The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed.

1. Solved Examples of
Thermodynamics
Class XIth
Subject Expert: Dr. Tanuja Nautiyal
Sharda Public School, Almora

2. 1. Question: Define Thermodynamics?
Answer: Thermodynamics is a physical science that deals
with quantitative relation between heat and mechanical
energy. It is mainly based on three laws of thermodynamics.
2. Question:Which law of thermodynamics deals with
equivalence of different forms of energies?
Answer: The first law of thermodynamics:
3. Question:What is the first law of thermodynamics?
Answer: The first law of thermodynamics: It states that
energy can neither be created nor destroyed. Energy,
however, can be transferred from one form to another.
Mathematically, when work is done on the system,
U = q + w
And when work is done on by the system, U = q - w

3. 4. Question: Define the Second Law of
Thermodynamics?
Answer: The Second Law of Thermodynamics can be
rephrased in several ways. Fundamentally, it says that heat
always flows from hot objects to cold objects (unless work is
exerted to make it flow the other direction). It can also be
expressed using the concept of entropy as saying that the
system's entropy will always naturally increase if no work is
exerted to decrease it, because heat deals with kinetic
energy and increasing a system's kinetic energy will increase
the system's entropy.
5. Question: Which law of thermodynamics evaluate
thermodynamic parameters?
Answer: Third law of thermodynamics.
6. Question: State the Zeroth law of
Thermodynamics?
Answer: The Zeroth law of thermodynamics states that “Two
systems separately in thermal equilibrium with the third
system are said to be in thermal equilibrium with each
other”. If system X and system Y separately are in thermal

4. equilibrium with another system Z, then system X and System
Y are also in thermal equilibrium with each other.
7. Question: Define Entropy?
Answer: Entropy measures the amount of disorder in a
system. Nature tends towards disorder, so as time elapses,
entropy naturally increases. Energy is required in order to
decrease entropy.
8. Question: Which law of thermodynamics describes
entropy?
Answer: The second law of thermodynamics describes
entropy.
9. Question: Which law of thermodynamics is about
the absolute zero temperature?
Answer: The third law of thermodynamics

5. 10. Question: What is Thermochemistry?
Answer: Thermochemistry is a branch of thermodynamics
which deals with the relationships between chemical
reactions and corresponding energy changes. It is based on
first law of thermodynamics.
11. Question: What is a system? What are its different
types?
Answer: The part of universe under thermodynamic study is
called a system. System is isolated from the rest of the
universe by with a bounding surface. The different types of
systems are (a). Open System (b). Closed System (c). Isolated
System.
12. Question: Give examples of different types of
Systems?
Answer: (a). Closed System: Ice in a closed beaker
(b). Open System: Ice in a glass
(c). Isolated System: Ice in thermos flask

6. 13. Question: Define Intensive property? Give
examples of intensive property?
Answer: Intensive Property: A property, which is
independent of the amount of substances present in the
system, is called intensive property.
Intensive property: temperature, viscosity, density, refractive
index, concentration, specific heat, surface tension.
14. Question: What is extensive property? Give
example?
Answer: A property, which is dependent of the amount of
substances present in the system, is called extensive
property. Example: internal energy
15. Question: Define Entropy?
Answer: The thermodynamic parameter, which depend on
the initial and final states of the system and are independent
of how the change is accomplished are called state functions.
Examples: Entropy, Pressure, Temperature, Volume, Internal
Energy, Enthalpy.

7. 16. Question: What are path functions?
Answer: The system variables, which depend upon the path
of the system, are called path functions. Examples: Heat,
Work and Length.
17. Question: Define the Thermodynamic process?
What are the different types of thermodynamic
processes?
Answer: Thermodynamic process is the conversion of state of
a system from one to another state. Various processes are:
Isothermal, Isobaric, Adiabatic, Isochoric and Cyclic.
18. Question: Define Isothermal process and adiabatic
process?
Answer: Isothermal process: The processes in which the
temperature remains fixed are termed isothermal processes.
In isothermal process T = 0 and U = 0
Adiabatic process: The process in which no heat can flow
into or out of the system is called adiabatic process. In
adiabatic process q = 0

8. 19. Question: What is internal energy?
Answer: The internal energy (U or E) of a system is the total
of all kinds of energy possessed by the particles that make up
the system. The internal energy is the sum of translational,
vibrational, rotational energies. It is a state property and its
absolute value cannot be determined.
20. Question: What are the factors an internal energy
of the system depends upon?
Answer: The factors are: (a). quantity of the gas or substance
(b). Chemical nature of gas (c). Temperature, Pressure and
Volume.
21. Question: Is internal energy intensive or extensive
property?
Answer: The value of internal energy of a system depends on
the mass of the matter contained in a system, it is an
extensive property.

9. 22. Question: What are the sign conversions used for
heat and work?
Answer: (a). q = -ve; Heat flows out of the system.
(b). q = +ve; Heat flows into the system.
(c). w = -ve; Work is done by the system.
(d). w = +ve; Heat is done on the system.
23. Question: Define Hess law of heat summation?
Hess Law: This law states that the amount of heat evolved or
absorbed in a process, including a chemical change is the
same whether the process takes place in one or several
steps.
Suppose in a process the system changes from state A to
state B in one step and the heat exchanged in this change is
q. Now suppose the system changes from state A to state B
in three steps involving a change from A to C, C to D and
finally from D to B. If q1, q2 and q3 are the heats exchanged in
the first, second and third step, respectively then according
to Hess’s law

10. q1 + q2 + q3 = q
Hess’s law is simply a corollary of the first law of
thermodynamics. It implies that enthalpy change of a
reaction depends on the initial and final state and is
independent of the manner by which the change is brought
about.
Choose the correct answer and explain.
24. Question: For the process to occur under adiabatic
conditions, the correctcondition is:
(i) 'T = 0
(ii) 'p = 0
(iii) q = 0
(iv) w = 0
Answer : (iii)
A system is said to be under adiabaticconditionsif there is
no exchange of heat between the system and its
surroundings. Hence, under adiabaticconditions,q = 0.
25. Question: A thermodynamic state function is a
quantity:

11. (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: (ii)
A thermodynamic state function is a quantity whose value is
independentof a path. Functions like p, V, T etc. depend only
on the state of a system and not on the path.
25. Question: What is true of Isothermal process:
(a). ΔT >0
(b). ΔU=0
(c). ΔQ=ΔW
(d). PV=constants
Solution:
In an Isothermal Process
Temperature remains constant ΔT =0
Since Internal energy dependson the temperature; ΔU=0
From first law of Thermodynamics
ΔU=ΔQ-ΔW
Since ΔU=0
ΔQ=ΔW

12. Also PV=nRT
As T is constant
PV= constant
26. Question: At 27°C, two moles of an ideal
monoatomicgas occupy a volume V. The gas is
adiabatically expanded to a volume 2V.
(a). Calculatethe ratio of finalpressure to the intialpressure
(b). Calculatethe final temperature
(c). Change in internal energy
(d). Calculatethe molarspecific heat capacity of the process
Solution
Given: n=2, T=27°C=300 K , V1=V, V2=2V
Now PVy=constant
P1V1y=P2V2y
P2/P2 =(V1/V2)y =.55/3
Also
T1V1y-1=T2V2y-1 or T2=300/25/3 =189K
Change in internalenergy=nCvΔT
For monoatomic gas Cv=3R/2

13. Substituting all the values
Change in internalenergy==-2764J
As in adiabaticprocess ΔQ=0,molarspecific heat capacity=0
27. Question: A gas is contained in a cylinder with a
moveable piston on which a heavy block is placed.
Suppose the region outside the chamber is evacuated
and the total mass of the block and the movable
piston is 150 kg. When 2400 J of heat flows into the
gas, the internal energy of the gas increases by 1500 J.
What is the distance through which the piston rises?
Solution:
Total heat supplied = Workdone + Change in internalenergy
So work done=2400-1500=900 J
Lets be the distance moved then the workdone is given by
=Fs
Fs=900, s = 900/F , s= 900/150 X 10
s=0.6 m

14. 28. Question: A container holds a mixture of three
non-reacting gases: n1 moles of the first gas with
molar specific heat at constant volume C1, and so on.
Find the molar specific heat at constant volume of the
mixture, in terms of the molar specific heats and
quantities of the three separate gases.
Answer: Heat capacity C of a body as the ratio of the amount
of heat energy Q transferred to a body in any process to its
corresponding temperature change ΔT.
C = Q/ΔT So, Q = C ΔT
Each species will experience the equaltemperature change.
If the gas has n molecules, then Q will be,
Q = nC ΔT
If the gas has n1 moles, then the amount of heat
energy Q1 transferred to a body having heat capacity C1 will
be,
Q1 = n1C1 ΔT
Similarly,if the gas has n2 moles, then the amount of heat
energy Q2 transferred to a body having heat capacity C2 will
be,
Q2 = n2C2 ΔT

15. And if the gas has n3 moles, then the amount of heat
energy Q3 transferred to a body having heat capacity C3 will
be,
Q3 = n3C3 ΔT
As, each species will experience the same temperature
change, thus,
Q = Q1 + Q2 + Q3
= n1C1 ΔT + n2C2 ΔT + n3C3 ΔT
Dividingboth the sides by n = n1 + n2 + n3 and ΔT, then we
will get,
Q/n.ΔT = (n1C1 ΔT + n2C2 ΔT + n3C3 ΔT)/ n.ΔT
As, Q/nΔT = C, thus,
C=ΔT (n1C1 + n2C2 + n3C3)/ n. ΔT
= (n1C1 + n2C2 + n3C3)/ n
= (n1C1 + n2C2 + n3C3 )/( n1 + n2 + n3)
From the above observationwe conclude that, the molar
specific heat at constant volume of the mixture would
be (n1C1 + n2C2 + n3C3) / ( n1 + n2 + n3)

16. 29. Question: An ideal gas heat engine operates in
Carnot cycle between 227°C and 127°C. It absorbs 8
X102
cal of heat at the higher temperature. Calculate
the amount of heat supplied to the engine from the
source in each cycle
Solutions:
T1 = 227°C =500K, T2 = 127°C =400K
Efficiency of the carnot cycle is given by = 1-(T2/T1) = 1/5
Now also efficiency = Heat supplied from source/Heat
absorbed at high temperature
so Heat supplied from source=8X102
X (1/5) =1.6 X102
cal
30. QuestionTwo absolute scales A and B have triple
points of water definedas 200A and 350A. What is
the relationbetween TA and TB
Answer: Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalentto

17. 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A
= (273.16/200) K Or,
Value of temperature TA on absolutescale A
= (273.16XTA) /200
Similarlyvalue of temperature TB on absolute scale B
= (273.16XTB)/350
Since TA and TB represent the same temperature
273.16×TA/200 = 273.16×TB/350
Or, TA = 200TB/350 = 4TB/ 7
31. Question: A thermometer of mass 0.055 kg and
heat capacity 46.1 J/K reads 15.0°C. It is then
completely immersed in 0.300 kg of water and it
comes to the same final temperature as the water. If
the thermometer reads 44.4°C, what was the
temperature of the water before insertion of the
thermometer, neglecting other heat losses?

18. Answer: In accordance to the law of conservationof energy,
for a thermodynamicsystem, in which internalis the only
type of energy the system may have, the law of conservation
of energy may be expressed as,
Q + W = ΔE
Here Q is the energy transferred between the system and its
environment,
W is the work done on or by the system and ΔE is the change
in the internalenergy of the system.
The heat capacity C of a body as the ratio of amount of heat
energy Q transferred to a body in any process to its
corresponding temperature change ΔT.
C = Q/ΔT So, Q = C ΔT
The heat capacity per unit mass of a body, called specific heat
capacity or usually just specific heat, is characteristic of the
material of which the body is composed.
c = C/m = Q/mΔT
So, Q = c mΔT
The heat transfers for the water Qw is,
Qw = mwcw (Tf –Ti )
Here, mass of water is mw, specific heat capacity of water
is cw, final temperature is Tf and initialtemperature is Ti.
The heat transfers for the thermometer Qt is

19. Qt = Ct. ΔTt
Here, heat capacity of thermometer is Ct and ΔTt is the
temperature difference.
As the internalenergy of the system is zero and there is no
work is done, therefore substitute ΔEint = 0 and W = 0 in the
equationQ + W = ΔEint,
Q + W = ΔEint
Q + 0= 0
So, Q = 0
Or, Qw + Qt = 0
mwcw (Tf –Ti)+ CtΔTt = 0
So, Ti = (mwcw Tf + CtΔTt )/ mwcw
Here ΔTt = 44.4 ̊ C - 15.0 ̊ C
= 29.4 ̊ C
To obtainthe temperature of the water before insertion Ti of
the thermometer, substitute 0.3 kg for mw, 4190 J/kg.m
for cw, 44.4 ̊ C for Tf, 46.1 J/K for Ct and 29.4 ̊ C for ΔTt in the
equationTi = (mwcw Tf + CtΔTt )/ mwcw,
Ti = (mwcw Tf + CtΔTt )/ mwcw
= [(0.3 kg) (4190 J/kg.m) (44.4 ̊ C) + (46.1 J/K) (29.4 ̊ C)] /[(0.3
kg) (4190 J/kg.m)] = 45.5 ̊ C

20. From the above observation we conclude that, the
temperature of the water before insertion of the
thermometer was 45.5 ̊ C.
32. Question: A mixture of 1.78 kg of water and 262 g
of ice at 0°C is, in a reversible process, brought to a
final equilibrium state where the water / ice ratio, by
mass 1:1 at 0°C.
(a) Calculate the entropy change of the system during
this process. (b) The system is then returned to the
first equilibrium state, but in an irreversible way (by
using a Bunsen burner, for instance). Calculate the
entropy change of the system during this process. (c)
Show that your answer is consistent with the second
law of thermodynamics.
Answer: The entropy change ΔS for a reversible isothermal
process is defined as,
ΔS = Q/T = -mL/T
Here m is the mass, L is the latent heat and T is the
temperature.
(a) Mass of water = 1.78 kg
Mass of ice = 262 g

21. So the total mass of ice and water mixture will be,
Mass of ice-water mixture = (Mass of water) + (Mass of ice)
= (1.78 kg) + (262 g)
= (1.78 kg) + (262 g×10-3 kg/1 g)
= 1.78 kg + 0.262 kg
= 2.04 kg
If eventuallythe ice and water have the same mass, then the
final state will have 1.02 kg (2.04 kg/2) of each.
Thus the mass of the water that changed into ice m will be
the difference of mass of water mw and mass of final
state ms.
So, m = mw - ms
To obtainmass of water that changed into ice m, substitute
1.78 kg for mass of water mw and 1.02 kg for mass of final
state ms in the equation m = mw - ms,
m = mw - ms
= 1.78 kg – 1.02 kg
= 0.76 kg
The change of water at 0̊0
C to ice at 0̊0
C is isothermal.
To obtain the change in entropy ΔS of the system during this
process, substitute 0.76 kg for mass m, 333×103
J/kg for heat

22. of fusion of water L and 273 K for T in the equation ΔS = -
mL/T,
ΔS = -mL/T
= -(0.76 kg) (333×103 J/kg )/(273 K)
= -927 J/K
From the above observationwe conclude that, the change in
entropy ΔS of the system during this process will be -927 J/K.
(b) Now the system is returned to the first equilibriumstate,
but in an irreversible way. Thusthe change in entropy ΔS of
the system during this process is equal to the negative of
previouscase.
So, ΔS = -(- 927 J/K)
= 927 J/K
From the above observationwe conclude that, the change in
entropy ΔS of the system would be 927 J/K.
(c) In accordance to second law of thermodynamics, entropy
change ΔS is always zero.
The total change in entropy will be,
ΔS = (-927 J/K) + (927 J/K) = 0
From the above observationwe conclude that, our answer is
consistent with the second law of thermodynamics.

23. 33. Question: Apparatus that liquefies helium is in a
laboratory at 296 K. The helium in the apparatus is at
4.0 K. If 150 mJ of heat is transferredfromthe helium,
find the minimum amount of heat delivered to the
laboratory.
Answer:
Coefficient of performance K of a Carnot refrigerator is
defined as,
K = TL / TH - TL …… (1)
Here TL is the lower temperature of sink and TH is the higher
temperature of source.
A refrigerator would like to extract as much heat QL as
possible from the low-temperature reservoir for the least
amount of work W. So the efficiency of a refrigerator is
defined as,
K = QL/W
and this is called coefficient of performance. The larger the
value of K, the more efficient is the refrigerator.
Thus, W = QL/K …… (2)
Substitute the value of K from equation(1) in the
equationW = QL/K,
W = QL/K

24. = QL/( TL / TH - TL)
= QL (TH/ TL – 1) …… (3)
The first law of thermodynamics, appliedto the working
substance of the refrigerator, gives,
W = QH – QL
Here QH is the exhausted heat.
Thus exhausted heat will be, QH = W + QL …… (4)
Substitute the value of W from equation(3) in the
equationQH = W + QL,
QH = W + QL
= QL (TH/ TL – 1) + QL
= QL (TH/ TL)
To obtainthe minimum amount of heat delivered to the
laboratory, substitute 150 mJ for QL, 296 K for TH and 4.0 K
for TL in the equation QH = QL (TH/ TL),
QH = QL (TH/ TL)
= ((150 mJ) (10-3 J/1 mJ)) (296 K/4.0 K) = 11 J
From the above observationwe conclude that, the minimum
amount of heat delivered to the laboratorywould be 11 J.