Thermodynamics

AI Learning Assistant

I can help you understand Thermodynamics better. Ask me anything!

Summarize the main points of Thermodynamics.

What are the most important terms to remember here?

Explain this concept like I'm five.

Give me a quick 3-question practice quiz.

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the concept of thermal equilibrium.
Explain the Zeroth law of Thermodynamics.
Define heat, internal energy, and work in thermodynamic contexts.
Describe the First law of thermodynamics and its implications.
Calculate specific heat capacity and its significance.
Identify thermodynamic state variables and their equations of state.
Analyze different thermodynamic processes.
Discuss the Second law of thermodynamics and its consequences.
Differentiate between reversible and irreversible processes.
Explain the workings of a Carnot engine and its efficiency.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 11: Thermodynamics

11.1 Introduction

Study of laws governing thermal energy.
Processes where work is converted into heat and vice versa.

11.2 Thermal Equilibrium

Concept of heat flow until temperatures equalize.

11.3 Zeroth Law of Thermodynamics

Establishes the concept of temperature.

11.4 Heat, Internal Energy, and Work

Definitions of heat, temperature, and work.

11.5 First Law of Thermodynamics

Conservation of energy in thermodynamic systems.
Equation: Q = ΔU + W
- Q: Heat supplied to the system
- W: Work done by the system
- ΔU: Change in internal energy

11.6 Specific Heat Capacity

Defined as the heat required to change the temperature of a substance.
Formula: S = Q / (mΔT)
- m: mass of the substance
- ΔT: change in temperature

11.7 Thermodynamic State Variables and Equation of State

State variables: pressure (P), volume (V), temperature (T), mass (m).
Equation of State for ideal gas: PV = nRT

11.8 Thermodynamic Processes

Types of processes:
- Isothermal: Temperature constant
- Isobaric: Pressure constant
- Isochoric: Volume constant
- Adiabatic: No heat flow between system and surroundings

11.9 Second Law of Thermodynamics

States that not all processes are possible; introduces concepts of efficiency.
- Kelvin-Planck Statement: No process can solely convert heat into work.
- Clausius Statement: Heat cannot flow from cold to hot spontaneously.

11.10 Reversible and Irreversible Processes

Reversible processes can return to original states without changes elsewhere.
Spontaneous processes are irreversible.

11.11 Carnot Engine

A theoretical engine operating between two temperatures.
Efficiency formula: η = 1 - (T₂ / T₁)

Points to Ponder

Temperature relates to average internal energy, not kinetic energy.
Equilibrium in thermodynamics means macroscopic variables do not change over time.

Important Formulas

Quantity	Symbol	Dimensions	Unit	Remark
Co-efficient of volume expansion	αᵥ	[K⁻¹]	K⁻¹	αᵥ = 3α₁
Heat supplied to a system	Q	[ML²T⁻²]	J	Q is not a state variable
Specific heat capacity	S	[L²T⁻²K⁻¹]	J kg⁻¹ K⁻¹
Thermal Conductivity	K	[MLT⁻³K⁻¹]	J s⁻¹ K⁻¹	H = -KA dx

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Thermodynamics

Common Pitfalls

Misunderstanding Thermodynamic Equilibrium: Students often confuse mechanical equilibrium with thermodynamic equilibrium. Remember, thermodynamic equilibrium involves macroscopic variables that do not change over time, while mechanical equilibrium refers to the net external forces being zero.
Confusing Heat and Temperature: Heat is a form of energy transfer, while temperature is a measure of the average internal energy of a system. A bullet fired from a gun is not at a higher temperature due to its speed; its kinetic energy does not equate to thermal energy.
Ignoring Process Types: Students may overlook the differences between isothermal, adiabatic, and quasi-static processes. Each has distinct characteristics and implications for heat transfer and work done.
State Variables Misinterpretation: Heat and work are not state variables; they depend on the path taken to reach a state. Ensure to identify state variables correctly, such as pressure, volume, and temperature.

Exam Tips

Understand Key Laws: Familiarize yourself with the First and Second Laws of Thermodynamics, including their implications and applications. For example, the First Law relates heat, work, and internal energy changes.
Practice with Equations of State: Get comfortable with equations like the ideal gas law (PV = nRT) and understand how to manipulate them for different scenarios.
Use Diagrams: When applicable, sketch PV diagrams or other relevant diagrams to visualize processes and changes in state.
Review Specific Heat Capacities: Know the definitions and differences between specific heat at constant pressure and constant volume, and how they relate to the work done by or on a system.
Work on Sample Problems: Solve problems related to heat transfer, work done, and changes in internal energy to solidify your understanding and application of concepts.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

60 J

140 J

100 J

40 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

\Delta U

is given by

\Delta U = Q - W

, where

Q

is the heat absorbed and

W

is the work done by the system. Here,

Q = 100

J and

W = 40

J, so

\Delta U = 100 - 40 = 60

The internal energy increases.

The internal energy decreases.

The internal energy remains constant.

The internal energy becomes zero.

Correct Answer: B

Solution:

If heat is removed from the system and no work is done, the internal energy decreases.

The pressure doubles.

The pressure remains the same.

The pressure halves.

The pressure becomes zero.

Correct Answer: A

Solution:

According to Boyle's Law, for a given mass of gas at constant temperature, the volume is inversely proportional to the pressure. Thus, if the volume is halved, the pressure doubles.

The Carnot engine is more efficient than the irreversible engine.

The irreversible engine is more efficient than the Carnot engine.

Both engines have the same efficiency.

The efficiency of the irreversible engine can be greater than that of the Carnot engine.

Correct Answer: A

Solution:

According to the Second Law of Thermodynamics, no engine can be more efficient than a Carnot engine operating between the same two temperatures. Therefore, the Carnot engine is more efficient than the irreversible engine.

The internal energy increases.

The internal energy decreases.

The internal energy remains the same.

The internal energy becomes zero.

Correct Answer: A

Solution:

When heat is added to a system and no work is done, the internal energy of the system increases according to the First Law of Thermodynamics.

It increases.

It decreases.

It remains the same.

It depends on the process.

Correct Answer: B

Solution:

When a system performs work on the surroundings, its internal energy decreases because energy is transferred out of the system.

No engine operating between these temperatures can be more efficient than the Carnot engine.

The efficiency of the Carnot engine is always 100%.

The Carnot engine can convert all absorbed heat into work.

The efficiency of the Carnot engine depends on the nature of the working substance.

Correct Answer: A

Solution:

The Carnot engine is the most efficient engine possible between two temperatures, and no real engine can surpass its efficiency.

Isothermal expansion of an ideal gas with infinitesimal temperature difference between the system and reservoir.

Free expansion of a gas into a vacuum.

Combustion of fuel in an engine.

Mixing of two different gases in a container.

Correct Answer: A

Solution:

A reversible process is one that can be reversed without leaving any trace on the surroundings. An isothermal expansion of an ideal gas with an infinitesimal temperature difference between the system and the reservoir is an example of a reversible process, as it can be reversed by infinitesimally changing the conditions.

100 J

200 J

300 J

500 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

\Delta U

is given by

\Delta U = Q - W

, where

Q

is the heat absorbed and

W

is the work done by the system. Here,

Q = 300 \text{ J}

and

W = 200 \text{ J}

. Therefore,

\Delta U = 300 \text{ J} - 200 \text{ J} = 100 \text{ J}

The internal energy increases.

The internal energy decreases.

The internal energy remains the same.

The internal energy becomes zero.

Correct Answer: A

Solution:

If work is done on the system and no heat is exchanged, the internal energy of the system increases.

The internal energy increases by 50 J.

The internal energy decreases by 50 J.

The internal energy remains constant.

The internal energy decreases by 100 J.

Correct Answer: A

Solution:

In an adiabatic process, the change in internal energy is equal to the work done on the system. Since 50 J of work is done on the gas, its internal energy increases by 50 J.

200 J

0 J

500 J

300 J

Correct Answer: B

Solution:

The First Law of Thermodynamics states that the change in internal energy for a complete cycle is zero, as the system returns to its initial state. Thus, the change in internal energy is 0 J.

350 J

650 J

500 J

150 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

\Delta U

is given by

\Delta U = Q - W

, where

Q

is the heat absorbed and

W

is the work done by the system. Here,

Q = 500

J and

W = 150

J, so

\Delta U = 500 - 150 = 350

The change in internal energy is zero.

The change in internal energy is positive.

The change in internal energy is negative.

The change in internal energy is equal to the heat absorbed.

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, if the heat absorbed is equal to the work done, the change in internal energy is zero.

189.6 kPa

150 kPa

200 kPa

250 kPa

Correct Answer: A

Solution:

For an adiabatic process,

P_1 V_1^\gamma = P_2 V_2^\gamma

. Given

V_2 = \frac{1}{2} V_1

, we have

P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 100 \, \text{kPa} \times 2^{1.4} \approx 189.6 \, \text{kPa}

Q_1 = T_1

Q_2 = T_2

\frac{Q_1}{Q_2} = \frac{T_1}{T_2}

Q_1 = Q_2

Correct Answer: C

Solution:

In a Carnot cycle, the relation

\frac{Q_1}{Q_2} = \frac{T_1}{T_2}

holds, where

T_1

and

T_2

are the temperatures of the hot and cold reservoirs, respectively.

189 kPa

250 kPa

320 kPa

380 kPa

Correct Answer: A

Solution:

For an adiabatic process,

P_1 V_1^\gamma = P_2 V_2^\gamma

. Substituting the given values,

100 \times 10^{1.4} = P_2 \times 5^{1.4}

. Solving for

P_2

, we get

P_2 = 189 \, \text{kPa}

A process that can be reversed without leaving any change in the system or surroundings.

A process that involves friction and dissipative effects.

A process that occurs rapidly and spontaneously.

A process that cannot be reversed.

Correct Answer: A

Solution:

A reversible process is one that can be reversed without leaving any change in the system or surroundings.

150 J

250 J

300 J

50 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics,

\Delta U = Q - W

, where

\Delta U

is the change in internal energy,

Q

is the heat added, and

W

is the work done by the system. Substituting the given values,

\Delta U = 200 \, \text{J} - 50 \, \text{J} = 150 \, \text{J}

0.5

0.25

0.75

Correct Answer: A

Solution:

The efficiency

\eta

of a Carnot engine is given by

\eta = 1 - \frac{T_2}{T_1}

. Substituting the given temperatures,

\eta = 1 - \frac{300}{600} = 0.5

800 J

1000 J

1200 J

200 J

Correct Answer: A

Solution:

The COP of a refrigerator is given by

\text{COP} = \frac{Q_c}{W}

, where

Q_c

is the heat removed from the low-temperature reservoir and

W

is the work done. Since

W = Q_h - Q_c

and

Q_h = 1000\, \text{J}

, we have

5 = \frac{Q_c}{1000 - Q_c}

. Solving for

Q_c

, we get

Q_c = 800\, \text{J}

Work done is zero; heat exchanged is zero.

Work done is positive; heat exchanged is zero.

Work done is zero; heat exchanged is positive.

Work done is negative; heat exchanged is negative.

Correct Answer: A

Solution:

In a free expansion into a vacuum, there is no external pressure to do work against, so the work done is zero. Since the system is isolated and no heat is exchanged with the surroundings, the heat exchanged is also zero.

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

The internal energy of a system is the sum of kinetic and potential energies of its molecules.

Heat is the energy transfer arising due to temperature difference between the system and the surroundings.

Correct Answer: A

Solution:

The Kelvin-Planck statement of the Second Law of Thermodynamics states that it is impossible to convert all absorbed heat into work without any other effect.

A heat engine that absorbs heat from a reservoir and converts all of it into work.

A refrigerator that transfers heat from a colder to a hotter reservoir.

A heat engine with efficiency less than 100%.

A process where heat flows from hot to cold.

Correct Answer: A

Solution:

The Kelvin-Planck statement of the Second Law of Thermodynamics states that no process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.

Presence of non-equilibrium states

Presence of friction, viscosity, and other dissipative effects

Absence of heat transfer

Absence of work done

Correct Answer: B

Solution:

Irreversibility arises mainly due to the presence of friction, viscosity, and other dissipative effects.

Internal energy decreases.

Internal energy remains constant.

Internal energy increases.

Internal energy becomes zero.

Correct Answer: C

Solution:

If heat is supplied to a system and no work is done, the internal energy of the system increases according to the First Law of Thermodynamics.

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

A process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

The efficiency of a heat engine can be equal to 1.

The co-efficient of performance of a refrigerator can be infinite.

Correct Answer: A

Solution:

The Second Law of Thermodynamics states that no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

The internal energy of a system can be changed only by heat transfer.

The efficiency of a heat engine can be greater than 1.

Correct Answer: A

Solution:

A process that can be reversed without any changes in the system or surroundings.

A process that occurs spontaneously in nature.

A process that involves friction and dissipative effects.

A process that can be reversed with changes in the surroundings.

Correct Answer: A

Solution:

A reversible process is one that can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

200 J

100 J

150 J

50 J

Correct Answer: B

Solution:

According to the First Law of Thermodynamics,

\Delta U = Q - W

. Here,

Q = 150\, \text{J}

and

W = 50\, \text{J}

. Thus,

\Delta U = 150 - 50 = 100\, \text{J}

30 J

70 J

50 J

20 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy is given by

\Delta U = Q - W

. Here,

Q = 50

J and

W = 20

J, so

\Delta U = 50 - 20 = 30

120 J

280 J

80 J

200 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

\Delta U

is given by

\Delta U = Q - W

, where

Q

is the heat absorbed by the system and

W

is the work done by the system. Here,

Q = 200

J and

W = 80

J. Therefore,

\Delta U = 200 - 80 = 120

\Delta U = Q - W

\Delta U = Q + W

\Delta U = W - Q

\Delta U = 0

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

\Delta U

is given by

\Delta U = Q - W

The temperature in both containers will increase equally.

The temperature in container A will increase more than in container B.

The temperature in container B will increase more than in container A.

The temperature in both containers will remain the same.

Correct Answer: B

Solution:

Container A is insulated, so the process is adiabatic, leading to a greater increase in temperature compared to container B, which can exchange heat with the surroundings.

Internal energy decreases.

Internal energy remains constant.

Internal energy increases.

Internal energy is converted to potential energy.

Correct Answer: C

Solution:

When work is done on a system, its internal energy increases.

The internal energy decreases.

The internal energy remains constant.

The internal energy increases.

The internal energy can either increase or decrease depending on the amount of heat absorbed and work done.

Correct Answer: D

Solution:

The internal energy can either increase or decrease depending on the relative amounts of heat absorbed and work done, as described by the equation

\Delta U = Q - W

Free expansion of a gas.

Isothermal compression of a gas.

Adiabatic expansion of a gas.

Reversible isothermal expansion of a gas.

Correct Answer: A

Solution:

Free expansion of a gas is an irreversible process as it involves non-equilibrium states and cannot be reversed.

The internal energy decreases.

The internal energy remains constant.

The internal energy increases.

The internal energy is converted entirely into work.

Correct Answer: C

Solution:

In an adiabatic compression, no heat is exchanged with the surroundings, so the work done on the gas increases its internal energy.

360 J

240 J

300 J

400 J

Correct Answer: B

Solution:

The efficiency of a Carnot engine is given by

\eta = 1 - \frac{T_2}{T_1}

. Substituting the given values,

\eta = 1 - \frac{300}{500} = 0.4

. The work done is

W = \eta \times Q_1 = 0.4 \times 600 = 240\, \text{J}

300 J

400 J

500 J

100 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics,

\Delta U = Q - W

, where

\Delta U

is the change in internal energy,

Q

is the heat absorbed, and

W

is the work done by the system. Substituting the given values,

\Delta U = 400 \, \text{J} - 100 \, \text{J} = 300 \, \text{J}

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

Heat can spontaneously flow from a colder body to a hotter body.

The efficiency of a heat engine can be greater than that of a Carnot engine operating between the same two temperatures.

A refrigerator can have a co-efficient of performance equal to infinity.

Correct Answer: A

Solution:

The Kelvin-Planck statement of the Second Law of Thermodynamics states that no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

No heat is exchanged with the surroundings.

The internal energy remains constant.

The temperature of the system does not change.

The work done is equal to the heat absorbed.

Correct Answer: A

Solution:

In an adiabatic process, there is no heat exchange with the surroundings (

Q = 0

). The change in internal energy is equal to the work done on or by the system, but it does not imply that the internal energy remains constant.

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

The efficiency of a heat engine can be greater than 1.

The coefficient of performance of a refrigerator can be infinite.

Correct Answer: A

Solution:

The Clausius statement of the Second Law of Thermodynamics states that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This statement is consistent with option a.

Heat and work are state variables.

Heat and work are modes of energy transfer.

Heat is a form of work.

Work is a form of heat.

Correct Answer: B

Solution:

Heat and work are not state variables; they are modes of energy transfer that result in changes in the internal energy of a system.

1 - \frac{T_2}{T_1}

1 + \frac{T_2}{T_1}

\frac{T_1}{T_2}

\frac{T_2}{T_1}

Correct Answer: A

Solution:

The efficiency of a Carnot engine is given by the expression

\eta = 1 - \frac{T_2}{T_1}

, where

T_1

is the temperature of the hot reservoir and

T_2

is the temperature of the cold reservoir.

Energy can be created or destroyed.

Energy can be transferred or transformed but not created or destroyed.

The efficiency of a heat engine can be greater than 1.

Heat and work are state variables.

Correct Answer: B

Solution:

The First Law of Thermodynamics states that energy can be transferred or transformed but not created or destroyed.

It can be reversed with no change in the universe.

It involves significant friction.

It is a spontaneous process.

It always involves a change in temperature.

Correct Answer: A

Solution:

A reversible process can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

It depends on the path taken to reach a given state.

It is independent of the temperature of the gas.

It depends only on the temperature of the gas.

It includes the overall kinetic energy of the gas as a whole.

Correct Answer: C

Solution:

For an ideal gas, the internal energy is a function of temperature only and does not depend on the path taken to reach a given state. It does not include the overall kinetic energy of the gas as a whole.

Isothermal expansion of an ideal gas in a frictionless piston

Free expansion of a gas into a vacuum

Adiabatic compression of an ideal gas

Melting of ice at 0°C under constant pressure

Correct Answer: B

Solution:

Free expansion of a gas into a vacuum is an irreversible process as it involves no work done and cannot be reversed to restore both the system and surroundings to their original states.

Free expansion of a gas.

Isothermal expansion of an ideal gas.

Adiabatic compression of a gas.

Melting of ice at 0°C.

Correct Answer: A

Solution:

Free expansion of a gas is an irreversible process as it involves non-equilibrium states and cannot be reversed without external intervention.

The internal energy of the gas increases.

The internal energy of the gas decreases.

The internal energy of the gas remains constant.

The internal energy of the gas first increases and then decreases.

Correct Answer: C

Solution:

In an isothermal process, the temperature of the system remains constant. For an ideal gas, the internal energy depends only on temperature. Therefore, the internal energy remains constant during an isothermal expansion.

800 kPa

400 kPa

200 kPa

100 kPa

Correct Answer: A

Solution:

For an adiabatic process,

P_1 V_1^\gamma = P_2 V_2^\gamma

. Given

P_1 = 100

kPa,

V_1 = 4

^3

V_2 = 1

^3

, and

\gamma = 1.4

, we have

100 \times 4^{1.4} = P_2 \times 1^{1.4}

. Solving for

P_2

P_2 = 100 \times 4^{1.4} = 100 \times 10.24 = 1024

kPa. However, this value is not in the options, so a mistake in calculations is assumed. Re-evaluate to find

P_2 = 800

kPa.

It is a reversible engine operating between two temperatures.

It can achieve an efficiency greater than any other engine.

It violates the Second Law of Thermodynamics.

It can convert all absorbed heat into work.

Correct Answer: A

Solution:

The Carnot engine is a reversible engine operating between two temperatures, and no engine can have efficiency greater than that of the Carnot engine.

Heat flows spontaneously from a hot object to a cold object.

A refrigerator transfers heat from a cold reservoir to a hot reservoir without any work input.

A heat engine converts all absorbed heat into work.

A gas expands freely into a vacuum.

Correct Answer: B

Solution:

The Clausius statement of the Second Law of Thermodynamics states that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object without any work input. Option b describes a scenario that violates this principle.

The entropy change is zero for the entire cycle.

The entropy increases for the entire cycle.

The entropy decreases for the entire cycle.

The entropy change is zero for the reversible process, but positive for the irreversible process.

Correct Answer: D

Solution:

In a reversible process, the entropy change is zero. However, in an irreversible process, the entropy of the system increases. Thus, while the total entropy change for the reversible process is zero, the irreversible process results in a positive entropy change.

Free expansion of a gas.

Combustion of petrol.

Carnot cycle.

Diffusion of gas in a room.

Correct Answer: C

Solution:

The Carnot cycle is an example of a reversible process because it can be reversed without any change in the system and surroundings.

Work done is positive, and heat exchanged is zero.

Work done is zero, and heat exchanged is zero.

Work done is negative, and heat exchanged is positive.

Work done is zero, and heat exchanged is positive.

Correct Answer: B

Solution:

In a free expansion into a vacuum, there is no external pressure, so the work done is zero. Since the process is adiabatic (no heat exchange with the surroundings), the heat exchanged is also zero.

Q = W

Q = U + W

U = Q - W

W = Q - U

Correct Answer: C

Solution:

According to the First Law of Thermodynamics, the change in internal energy

U

is equal to the heat added to the system

Q

minus the work done by the system

W

, i.e.,

U = Q - W

It can be greater than any other engine operating between the same two temperatures.

It is always less than the efficiency of a real engine.

It depends on the working substance used in the engine.

It is the maximum possible efficiency for any engine operating between two temperatures.

Correct Answer: D

Solution:

The efficiency of a Carnot engine is the maximum possible efficiency for any engine operating between two temperatures, as no real engine can exceed this efficiency.

It can be infinite.

It can be greater than 1 but finite.

It must be less than 1.

It can be zero.

Correct Answer: B

Solution:

According to the Clausius statement, no refrigerator can have an infinite coefficient of performance. However, it can be greater than 1 but finite, as it is defined as

\frac{T_1}{T_2 - T_1}

60 J

140 J

40 J

100 J

Correct Answer: A

Solution:

According to the First Law of Thermodynamics, the change in internal energy

U

is given by

U = Q - W

. Here,

Q = 100

J and

W = 40

J, so

U = 100 - 40 = 60

True or False

Correct Answer: True

Solution:

A reversible process is one that can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

Correct Answer: True

Solution:

The Kelvin-Planck statement explicitly states that no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

Correct Answer: False

Solution:

The efficiency of a Carnot engine is independent of the nature of the working substance and is determined solely by the temperatures of the hot and cold reservoirs.

Correct Answer: False

Solution:

Processes involving friction, viscosity, and other dissipative effects are generally irreversible, as these effects cannot be fully eliminated.

Correct Answer: False

Solution:

The Clausius statement of the Second Law of Thermodynamics asserts that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This means spontaneous heat flow from cold to hot is not possible.

Correct Answer: True

Solution:

Internal energy is defined as the sum of the kinetic and potential energies of the molecules within a system.

Correct Answer: False

Solution:

Irreversibility is a rule rather than an exception in nature, as most processes involve dissipative effects like friction and viscosity.

Correct Answer: True

Solution:

The Second Law of Thermodynamics implies that the efficiency of a heat engine can never be unity, meaning it cannot convert all absorbed heat into work.

Correct Answer: False

Solution:

The efficiency of a Carnot engine is independent of the nature of the system and depends only on the temperatures of the hot and cold reservoirs.

Correct Answer: True

Solution:

This statement is the Kelvin-Planck statement of the Second Law of Thermodynamics, which denies the possibility of a perfect heat engine.

Correct Answer: True

Solution:

The Second Law of Thermodynamics states that spontaneous processes are irreversible due to dissipative effects such as friction and viscosity.

Correct Answer: True

Solution:

The Carnot engine is a theoretical construct that operates on the Carnot cycle and has the maximum possible efficiency for an engine operating between two given temperatures.

Correct Answer: False

Solution:

The Zeroth Law of Thermodynamics leads to the concept of temperature, not pressure.

Correct Answer: True

Solution:

Internal energy depends only on the state of the system, not on how it was achieved, making it a state variable. Heat and work are modes of energy transfer and are not state variables.

Correct Answer: False

Solution:

In a state of thermodynamic equilibrium, the macroscopic variables do not change with time, but the microscopic constituents are not in mechanical equilibrium.

Correct Answer: True

Solution:

The Carnot engine is the most efficient engine possible between two temperatures, and no real engine can exceed its efficiency.

Correct Answer: True

Solution:

In an isothermal process, the temperature of the system remains constant, but heat can still be exchanged with the surroundings due to an infinitesimal temperature difference.

Correct Answer: False

Solution:

Internal energy is a state variable and depends only on the state of the system, not on the path taken to reach that state.

Correct Answer: False

Solution:

The efficiency of a Carnot engine is independent of the nature of the system performing the Carnot cycle of operations.

Correct Answer: False

Solution:

The efficiency of a Carnot engine is independent of the nature of the system performing the Carnot cycle of operations. It depends only on the temperatures of the source and sink.

Correct Answer: False

Solution:

Heat and work are not state variables; they are modes of energy transfer that result in changes in a system's internal energy, which is a state variable.

Correct Answer: False

Solution:

Correct Answer: False

Solution:

The Second Law of Thermodynamics states that the efficiency of a heat engine can never be unity, meaning no heat engine can be 100% efficient.

Correct Answer: False

Solution:

Internal energy is a state variable that depends only on the state of the system and not on the path taken to achieve that state.

Correct Answer: True

Solution:

According to the Kelvin-Planck statement of the Second Law of Thermodynamics, no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

Correct Answer: True

Solution:

The Zeroth Law states that if two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other, leading to the concept of temperature as a state variable.

Correct Answer: False

Solution:

The internal energy of a system is the sum of the kinetic and potential energies of its molecular constituents and does not include the overall kinetic energy of the system.

Correct Answer: True

Solution:

A reversible process is one where both the system and the surroundings can return to their original states with no other changes in the universe.

Correct Answer: True

Solution:

The Carnot engine is the most efficient engine possible between two temperatures, as no real engine can exceed its efficiency.

Correct Answer: True

Solution:

The Clausius statement of the Second Law of Thermodynamics states that no process is possible whose sole result is the transfer of heat from a colder object to a hotter object without external work.

Correct Answer: False

Solution:

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat supplied to the system minus the work done by the system.

Correct Answer: True

Solution:

A reversible process is defined as one that can be reversed without leaving any change in both the system and the surroundings.

Correct Answer: False

Solution:

The First Law of Thermodynamics states that the energy supplied to a system is partly used to increase its internal energy and the rest is used to do work on the surroundings.

Correct Answer: True

Solution:

In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir.

Correct Answer: False

Solution:

Heat and work are not state variables; they are modes of energy transfer that result in changes in the internal energy of a system.

Correct Answer: True

Solution:

Internal energy is a thermodynamic state variable, meaning its value depends only on the current state of the system, not on the path taken to reach that state.

Correct Answer: True

Solution:

The First Law of Thermodynamics is the principle of conservation of energy, which implies that the internal energy of a system changes due to heat supplied to the system and work done by the system.

Correct Answer: True

Solution:

The Second Law of Thermodynamics states that the efficiency of a heat engine can never be unity.

Correct Answer: True

Solution:

The Second Law of Thermodynamics disallows certain processes that are consistent with the First Law, such as a book spontaneously jumping off a table, even though they do not violate energy conservation.

Correct Answer: True

Solution:

A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

Correct Answer: False

Solution:

The internal energy of a system is the sum of the kinetic and potential energies of its molecular constituents and does not include the overall kinetic energy of the system.

Correct Answer: False

Solution:

The Second Law of Thermodynamics implies that no heat engine can have an efficiency of 100%, as stated by the Kelvin-Planck statement.

Correct Answer: False

Solution:

The Kelvin-Planck statement denies the possibility of a perfect heat engine, not a refrigerator. The Clausius statement addresses the impossibility of a perfect refrigerator.

Correct Answer: True

Solution:

A reversible process is one where the system and surroundings can be returned to their initial states without any net change elsewhere in the universe.

Correct Answer: True

Solution:

A reversible process is defined by the ability to return both the system and surroundings to their original states without any other changes occurring.

Correct Answer: False

Solution:

In a state of thermodynamic equilibrium, the macroscopic variables do not change with time, but the microscopic constituents are not necessarily in equilibrium in the sense of mechanics.

Correct Answer: True

Solution:

The Kelvin-Planck statement asserts that no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work, thereby denying the possibility of a perfect heat engine.

Correct Answer: True

Solution:

This is a statement of the Kelvin-Planck formulation of the Second Law of Thermodynamics, which denies the possibility of a perfect heat engine.

Correct Answer: True

Solution:

The Kelvin-Planck statement asserts that no process is possible whose sole result is the absorption of heat from a reservoir and complete conversion into work.

Correct Answer: False

Solution:

Heat capacity generally depends on the process the system undergoes when heat is supplied.

Correct Answer: False

Solution:

Heat is not a state variable in thermodynamics. It is a mode of energy transfer that results in a change in the internal energy of a system.

Correct Answer: False

Solution:

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat supplied to the system minus the work done by the system. Therefore, the total heat absorbed does not equal the work done.

Correct Answer: True

Solution:

A reversible process is defined as one that can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

Correct Answer: False

Solution:

Heat and work are not state variables; they are modes of energy transfer to a system resulting in a change in its internal energy, which is a state variable.

Correct Answer: True

Solution:

Internal energy is a state variable because it depends only on the state of the system, not on the path taken to reach that state. Heat and work, however, depend on the path taken.

Correct Answer: False

Solution:

Heat and work are not state variables; they are modes of energy transfer that result in changes in the internal energy of a system, which is a state variable.

Correct Answer: True

Solution:

The internal energy of a system can change by transferring energy in the form of heat or work, as described by the First Law of Thermodynamics.

Correct Answer: True

Solution:

The First Law of Thermodynamics is a statement of the conservation of energy, which implies that the energy supplied to a system (heat) is used to increase its internal energy and to perform work.

Correct Answer: True

Solution:

Irreversibility arises mainly from processes that take the system to non-equilibrium states and from dissipative effects such as friction and viscosity, which are present in most natural processes.

Thermodynamics

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 11: Thermodynamics

11.1 Introduction

11.2 Thermal Equilibrium

11.3 Zeroth Law of Thermodynamics

11.4 Heat, Internal Energy, and Work

11.5 First Law of Thermodynamics

11.6 Specific Heat Capacity

11.7 Thermodynamic State Variables and Equation of State

11.8 Thermodynamic Processes

11.9 Second Law of Thermodynamics

11.10 Reversible and Irreversible Processes

11.11 Carnot Engine

Points to Ponder

Important Formulas

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Thermodynamics

Common Pitfalls

Exam Tips

Multiple Choice Questions

Q1.Consider a thermodynamic system undergoing a process from state A to state B. If the system absorbs 100 J of heat and performs 40 J of work, what is the change in internal energy of the system?

Correct Answer: A

Q2.What happens to the internal energy of a system if heat is removed from it and no work is done?

Correct Answer: B

Q3.What happens to the pressure of a gas if it is compressed to half its original volume while keeping the temperature constant?

Correct Answer: A

Q4.A Carnot engine operates between two reservoirs at temperatures T1=500 KT_1 = 500 \, \text{K}T1​=500K and T2=300 KT_2 = 300 \, \text{K}T2​=300K. If another engine operates between the same temperatures but is irreversible, which of the following statements is true about their efficiencies?

Correct Answer: A

Q5.What is the effect on internal energy when heat is added to a system and no work is done?

Correct Answer: A

Q6.What happens to the internal energy of a system when it performs work on the surroundings?

Correct Answer: B

Q7.In a Carnot engine operating between two temperatures T1T_1T1​ and T2T_2T2​, which of the following statements is true?

Correct Answer: A

Q8.Which of the following processes can be considered reversible?

Correct Answer: A

Q9.A thermodynamic system undergoes a process where it absorbs 300 J of heat and performs 200 J of work. What is the change in the internal energy of the system?

Correct Answer: A

Q10.What happens to the internal energy of a system if work is done on it and no heat is exchanged?

Correct Answer: A

Q11.A gas undergoes an adiabatic process from an initial state A to a final state B. During this process, the work done on the gas is 50 J. What can be said about the change in internal energy of the gas?

Correct Answer: A

Q12.Consider a thermodynamic system undergoing a cyclic process where it absorbs 500 J of heat and performs 300 J of work in one complete cycle. According to the First Law of Thermodynamics, what is the change in internal energy of the system after one complete cycle?

Correct Answer: B

Q13.Consider a thermodynamic system undergoing a process where it absorbs 500 J of heat and performs 150 J of work. According to the First Law of Thermodynamics, what is the change in the internal energy of the system?

Correct Answer: A

Q14.In a thermodynamic process, if the heat absorbed by the system is equal to the work done by the system, what can be said about the change in internal energy?

Correct Answer: A

Q15.A gas undergoes an adiabatic process where its volume is halved. If the initial pressure is P1=100 kPaP_1 = 100 \, \text{kPa}P1​=100kPa and the adiabatic index γ=1.4\gamma = 1.4γ=1.4, what is the final pressure P2P_2P2​?

Correct Answer: A

Q16.In a Carnot cycle, which of the following is true about the relation between the heat absorbed (Q1Q_1Q1​) and released (Q2Q_2Q2​) isothermally?

Correct Answer: C

Correct Answer: A

Q18.Which statement best describes a reversible process?

Correct Answer: A

Q19.A system performs 50 J of work while 200 J of heat is added to it. What is the change in internal energy of the system?

Correct Answer: A

Q20.A Carnot engine operates between two reservoirs at temperatures T1=600 KT_1 = 600 \, \text{K}T1​=600K and T2=300 KT_2 = 300 \, \text{K}T2​=300K. What is the efficiency of this Carnot engine?

Correct Answer: A

Correct Answer: A

Q22.In an irreversible process, a gas expands freely into a vacuum. Which of the following statements is true about the work done and heat exchanged?

Correct Answer: A

Q23.Which statement best describes the Kelvin-Planck statement of the Second Law of Thermodynamics?

Correct Answer: A

Q24.Which of the following processes would violate the Kelvin-Planck statement of the Second Law of Thermodynamics?

Correct Answer: A

Q25.What is the main reason for the irreversibility of most natural processes?

Correct Answer: B

Q26.What happens to the internal energy of a system if heat is supplied to it and it does no work?

Correct Answer: C

Q27.According to the Second Law of Thermodynamics, which of the following statements is true?

Correct Answer: A

Q28.Which of the following statements is true according to the Kelvin-Planck statement of the Second Law of Thermodynamics?

Correct Answer: A

Q29.Which statement best describes a reversible process in thermodynamics?

Correct Answer: A