ^{ } ^{1} = ê 2 ú
= ê 2 ú
= ê 2 ú .
1 2 T2
ë V1 û
ë L1 A û
ë L1 û
Problem 34. Four curves A, B, C and D are drawn in the adjoining figure for a given amount of gas. The curves which represent adiabatic and isothermal changes are [CPMT 1986; UPSEAT 1999]
 C and D respectively
 D and C respectively
 A and B respectively
 B and A respectively
Solution : (c) As we know that slope of isothermal and adiabatic curves are always negative and slope of adiabatic curve is always greater than that of isothermal curve so is the given graph curve A and curve B represents adiabatic and isothermal changes respectively.
Problem 35. A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are P and T for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will [RPET 1999]
(a) Decrease (b) Increase (c) Remain same (d) None of these
Solution : (c) In second part there is a vacuum i.e. P = 0. So work done in expansion = PDV = 0
Problem 36. Two samples A and B of a gas initially at the same pressure and temperature are compressed from volume V
to V/2 (A isothermally and B adiabatically). The final pressure of A is [MP PET 1996, 99; MP PMT 1997, 99]
(a) Greater than the final pressure of B (b) Equal to the final pressure of B
(c) Less than the final pressure of B (d) Twice the final pressure of B
Solution : (c) For isothermal process P1V = P^{‘} V Þ P^{‘} = 2P
…..(i)
2 2 2 1
_{g} æ V ö^{g} _{g}
For adiabatic process P_{1}V = P_{2}ç 2 ÷ Þ P_{2} = 2 P_{1}
…..(ii)
è ø

Since g > 1 . Therefore P2 > P^{‘}
Problem 37. A gas has pressure P and volume V. It is now compressed adiabatically to 1
32
(32)^{1.4} = 128 , the final pressure is
times the original volume. If
(a) 32 P (b) 128 P (c)
P (d) P
128 32
P æ V ö^{g}
æ V ö1.4
Solution : (b) For an adiabatic process P1V ^{g}
= P2V ^{g}
Þ 2 = ç 1 ÷ = ç
÷ = (32)^{1.4}
= 128
1 2 P1 ç V2 ÷
è V / 32 ø
è ø
\ Final pressure = 128 P.
Isobaric Process.
When a thermodynamic system undergoes a physical change in such a way that its pressure remains constant, then the change is known as isobaric process.
In this process V and T changes but P remains constant. Hence Charle’s law is obeyed in this process.
 Equation of state : From ideal gas equation PV = mRT
If pressure remains constant V µ T or
V1 = V2
= constant
T1 T2
 Indicator diagram : Graph I represent isobaric expansion, graph II represent isobaric
Slope of indicator diagram
dP = 0
dV
 Specific heat : Specific heat of gas during isobaric process C
= æ f + 1öR
(4) Bulk modulus of elasticity :
K = DP = 0
 DV / V
_{P} ç ÷

è ø
[As DP = 0]
(5) Work done in isobaric process :
DW =
Vf P dV = P

Vi
Vf dV = P[V

Vi f
 Vi ]
[As P = constant]
\ DW = P(Vf – Vi ) = mR[Tf – Ti ] = mR DT
(6) FLTD in isobaric process :
DU = m CV
DT = m
R DT
(g – 1)
and
DW = mR DT
From FLTD
\
DQ = DU + DW

DQ = m R DT + mR DT = mR DT
(g – 1)
DQ = m CP DT
+ 1ú = mR DT
R DT
(7)

Examples of isobaric process
 Conversion of water into vapour phase (boiling process) :
When water gets converted into vapour phase, then its volume increases. Hence some part of absorbed heat is used up to increase the volume against external pressure and remaining amount of heat is used up to increase the internal potential energy of the molecules (because interatomic forces of attraction takes place between the molecules of system and when the distance between them increases, then its potential energy increases. It must be noted that during change of state since temperature remains constant, hence there will be no increase in internal kinetic energy of the molecules).
From first law of thermodynamics DQ = DU + DW
= DUK + DUP + DW
= 0 + DUP + PDV
\ DUP
or DUP
= DQ – P[Vf – Vi ]
= mL – P[Vf – Vi ]
[As DQ = mL]
 Conversion of ice into water
DQ = DU + DW = DUP + DUK + DW
mL = DUP + 0 + P[Vf – Vi ]
[ DUK = 0
as there is no change in temperature]
DUP = mL
[when ice convert into water then change in volume is negligible]
Note : @ In isobaric compression, temperature increases and internal energy flows out in the form of heat energy, while in isobaric expansion, temperature increases and heat flows into the system.
@Isobaric expansion of the volume of a gas is given by Vt = V0 (1 + g _{V} t)
where g
= 1 per°C = coefficient of volume expansion.
^{V} 273
Problem 38. 1 cm^{3} of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm^{3} .
If the atmospheric pressure is 1.013 ´ 10^{5} N / m^{2}
and the mechanical equivalent of heat = 4.19 J/calorie, the
energy spent in this process in overcoming intermolecular forces is [MP PET 1999, 2001; Orissa JEE 2002]
 540 calorie (b) 40 calorie (c) 500 calorie (d) Zero
Solution : (c) Energy spent in overcoming inter molecular forces DU = DQ – DW
= DQ – P(V
– V ) = 540 – 1.013 ´ 10^{5} (1671 – 1) ´ 10 ^{–}^{6}
» 500 calorie
^{2} ^{1} 4.2
Problem 39. A gas expands 0.25 m^{3} at constant pressure 10^{3} N / m^{2} , the work done is
[CPMT 1997; UPSEAT 1999; JIPMER 2001, 2002]
(a) 2.5 ergs (b) 250 J (c) 250 W (d) 250 N
Solution : (b) As we know, work done = PDV = 10^{3} ´ 0.25 = 250 J
Problem 40. 5 mole of hydrogen gas is heated from 30°C to 60°C at constant pressure. Heat given to the gas is (given
R = 2 cal/mole degree) [MP PET 2002]
(a) 750 calorie (b) 630 calorie (c) 1050 calorie (d) 1470 calorie
æ g ö
Solution : (c)
(DQ) p = m Cp DT = m ç g – 1 ÷ R DT
è ø
æ 7 ö
ç 5 ÷ 7 5 7

\ (DQ) p = 5 ´ ç 7

ç
è
÷ ´ 2 ´ 30 = 5 ´ 2 ´ ´

– 1 ÷ 5
ø
2 ´ 30 = 1050 calorie
[As m = 5 mole and g =
5
for H_{2}]
Problem 41. The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1g is 168 J, then increase in internal energy is [CPMT 2000]
(a) 2408 J (b) 2240 J (c) 2072 J (d) 1904 J
Solution : (c)
DQ = DU + DW
Þ 2240 = DU + 168 Þ DU = 2072 J
Problem 42. When an ideal gas (g = 5/3) is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work [RPET 1999]
(a) 40% (b) 30% (c) 60% (d) 20%
Solution : (a)
DW = DQ – DU = Cp – Cv = 1 – 1 = 1 – 1 = 2
DQ DQ Cp
g 1 – 5 5
3
i.e. percentage energy utilised in doing external work =
2 ´ 100 = 40%.
5
Problem 43. At 100°C the volume of 1kg of water is 10^{–}^{3} m^{3} and volume of 1 kg of steam at normal pressure is 1.671 m^{3} .
The latent heat of steam is 2.3 ´ 10^{6} J / kg and the normal pressure is 10^{5} N / m^{2} . If 5 kg of water at 100°C is
converted into steam, the increase in the internal energy of water in this process will be
(a)
8.35 ´ 10^{5} J (b)
10.66 ´ 10^{6} J (c)
11.5 ´ 10^{6} J (d) Zero
Solution : (b) Heat required to convert 5 kg of water into steam DQ = mL = 5 ´ 2.3 ´ 10^{6} = 11.5 ´ 10^{6} J
Work done in expanding volume, DW = PDV = 5 ´ 10^{5} [1.671 – 10 ^{–}^{3} ] = 0.835 ´ 10^{6} J
Now by first law of thermodynamics DU = DQ – DW
Þ DU = 11.5 ´ 10^{6} – 0.835 ´ 10^{6} = 10.66 ´ 10^{6} J
Isochoric or Isometric Process.
When a thermodynamic process undergoes a physical change in such a way that its volume remains constant, then the change is known as isochoric process.
In this process P and T changes but V = constant. Hence Gaylussac’s law is obeyed in this process.
(1) Equation of state
From ideal gas equation PV = mRT
If volume remains constant P µ T or
P1 = P2 T1 T2
= constant
 Indicator diagram : Graph I and II represent isometric decrease in
pressure at volume V_{1} and isometric increase in pressure at volume V_{2}
respectively and slope of indicator diagram dP = ¥
dV
 Specific heat : Specific heat of gas during isochoric process C
= f R
V 2
(4) Bulk modulus of elasticity :
K = DP = DP = ¥
– DV / V 0
 Work done in isobaric process : DW = PDV = P[Vf – Vi ]
[As V = 0]
\ DW = 0
(6) FLTD in isochoric process :
DQ = DU + DW
= DU
[As DW = 0]
DQ = m CV
DT = m R DT
g – 1
= Pf Vf – Pi Vi
g – 1
Note : @ Isometric expansion of the pressure of a gas is given by
Pt = P0 (1 + g _{P} t)
where g _{P}
= æ 1 öper°C = coefficient of pressure expansion.

273
è ø
Problem 44. In pressurevolume diagram given below, the isochoric, isothermal, and isobaric parts respectively, are
[Manipal MEE 1995]
 BA, AD, DC
 DC, CB, BA
 AB, BC, CD
 CD, DA, AB
Solution : (d) Process CD is isochoric as volume is constant, Process DA is isothermal as temperature constant and Process
AB is isobaric as pressure is constant.
Problem 45. Molar specific heat of oxygen at constant pressure
Cp = 7.2 cal / mol / °C
and R = 8.3 J/mol/K. At constant
volume, 5 mol of oxygen is heated from 10°C to 20°C, the quantity of heat required is approximately
[MP PMT 1987]
(a) 25 cal (b) 50 cal (c) 250 cal (d) 500 cal Solution : (c) By Mayer’s formula Cv = Cp – R = 7.2 – 2 » 5 cal/mol°C
At constant volume DQ = m cv DT = 5 ´ 5 ´ 10 = 250 cal
Reversible and Irreversible Process.
 Reversible process : A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite order and inverse sense and no change is left in any of the bodies taking part in the process or in the For example if heat is absorbed in the direct process, the same amount of heat should be given out in the reverse process, if work is done on the working substance in the direct process then the same amount of work should be done by the working substance in the reverse process. The conditions for reversibility are
 There must be complete absence of dissipative forces such as friction, viscosity, electric resistance
 The direct and reverse processes must take place infinitely
 The temperature of the system must not differ appreciably from its
Some examples of reversible process are
 All isothermal and adiabatic changes are reversible if they are performed very
 When a certain amount of heat is absorbed by ice, it melts. If the same amount of heat is removed from it, the water formed in the direct process will be converted into
 An extremely slow extension or contraction of a spring without setting up
 When a perfectly elastic ball falls from some height on a perfectly elastic horizontal plane, the ball rises to the initial
 If the resistance of a thermocouple is negligible there will be no heat produced due to Joule’s heating In such a case heating or cooling is reversible. At a junction where a cooling effect is produced due to Peltier effect when current flows in one direction and equal heating effect is produced when the current is reversed.
 Very slow evaporation or
It should be remembered that the conditions mentioned for a reversible process can never be realised in practice. Hence, a reversible process is only an ideal concept. In actual process, there is always loss of heat due to friction, conduction, radiation etc.
 Irreversible process : Any process which is not reversible exactly is an irreversible process. All natural processes such as conduction, radiation, radioactive decay etc. are irreversible. All practical processes such as free expansion, JouleThomson expansion, electrical heating of a wire are also irreversible. Some examples of irreversible processes are given below
 When a steel ball is allowed to fall on an inelastic lead sheet, its kinetic energy changes into heat energy by The heat energy raises the temperature of lead sheet. No reverse transformation of heat energy occurs.
 The sudden and fast stretching of a spring may produce vibrations in it. Now a part of the energy is This is the case of irreversible process.
 Sudden expansion or contraction and rapid evaporation or condensation are examples of irreversible
 Produced by the passage of an electric current through a resistance is
 Heat transfer between bodies at different temperatures is also
 JouleThomson effect is irreversible because on reversing the flow of gas a similar cooling or heating effect is not
Cyclic and Noncyclic Process.
A cyclic process consists of a series of changes which return the system back to its initial state.
In noncyclic process the series of changes involved do not return the system back to its initial state.
 In case of cyclic process as Uf= Ui
\ DU = Uf – Ui = 0
i.e. change in internal energy for cyclic process is zero and also DU µ DT
\ DT = 0 i.e. temperature of system remains constant.
 From first law of thermodynamics DQ = DU + DW
DQ = DW i.e. heat supplied is equal to the work done by the system. [As DU = 0]
 For cyclic process PV graph is a closed curve and area enclosed by the closed path represents the work If the cycle is clockwise work done is positive and if the cycle is anticlockwise work done is negative.
 Work done in non cyclic process depends upon the path chosen or the series of changes involved and can be calculated by the area covered between the curve and volume axis on PV
Problem 46. The PV diagram of a system undergoing thermodynamic transformation is shown in figure. The work done on the system in going from A ® B ® C is 50 J and 20 cal heat is given to the system. The change in internal energy between A and C is [UPSEAT 2002]
 34 J
 70 J
 84 J
 134 J
Solution : (d) Heat given DQ = 20 cal = 20 ´ 4.2 = 84 J . Work done DW = – 50 J [As process is anticlockwise] By first law of thermodynamics Þ DU = DQ – DW = 84 – ( 50) = 134 J
Problem 47. An ideal gas is taken through the cycle A ® B ® C ® A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C ® A is [IITJEE (Screening) 2002]
 – 5 J
 – 10 J
 – 15 J
 – 20 J
Solution : (a) For a cyclic process. Total work done = WAB + WBC + WCA
Þ 1 ´ 1.0 ´ 10 = 10 ´(2 – 1) + 0 + W [W = 0 since there is no change in volume along BC]
2 CA BC
Þ 5 J = 10 J + WCA Þ WCA = 5 J
Problem 48. In the following indicator diagram, the net amount of work done will be
 Positive
 Negative
 Zero
 Infinity
Solution : (b) Work done during process 1 is positive while during process 2 it is negative. Because process 1 is clockwise while process 2 is anticlockwise. But area enclosed by PV graph (i.e. work done) in process 1 is smaller so, net work done will be negative.
Problem 49. A cyclic process for 1 mole of an ideal gas is shown in figure in the V–T, diagram. The work done in AB, BC
and CA respectively
æ V1 ö
(a) 0, RT2 lnç V ÷ , R (T1 – T2)
è 2 ø
(b)
R(T1 – T2), 0, RT1 ln V1
V2
æ V ö
(c)
0, RT2 lnç ^{2} ÷ , R (T1 – T2)
è V1 ø
æ V2 ö
(d)
0, RT2 lnç V ÷ , R (T2 – T1)
è 1 ø
Solution : (c) Process AB is isochoric, \ WAB = P DV = 0
æ V2 ö
Process BC is isothermal \ WBC = RT2. lnç V ÷
è 1 ø
Process CA is isobaric \ WCA = PDV = RDT = R(T2 – T1)
Problem 50. A cyclic process ABCD is shown in the figure PV diagram. Which of the following curves represent the same process
 P A B C
 P A B C
 P B
A
 P
A B
D D D C D C
T T T T
Solution : (a) AB is isobaric process, BC is isothermal process, CD is isometric process and DA is isothermal process These process are correctly represented by graph (a).
Graphical Representation of Various Processes.
Heat Engine.
Heat engine is a device which converts heat into work continuously through a cyclic process. The essential parts of a heat engine are
Source : It is a reservoir of heat at high temperature and infinite thermal capacity. Any amount of heat can be extracted from it.
Working substance : Steam, petrol etc.
Sink : It is a reservoir of heat at low temperature and infinite thermal capacity. Any amount of heat can be given to the sink.
The working substance absorbs heat Q_{1} from the source, does an amount of work W, returns the remaining amount of heat to the sink and comes back to its original state and there occurs no change in its internal energy.
By repeating the same cycle over and over again, work is continuously obtained.
The performance of heat engine is expressed by means of “efficiency” h which is defined as the ratio of useful work obtained from the engine to the heat supplied to it.
h = Work done = W
Heat input Q_{1}
But by first law of thermodynamics for cyclic process DU = 0. \ DQ = DW so
W = Q1 – Q2
\ h = Q1 – Q2 Q1
= 1 – Q2
Q1
A perfect heat engine is one which converts all heat into work i.e. W = Q1
But practically efficiency of an engine is always less than 1.
Refrigerator or Heat Pump.
so that Q2 = 0 and hence h = 1 .
A refrigerator or heat pump is basically a heat engine run in reverse direction. It essentially consists of three parts
Source : At higher temperature T_{1}.
Working substance : It is called refrigerant liquid ammonia and freon works as a working substance.
Sink : At lower temperature T_{2}.
The working substance takes heat Q_{2} from a sink (contents of refrigerator) at lower temperature, has a net amount of work done W on it by an external agent (usually compressor of
refrigerator) and gives out a larger amount of heat Q_{1} to a hot body at temperature T_{1} (usually atmosphere). Thus, it transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold body is thus cooled more and more.
The performance of a refrigerator is expressed by means of “coefficient of performance” b which is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.
i.e.
b = Heat extracted = Q_{2} = Q_{2}
\ b =
work done
Q2 Q1 – Q2
W Q1 – Q2
A perfect refrigerator is one which transfers heat from cold to hot body without doing work
i.e. W = 0 so that Q_{1} = Q_{2}
(1) Carnot refrigerator
and hence
b = ¥
For Carnot refrigerator Q1
Q2
= T1 T2
\ Q1 – Q2
Q2
= T1 – T2
T2
or Q2 Q1 – Q2
= T2 T1 – T2
So coefficient of performance
b = T2 T1 – T2
where T_{1} = temperature of surrounding, T_{2} = temperature of cold body It is clear that b = 0 when T_{2} = 0
i.e. the coefficient of performance will be zero if the cold body is at the temperature equal to absolute zero.
(2) Relation between coefficient of performance and efficiency of refrigerator
We know
b = Q2 Q1 – Q2
or b =
Q2 / Q1
1 – Q2 / Q1
….. (i)
But the efficiency h = 1 – Q2
Q1
or Q2
Q1
= 1 – h
…..(ii)
From (i) and (ii) we get
b = 1 – h
h
Second Law of Thermodynamics.
First law of thermodynamics merely explains the equivalence of work and heat. It does not explain why heat flows from bodies at higher temperatures to those at lower temperatures. It cannot tell us why the converse is possible. It cannot explain why the efficiency of a heat engine is always less than unity. It is also unable to explain why cool water on stirring gets hotter whereas there is no such effect on stirring warm water in a beaker. Second law of thermodynamics provides answers to these questions. Statement of this law is as follows
 Clausius statement : It is impossible for a self acting machine to transfer heat from a colder body to a hotter one without the aid of an external
From Clausius statement it is clear that heat cannot flow from a body at low temperature to one at higher temperature unless work is done by an external agency. This statement is in fair agreement with our experiences in different branches of physics. For example, electrical current cannot flow from a conductor at lower electrostatic potential to that at higher potential unless an external work is done. Similarly, a body at a lower gravitational potential level cannot move up to higher level without work done by an external agency.
 Kelvin’s statement : It is impossible for a body or system to perform continuous work by cooling it to a temperature lower than the temperature of the coldest one of its surroundings. A Carnot engine cannot work if the source and sink are at the same temperature because work done by the engine will result into cooling the source and heating the surroundings more and
 KelvinPlanck’s statement : It is impossible to design an engine that extracts heat and fully utilises into work without producing any other
From this statement it is clear that any amount of heat can never be converted completely into work. It is essential for an engine to return some amount of heat to the sink. An engine essentially requires a source as well as sink. The efficiency of an engine is always less than unity because heat cannot be fully converted into work.
Carnot Engine.
Carnot designed a theoretical engine which is free from all the defects of a practical engine. This engine cannot be realised in actual practice, however, this can be taken as a standard against which the performance of an actual engine can be judged.
It consists of the following parts
 A cylinder with perfectly nonconducting walls and a perfectly conducting base containing a perfect gas as working substance and fitted with a nonconducting frictionless piston
 A source of infinite thermal capacity maintained at constant higher temperature T
 A sink of infinite thermal capacity maintained at constant lower temperature T_{2}.
 A perfectly nonconducting stand for the
 Carnot cycle : As the engine works, the working substance of the engine undergoes a cycle known as Carnot
The Carnot cycle consists of the following four strokes
 First stroke (Isothermal expansion) (curve AB) :
The cylinder containing ideal gas as working substance allowed to expand slowly at this constant temperature T_{1}.
Work done = Heat absorbed by the system
V2 æ V_{2} ö
W1 = Q1 = òV P dV = RT1 log e ç V ÷ = Area ABGE
1 è 1 ø
 Second stroke (Adiabatic expansion) (curve BC) :
The cylinder is then placed on the non conducting stand and the gas is allowed to expand adiabatically till the temperature falls from T_{1} to T_{2}.
W = V3 P dV =
R [T
 T ] = Area BCHG

2 òV
(g – 1) ^{1} ^{2}
 Third stroke (Isothermal compression) (curve CD) :


The cylinder is placed on the sink and the gas is compressed at constant temperature T_{2}. Work done = Heat released by the system
V

W = Q = – 4 P dV = –RT
log V4 = RT
log V3 = Area CDFH
3 2 òV
2 e
3
2 e
4
 Fourth stroke (adiabatic compression) (curve DA) : Finally the cylinder is again placed on nonconducting stand and the compression is continued so that gas returns to its initial
V R
W = – 1 P dV = – (T
 T ) = R (T – T ) = Area ADFE

4 òV
g – 1 ^{2}
1 g – 1 1 2
 Efficiency of Carnot cycle : The efficiency of engine is defined as the ratio of work done to the heat
supplied i.e.
h = work done = W
Heat input Q_{1}
Net work done during the complete cycle
W = W1 + W2 + (W3 ) + (W4 ) = W1 – W3 = Area ABCD
[As W2 = W4 ]
\ h = W = W_{1} – W_{3}
= Q1 – Q2
= 1 – W3
= 1 – Q2
Q1 W1 Q1 W1 Q1
or h = 1 – RT2 log e (V3 / V4 )
RT1 log e (V2 / V1 )
T æ V
ög 1
Since points B and C lie on same adiabatic curve \
T1 V g 1 = T2 V g 1 or 1 = ç 3 ÷
…..(i)
2 3 T2
è V2 ø
T æ V
ög 1
Also point D and A lie on the same adiabatic curve \ T_{1}V ^{g} ^{1} = T_{2}V ^{g} ^{1} or^{ } ^{1} = ç ^{4} ÷
…..(ii)
V3 V4
V3 V2
1 4
æ V3 ö æ V2 ö
T2 è V1 ø
From (i) and (ii) V = V
or =
V V
Þ log e ç V ÷ = log e ç V ÷
2 1 4 1
è 4 ø
è 1 ø
So efficiency of Carnot engine h = 1 – T2
T1
 Efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other
 All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal).
 As on Kelvin scale, temperature can never be negative (as 0 K is defined as the lowest possible temperature) and T_{l} and T_{2} are finite, efficiency of a heat engine is always lesser than unity, e., whole of heat can never be converted into work which is in accordance with second law.
Note : q The efficiency of an actual engine is much lesser than that of an ideal engine. Actually the practical efficiency of a steam engine is about (815)% while that of a petrol engine is 40%. The efficiency of a diesel engine is maximum and is about (5055)%.
 Carnot theorem : The efficiency of Carnot’s heat engine depends only on the temperature of source (T_{1})

and temperature of sink (T ), i.e., h = 1 – T2 .
T1
Carnot stated that no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures. Carnot’s reversible engine working between two given temperatures is considered to be the most efficient engine.
Difference Between Petrol Engine And Diesel Engine.
Problem 51. If the door of a refrigerator is kept open, then which of the following is true
[BHU 2001; JIPMER 2002; AIEEE 2002; CPMT 2003]
 Room is cooled (b) Room is heated
(c) Room is either cooled or heated (d) Room is neither cooled nor heated
Solution : (b) In a refrigerator, the working substance takes heat Q_{2} from the sink at lower temperature T_{2} and gives out a
larger amount of heat Q_{1} to a hot body at higher temperature T_{1} . Therefore the room gets heated if the door of a refrigerator is kept open.
Problem 52. The coefficient of performance of a Carnot refrigerator working between 30^{o}C and 0^{o}C is [UPSEAT 2002]
(a) 10 (b) 1 (c) 9 (d) 0
Solution : (c) Coefficient of performance of a Carnot refrigerator working between 30°C and 0°C is
b = T2
= 273°C = 273°C » 9
T1 – T2
303°C – 273°C 30°C
Problem 53. A Carnot engine working between 300 K and 600 K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycle [Pb. PMT 2002]
(a) 1800 J/cycle (b) 1000 J/cycle (c) 2000 J/cycle (d) 1600 J/cycle
Solution : (d) Efficiency of Carnot engine
h = T_{1} – T_{2} 600 – 300 = 1
T_{1} 600 2
Again
h = Work done
Heat input
\ Heat input = Work done
h
= 800 = 1600 J.
1 / 2
Problem 54. Carnot cycle (reversible) of a gas represented by a PressureVolume curve is shown in the diagram
Consider the following statements
 Area ABCD = Work done on the gas
 Area ABCD = Net heat absorbed
 Change in the internal energy in cycle = 0 Which of these are correct
(a) I only (b) II only
 II and III
(d)
[AMU (Med.) 2001]
I, II and III
Solution : (c) Work done by the gas (as cyclic process is clockwise) \ DW = Area ABCD
So from the first law of thermodynamics DQ (net heat absorbed) = DW = Area ABCD
As change in internal energy in cycle DU = 0.
Problem 55. A Carnot engine takes 103 kcal of heat from a reservoir at 627°C and exhausts it to a sink at 27°C. The efficiency of the engine will be
(a) 22.2% (b) 33.3% (c) 44.4% (d) 66.6%
Solution : (d) Efficiency of Carnot engine = T1 – T2
= 900 – 300 = 6
or 66.6%.
T1 900 9
Entropy.
Entropy is a measure of disorder of molecular motion of a system. Greater is the disorder, greater is the entropy.
The change in entropy i.e. dS = Heat absorbed by system
Absolute temperature
or dS = dQ
T
The relation is called the mathematical form of Second Law of Thermodynamics.
(1) For solids and liquids
 When heat given to a substance changes its state at constant temperature, then change in entropy
dS = dQ = ± mL
T T
where positive sign refers to heat absorption and negative sign to heat evolution.
 When heat given to a substance raises its temperature from T_{1} to T_{2}, then change in entropy
dQ T2 dT
æ T2 ö
dS = ò T = òT mc T = mc log e ç T ÷
1
æ T2 ö
è 1 ø
Þ DS = 2.303 mc log e ç T ÷ .
è 1 ø
 For a perfect gas : Perfect gas equation for n moles is PV = nRT
DS = ò dQ = ò nC_{V} dT + P dV
[As dQ = dU + dW}
Þ DS =
T
nCV
T
dT + nRT dV
V
= nC
T2 dT + nR
V2 dV
[As PV = nRT]
ò _{T}

æ T2 ö
V òT T

æ V2 ö
òV V
\ DS = nCV log e ç T ÷ + nR log e ç V ÷
è 1 ø è 1 ø
æ T2 ö æ P2 ö
Similarly in terms of T and P
DS = nCP log e ç T ÷ – nR log e ç P ÷
è 1 ø è 1 ø
æ P2 ö æ V2 ö
and in terms of P and V
DS = nCV log e ç P ÷ + nCP log e ç V ÷
è 1 ø è 1 ø
Problem 56. An ideal gas expands in such a manner that its pressure and volume can be related by equation PV ^{2} = constant.
During this process, the gas is  [UPSEAT 2002]  
(a) Heated
(c) Neither heated nor cooled 
(b)
(d) 
Cooled
First heated and then cooled 
Solution : (b) For an adiabatic expansion PV ^{g}
= constant
and for the given process PV ^{2} = constant
\ It is also an adiabatic expansion and during adiabatic expansion the gas is cooled.
Problem 57. A cyclic process ABCA is shown in the VT diagram. Process on the PV diagram is
 P
 P
A
V
B (c) P
A B

C
V
 P
V V
Solution : (c) From the given VT diagram, we can see that
In process AB, V µ T \ Pressure is constant (As quantity of the gas remains same) In process BC, V = constant and in process CA, T = constant
\ These processes are correctly represented on PV diagram by graph (c).