Thermodynamics
Class 11th Physics NCERT Exemplar Solution
Mcq I
Question 1.An ideal gas undergoes four different processes from the same initial state (Fig. 12.1). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic.
A. 4
B. 3
C. 2
D. 1
Answer:
During an adiabatic process, no heat is transferred to the gas, but the temperature, pressure and volume of the gas changes.
The curve on a p-v diagram is equal to the work performed by a gas during the process.
∴ curve 2 shows the adiabatic process.
Hence option is correct.
Question 2.
If an average person jogs, she produces 14.5 × 103 Cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 Cal for evaporation) is
A. 0.25 kg
B. 2.25 kg
C. 0.05 kg
D. 0.20 kg
Answer:
Given, calories produced per min = 
amount of sweat evaporated is equal to the rate of calories burned.
∴ 
Hence option(a) is correct.
Question 3.
Consider P-V diagram for an ideal gas shown in Fig 12.2.
Out of the following diagrams (Fig. 12.3), which represents the T-P diagram?
A. (iv)
B. (ii)
C. (iii)
D. (i)
Answer:
It is given PV= constant
which states that the process is isothermal process. From the graph, we can see that in process 1 & 2 temperature
is constant and the gas expands, while pressure decreases
i.e., P2<P1
∴ graph (iii) represents the correct T-P graph.
option (c) is correct.
Question 4.
An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram (Fig. 12.4).
The amount of work done by the gas is
A. 6Po Vo
B. –2 P0 Vo
C. + 2 Po Vo
D. + 4 Po Vo
Answer:
We know that total amount of work done = area under
P-V diagram.
According to the P-V diagram,
work done in process ABCDA = area of rectangle ABCDA
= AB× BC
= 
= 
=
∵ the process is anti-clockwise
∴ work done by the process is negative.
Hence, amount of work done by the gas = 
option (b) is correct.
Question 5.
Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is
A. 2y-1
B. 
C. 
D. 
Answer:
for container A (gas is compressed to half of its original
volume) Isothermal compression
⇒ 
Let original volume be 
then after compression it becomes 
⇒ 
⇒ 
…. (1)
For container B (gas is compressed to half of its original
volume) Adiabatic compression
⇒ 
Let original volume be then after compression it becomes
⇒ 
⇒ 
⇒ 
…… (2)
Ratio of final pressure of gas in B to that of gas in A will be
obtained by (2)/ (1)
=
=
option (a) is correct.
Question 6.
Three copper blocks of masses M1, M2 and M3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2, T3 (T1 > T2 > T3 ). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)
A. T = 
B. T = 
C. T = 
D. T = 
Answer:
It is given that there is no heat loss in the surrounding
and equilibrium temperature in the system is T.
It is also given s is the specific heat of the copper
Let us assume 
∴ heat loss by 
= heat gain by 
+ heat gain by 
⇒ 
⇒ 
∴ 
option (b) is correct.
Mcq Ii
Question 1.Which of the processes described below are irreversible?
A. The increase in temperature of an iron rod by hammering it.
B. A gas in a small container at a temperature T1 is brought in contact with a big reservoir at a higher temperature T2 which increases the temperature of the gas.
C. A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston.
D. An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of gas.
Answer:
We know that in thermodynamics, a reversible process is a
process whose direction can be reversed by including
infinitesimal changes to the property of the system through
its surroundings.
Now,
case(a) in which internal energy of the rod is increased
through external work done, this process is irreversible.
case(b) heat is transferred from one container to
another, process is irreversible.
case (c) process is reversible because cylinder is fitted with
frictionless piston.
case(d) process is irreversible as weight is added to the
cylinder arrangement in form of external pressure.
option (a, b, d) is correct.
Question 2.
An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives.
A. dU = 0
B. dQ= 0
C. dQ = dU
D. dQ = dW
Answer:
In isothermal process the temperature is constant therefore
internal energy is constant and thus net change in internal
energy is zero.
option (a, d) is correct.
Question 3.
Figure 12.5 shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
A. Change in internal energy is same in IV and III cases, but not in I and II.
B. Change in internal energy is same in all the four cases.
C. Work done is maximum in case I
D. Work done is minimum in case II.
Answer:
We know that internal energy of a system is equal to sum of
all kinetic and potential energies of the particles inside it.
Also, change in internal energy does not depend on path
it depends on initial and final states.
∴ internal energy is same for all the four paths.
Now,
work done is equal to the area under P-V curve
∴ work done is maximum for path 1.
option (b, c) is correct.
Question 4.
Consider a cycle followed by an engine (Fig. 12.6) 1 to 2 is isothermal 2 to 3 is adiabatic 3 to 1 is adiabatic Such a process does not exist because
A. heat is completely converted to mechanical energy in such a process, which is not possible.
B. mechanical energy is completely converted to heat in this process, which is not possible.
C. curves representing two adiabatic processes don’t intersect.
D. curves representing an adiabatic process and an isothermal process don’t intersect.
Answer:
If two(similar) curves eat each other it means one could transform from one state to another without changing energy(heat) which is not possible.
Heat cannot be converted completely to mechanical energy.
option (a, c) is correct.
Question 5.
Consider a heat engine as shown in Fig. 12.7. Q1 and Q2 are heat added to heat bath T1 and heat taken from T2 in one cycle of engine. W is the mechanical work done on the engine. If W > 0, then possibilities are:
A. Q1 > Q2 > 0
B. Q2 > Q1 > 0
C. Q2 < Q1 < 0
D. Q1 < 0, Q2 > 0
Answer:
From the above diagram, we get that 
it is given that W>0
∴ 
Thus, there arises two cases –
(a) when both 
are positive
⇒ 
(b) when both 
are negative
⇒ 
option (a, c) is correct.
Vsa
Question 1.Can a system be heated and its temperature remains constant?
Answer:
Given temperature remains constant
It states that heat supplied is used for the work in the
surroundings.
∴ it is possible that the heat applied to the system is
used in expansion.
Question 2.
A system goes from P to Q by two different paths in the P-V diagram as shown in Fig. 12.8. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?
Answer:
From first law of thermodynamics,
Given, 
We know that change in internal energy is same for two
different paths having same initial and final states.
Question 3.
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer:
If the refrigerator’s door is kept open the room will get
warmer because with the door open the temperature will start to
rise inside the refrigerator. The thermostat will kick in and try to
cool it back down. This means the motor is running which means heat is being added to the room.
Question 4.
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Answer:
In a closed system, in absence of external force field present, if internal energy of system is ‘U’ and amount of heat absorbed in the system is ‘q’ then internal energy will become U1+q. After this if ‘W’ is the amount of work is done on the system then final energy is given by U2,
According to 1st law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus work done by the system.
But here work is done on the system. So,
According to question, no heat is given or taken out of the system, so this is an adiabatic process.
if q=0
Then, 
In adiabatic compression, work done on the system is positive.
As 
is given by 
n = No. Of moles
Cv= Specific heat capacity at constant volume.
= Change in temperature
Temperature increases when internal energy increases.
Question 5.
Air pressure in a car tyre increases during driving. Explain.
Answer:
We know that when we drive there is static friction between the tire and the road which increases the temperature of the rubber of the tire and hence the air inside it.
According to ideal gas equation and Gay Lussac’s law which states that states that, for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.
Thus, increase in temperature causes increase in pressure too.
Sa
Question 1.Consider a Carnot’s cycle operating between T1 = 500K and T2=300K producing 1 k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
Answer:
Efficiency of any system, reversible cycle =
Now, Carnot theorem states that the efficiency of Carnot heat engine depends only on temperature of the source (T1) and temperature of the sink (T2) and is given by
According to question,
Temperature of source, T1=500K
Temperature of source, T2=300K
We can see from the figure that |W|=q2-q1
Now, since 
=
=
Question 2.
A person of mass 60 kg wants to lose 5kg by going up and down a 10m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?
Answer:
The calories he has to burn to lose 5 kg mass
Now, as we know 1 cal = 4.2 J
Work done=
Let, while going up work done = mgh
Therefore, while coming down,
According to the question, work done
Total work done 
Given,
Mass, m = 60kg
Height, h = 10m
let, g=10 m/s
While going up and coming down fat burnt 
no. of times he has to go up and down=
He must go up and down 16334 times to reduce his weight by 5 kg.
Question 3.
Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump ∆V (V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P1 to P2?
Answer:
Let’s understand the situation first.
Here, as air is filled adiabatically, so there is no involvement of heat in the work equation.
We know that for adiabatic compression or expansion
As according to question 
is filled every time in the tube , but the volume is not increasing, so P must be increasing.
Just for instance before affect the pressure we can write
.............(1)
But after it affects P,
.............(2)
Because the process is adiabatic equation 1 and 2 should be equal
(cancelling out
on both sides)
As 
applying binomial expansion in the right hand side, which states that,
Work done is given by, 
W
Question 4.
In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.
Answer:
One thing to be noted is that Carnot engine is the most efficient engine. Its efficiency is given by,
Here,
According to question, its efficiency is 50% of a perfect engine which is a Carnot engine.
Efficiency 
We know that,
As power of the motor is 1kW i.e. 1kJ/sec
Heat taken out of the refrigerator per second=19kJ.
Question 5.
If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature (27 °C), find the temperature inside the refrigerator.
Answer:
According to question, coefficient of performance=5
Coefficient of performance is given by 
Where, W
Now,
The temperature inside the refrigerator=
Question 6.
The initial state of a certain gas is (Pi, Vi ,Ti ). It undergoes expansion till its volume becomes Vf. Consider the following two cases:
(a) the expansion takes place at constant temperature.
(b) the expansion takes place at constant pressure. Plot the P-V diagram for each case. In which of the two cases, is the work done by the gas more?
Answer:
P vs. V for isothermal process is like hyperbola.
If initial and final conditions for an ideal gas is given and then for various processes the plot is given below
a. At constant temperature (isothermal process)
b. At constant pressure (isobaric process)
Work done is given by the area under the curve of the P Vs V plots given above.
Since area under the curve of the isobaric process is more than the isothermal process. Therefore, more work is done in isobaric process.
La
Question 1.Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig. 12.9.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature T1/T2, if V2 = 2V1?
(c) Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V2 = 2V1.
Answer:
a. We know that work done is given by an ideal gas is given by,
According to question,
Work done by the gas=
...............(1)
Since ,
Putting the value equation 1 becomes,
Work done=
c. To relate the temperature we have to modify the equation in terms of T.
From ideal gas equation we know,
PV=nRT
Where,
P=pressure
V=volume
T=absolute temperature
n=number of moles of gas
R=ideal gas constant
Here, n=1 (given)
But ,
So, comparing both we get,
Or 
=new constant=C
For state 1 let, 
For state 2 let, 
Equating both we get,
Now, givenV2 = 2V1
c.
Where, is change in internal energy given by 
n = No. Of moles
Cv= Specific heat capacity at constant volume.
= Change in temperature
...........(2)
W=
AlsoV2 = 2V1
W=
W=
W=
............(3)
Since,
and n=1
Putting the value in equation 3
W=
Now,
Question 2.
A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig. 12.10.
A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic
Vc = VD = 2VA = 2VB
(a) In which part of the cycle heat is supplied to the engine from outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c) What is the work done by the engine in one cycle? Write your answer in term of PA, PB, VA.
(d) What is the efficiency of the engine?
[ γ = 
for the gas], ( Cv = 
R for one mole)
Answer:
Given,
A to B: volume constant (isochoric)
B to C: adiabatic
C to D: volume constant (isochoric)
D to A: adiabatic
a.
We know that in adiabatic process there is no heat supplied or evolved from the system.
The part BC and AD should be eliminated as they are adiabatic process.
Now,
According to 1st law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus work done by the system.
For isochoric process ,
as 
= 0 Since, dv=0
.........(1)
Also we know internal energy of gas,
..........(2)
Where, n= No. of moles
= change in internal energy
= change in temperature
= specific heat of air at constant volume
Equating equation 1 and 2...
In Process AB we can clearly see, the pressure is increasing and volume is constant,
From Ideal gas eq. i.e.
We can say 
T
From A to B is positive
Therefore, Q is also positive. So, heat is supplied to the engine.
Heat is supplied to the engine from outside in the cycle from A to B.
b. As in part CD, the P is decreasing with V constant
is negative
Hence is negative
And as ,(when Volume is constant)
Therefore, Q is also negative.
Heat is being given to the surrounding by the engine in the cycle from C to D.
c. Work done by the system is given by
For isochoric process ,
(since from A to B and from C to D the processes are isochoric)
= 0
Similarly, 
.......(1)
We know that for adiabatic compression or expansion
...........(2)
Substituting equation 2 in 1 we get,
Putting C=
Similarly, 
Given, Vc = VD = 2VA = 2VB.........(3)
Putting the value in the above equation and solving we get,
...........(4)
Similarly,
...........(5)
Total work done= 
=
=
Putting the values from 3,4 and 5 and solving we get
Total work done=
=
=
Again given, γ = 
for the gas
Total work done 
c. We know 
We know internal energy of gas,
.......(6)
Where, n = No. Of moles
Cv= Specific heat capacity at constant volume.
= Change in temperature
Also heat supplied during process A to B,
........(7)
Equating equation 7 and 8
=
Since, PV=nRT
Question 3.
A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)
AB: constant volume
BC: constant pressure
CD: adiabatic
DA: constant pressure
Answer:
According to 1st law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus work done by the system.
As process AB is isochoric Work done by the system is given by and for isochoric process it is 
........(1)
And we know
Where, n = No. Of moles i.e. one in this case
Cv= Specific heat capacity at constant volume i.e. in this case
= Change in temperature
= 
........(2)
Using Ideal gas eq. i.e. PV=nRT
We can replace R
by PBVB and R
by PAVA
VA = VB (Given in the question)
Substituting the values in eq. 2 we get
= 
Using eq. 1
Q=
Now Process BC is isobaric i.e. Constant Pressure
Work done by the system is given by as P is constant
W = 
= 
(
)
We know by 1st law of thermodynamics
Q = + 
since,
Here n = 1 mole
Cv= 
R
= 
+(
)
Using Ideal gas eq. i.e. PV=nRT
We can replace R
by PCVC and R by PBVB
PB = PC (Given in the question)
=
+()
Q = 
Process CD is adiabatic
=0
Again Process AD is isobaric i.e Constant Pressure
Work done by the system is given by as P is constant
W = = 
(
)
We know by 1st law of thermodynamics
Q = + since,
Here n = 1 mole
Cv= R
= 
+()
Using Ideal gas eq. i.e. PV=nRT
We can replace R by PAVA and R
by PDVD
PA= PD (Given in the question)
=
+()
Q = 
Question 4.
Consider that an ideal gas (n moles) is expanding in a process given by P = f (V), which passes through a point (V0, P0). Show that the gas is absorbing heat at (P0, V0) if the slope of the curve P = f (V) is larger than the slope of the adiabatic passing through (P0, V0).
Answer:
If an adiabatic process is passing through the point P0, V0 we can write its equation as
As 
Slope at P0, V0= 
Now, coming to the process given,
P = f (V)
We can write the equation,
Since, 
Now, as we know PV=nRT by rearranging it we get,
T=
and P = f (V)
T=
Now,
dQ=ncv
we know,
=
and 
And P=f(V)
Now, the volume is increasing. dV=+ve
dQ should also be +ve if heat is given to the system.
at point (P0, V0) for heat to be absorbed.
Since, 
Hence proved.
Question 5.
Consider one mole of perfect gas in a cylinder of unit cross section with a piston attached (Fig. 12.12). A spring (spring constant k) is attached (unstretched length L) to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from Vo to V1.
(a) What is the initial pressure of the system?
(b) What is the final pressure of the system?
(c) Using the first law of thermodynamics, write down a relation between Q, Pa, V, Vo and 
k.
Answer:
a. As initially the spring was unstretched and there was only external pressure (Atmospheric pressure). So, internal pressure should be equal to external pressure.
=> Pa=Pin
b. In this part we will use simple mechanics and not thermodynamics.
We can find the relax length of spring by A x l= Vo
Where A=cross sectional area of plate
Therefore,
Similarly, in the stretched position
So, pressure due to stretched spring 
Pin(final pressure)=Pa +
c. Ideally, the mechanical energy cannot be generated by heat, but let’s suppose it is happening in this case.
Using 1st law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus work done by the system.
But here work is done on the system. So,
Now, the work done on the system is by the atmospheric pressure and the spring.
Work done by atmospheric pressure, (WPa)= F.d
WPa = F.dx
Since, Pa= 
F=Pa A
And 
WPa
Since volume has expanded and Force is in opposite direction so there will be negative sign in work done.
)
Work done by the spring=
=
Here, negative sign implies that displacement is against the force.
So, work done by the system,
And
n=1
We know that ,
(here n=1)
