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Unit 6: Heat and Thermodynamics — Long Questions

11th Class Physics · Unit 6: Heat and Thermodynamics

1.Describe the fundamental postulates of the kinetic theory of gases.

The kinetic theory of gases is a fundamental theory in physics and chemistry that explains the behaviour of gases based on the motion of their constituent particles. This theory provides a macroscopic understanding of gas properties such as pressure, temperature, and volume. Here are the key assumptions of the kinetic theory of gases:

i. Gas Particles are in Constant, Random Motion
Gas molecules are in perpetual, random motion. They move in straight lines until they collide with either another molecule or the walls of the container.

ii. Negligible Volume of Gas Particles
The volume of the individual gas molecules is negligible compared to the total volume of the gas. This means that the particles are considered point masses with no significant volume.

iii. No Intermolecular Forces
There are no attractive or repulsive forces between the gas molecules. The particles do not exert any force on each other except during collisions.

iv. Elastic Collisions
Collisions between gas molecules, and between molecules and the container walls, are perfectly elastic. This means that there is no net loss of kinetic energy during collisions. The total kinetic energy is conserved.

v. Large Number of Particles
A gas contains a large number of particles. This large number allows for the use of statistical methods to describe the properties of the gas.

vi. Average Kinetic Energy is Proportional to Temperature
The average kinetic energy of gas particles is directly proportional to the absolute temperature of the gas. This implies that as the temperature increases, the average speed of the gas particles also increases.

vii. Pressure Due to Particle Collisions
The pressure exerted by a gas on the walls of its container is due to the collisions of gas particles with the walls. The force exerted by the particles during collisions generates pressure.

viii. Time of Collisions is Negligible
The time taken for collisions between gas particles is extremely short compared to the time between collisions. This assumption simplifies the analysis of particle dynamics.

2.What are the limitations of Kinetic theory of gases?

Limitations

i. The assumptions of the kinetic theory hold true for ideal gases, but real gases exhibit deviations due to intermolecular forces and finite molecular volume, especially at high pressures and low temperatures.

ii. The kinetic theory of gases provides a macroscopic view of gas behaviour, linking macroscopic properties like pressure and temperature to the motion of gas particles, and serves as a foundational concept in understanding thermodynamics and statistical mechanics.

3.Define an ideal gas. Write down ideal gas equation. Explain Why real gas behaves like an ideal gas at low pressure and high temperature?

A gas that obeys kinetic theory of gases is termed as an ideal gas. Ideal gas equation is given by

PV = nRT ----------(1)

Here P represents pressure, V is volume, n is number of moles of the gas, R is universal gas constants R= 8.3145 J mol⁻¹ K⁻¹ and T is the absolute temperature.

Real gas to Behave Like an Ideal Gas

According to kinetic theory of gases, an ideal gas has no intermolecular interactions and its molecules are far apart from each other. For a real gas to behave like an ideal gas, some conditions must be satisfied. P.E. of the gas molecules is negligible and this have only K.E.

In Eq. (1) 'n' represents number of moles which can be given by

n = Mass of gas / Molar mass of gas = m / M

So, Eq. (1) becomes PV = (m/M) RT or PM = (m/V) RT

As density; ρ = m/V, So, ρ = PM/RT or ρ ∝ P/T, M/R is constant

The density of a gas will be low at low pressure and high temperature due to which molecules of the gas will be at large distance from each other and the intermolecular forces will be negligible. So, the real gas behaves like an ideal gas at low pressure and high temperature.

4.Derive an ideal gas equation using Boltzmann constant in the form PV = NkbT from general gas equations.

Ideal Gas equation in Terms of Boltzmann Constant

From ideal gas equation PV = nRT ----------(i)

Here n represents number of moles of the ideal gas. It can be defined as the number of atoms or molecules per unit Avogadro's number ( NA = 6.02 × 10²³).

Mathematically;
n = N / NA ----------(ii)

Substituting Eq. (ii) in Eq. (i), we have
PV = (N / NA) RT ---(iii)

The term R / NA is termed as Boltzmann constant kb,

Mathematically;
kb = R / NA ----------(iv)

Substituting the values of R and NA, kb = 8.3145 / (6.02 × 10²³) = 1.38 × 10⁻²³ JK⁻¹

PV = N kb T ----------(v)

Equation (v) gives ideal gas equation in terms of Boltzmann constant kb.

5.State and explain briefly: (i) Gas laws (ii) Boyle's law (iii) Charle's law (iv) Gay-Lussac's law.

(i) Gas Laws
There are some variables (state functions) that describe quantity of gas which includes pressure, volume, and temperature (P, V and T) with change in one variable, the second variable changes while the third is kept constant. The laws that relate these variables mutually for an ideal gas are termed as gas laws.

(ii) Boyle's Law
This law was introduced by Robert Boyle in 1662, and it provides a relationship between pressure and volume of a gas at constant temperature. It is sated that for a fixed mass, the pressure P exerted by a gas varies inversely with volume V occupied by the gas at constant temperature.

Mathematically;
P ∝ 1/V at constant T
P = constant (1/V) or PV = constant
Or P₁V₁ = P₂V₂
Boyle's law is shown graphically in Fig.

(iii) Charle's Law
Charle's law relates volume and temperature of an ideal gas for a fixed mass at constant pressure This law was formulated in 1870 by a French Physicist Jacques Charles. It is stated that the volume of given mass of gas at constant pressure is directly proportional to the absolute temperature.

Mathematically;
V ∝ T at constant pressure
Or V / T = constant
Or V₁ / T₁ = V₂ / T₂
Graphically, it can be shown in Fig.

(iv) Gay-Lussac's Law
It states that for a fixed mass of an ideal gas, the pressure exerted by a gas varies directly with the absolute temperature of the gas at constant volume.

Mathematically;
P ∝ T at constant volume
Or P = constant T
Or P / T = constant
Or P₁ / T₁ = P₂ / T₂
Graphically, Gay-Lussac's Law is shown in Fig.

6.What is thermal equilibrium? Explain with an example.

Thermal Equilibrium

When two bodies are at the same temperature, the thermal energy (which is related to the kinetic energy of particles) of each body is equal. As a result, there is no driving force for heat transfer between them, and thus they remain in thermal equilibrium

Example

When we put a metal spoon into a hot cup of coffee
(i) Initially, the coffee is hotter than the spoon.
(ii) Over time, heat flows from the coffee to the spoon.
(iii) Eventually, the coffee and spoon reach the same temperature.
Thermal equilibrium is achieved at this point, there is no net heat flow between the coffee and the spoon, and they are said to be in thermal equilibrium.

7.Define and explain internal energy.

Internal Energy

The sum of all forms of molecular energies (kinetic and potential) of a substance in a system is termed as its internal energy. The molecules of an ideal gas are mere mass points which exert no forces on one another. So, the internal energy of an ideal gas system is generally the translational K.E. of its molecules.

Dependence on temperature

Since the temperature of a system is defined as the average K.E. of its molecules. Thus for an ideal gas system, the internal energy is directly proportional to its temperature.

< (1/2) m v² >= (3/2) kb T

Where kb is Boltzmann constant. Therefore, the rise in temperature of an object reflects an increase in the internal kinetic energy of its particles. This increase in internal energy can occur due to the absorption of heat energy, which raises the average kinetic energy of the particles and thus increases the temperature of the object.

8.Describe the transfer of energy into work and heat. Calculate the work done by a thermodynamic system. (OR) Justify how work and heat are similar?

Work and heat

We know that both heat and work are related to the transfer of energy by some means. This idea was applied first to the construction of steam engine which converts heat into the work. Both heat in and work out are taken as positive quantities. Hence work done by the system (gas) on its environment is taken positive while work done on the system by the environment is taken negative.

When an amount of heat "Q" enters the system, it either appears as an increase in internal energy of the system or is used up on doing work by the system on its surroundings.

Expression for Work

Consider a gas enclosed in a cylinder, which is fitted with a freely movable piston. Let the cross-sectional area of the piston is "A" as shown in Fig. The system is in equilibrium and occupies volume "V" and exerts a pressure "P" on walls of cylinder and piston.

P = F / A

Or F = PA

Let the piston moves upward through small distance "Δy", the work done by the system is

W = Fd Cos θ where θ = 0°
W = Fd Cos 0°
W = Fd (1)
Putting F = PA and d = Δy, we get
W = PA Δ Y

But A Δy = ΔV = Change in volume of gas.
W = P ΔV
∴ W = P ΔV

Where ΔV = V₂ - V₁

∴ W = P(V₂ - V₁) ----------(1)

When gas expands V₂ > V₁ then by equation (1) the work done by the system will be positive.

When gas is compressed (V₂ < V₁) then the work done on the system is negative.

9.State and explain first law of thermodynamics.

First Law of Thermodynamics

Statement

In any thermodynamic process, when heat Q is added to a system, this energy appears as an increase in the internal energy ΔU stored in the system plus the work W done by the system on its surroundings.

Mathematically

Q = ΔU + W

Conservation Principle

Explanation The underlying principle of the first law of thermodynamics is the conservation of energy. It asserts that although energy can change from one form to another (such as from chemical potential energy to thermal energy), the total amount of energy in an isolated system remains constant over time.

Wide Applicability

Beyond mechanical systems, the first law of thermodynamics applies universally to all forms of energy and all types of processes, including chemical reactions, electrical systems, and nuclear reactions. It provides a foundational understanding that allows scientists and engineers to predict and understand energy transformations in various contexts.

The first law of thermodynamics expresses the law of conservation of energy by affirming that energy is a conserved quantity in isolated systems. It provides a framework to understand how energy is transferred and transformed within systems without violating the fundamental principle that energy cannot be created nor destroyed. This alignment underscores the broader applicability and importance of the first law in understanding the behaviour of energy in the universe.

Examples of First Law of Thermodynamics

i. Bicycle Pump

A bicycle pump is a good example. When we pump on the handle rapidly, it becomes hot due to mechanical work done on the gas, raising thereby its internal energy. One such simple arrangement is shown in Fig. It consists of a bicycle pump with a blocked outlet. A thermocouple connected through a blocked outlet allows the air temperature to be monitored. When piston is rapidly pushed, thermometer shows a temperature rise due to increase of internal energy of the air. The push force does work on the air, thereby, increasing its internal energy, which is observed as a rise in temperature.

ii. Human Metabolism

Human metabolism also provides an example of energy conservation. Human beings and other animals do work when they walk, run, or move. Work requires energy. Energy is also needed for growth to make new cells and to replace old cells that have died. Energy transforming processes that occur within an organism are named as metabolism. We can apply the first law of thermodynamics.

ΔU = Q - W

To an organism of the human body. Work done will result in the decrease in internal energy of the body. Consequently, the body temperature or in other words internal energy is maintained by the food we eat.

10.Discuss the applications of first law of thermodynamics. (OR) Discuss the following processes and draw P-V diagram in each case. (i) Isothermal process (ii) Adiabatic process

Application of First law of thermodynamics

i. Isothermal Process:

"The process during which the temperature of the system remains constant is known as isothermal process".

For gaseous system during isothermal process product of pressure and volume remain constant i.e. the Boyle's law is fulfilled when a gas expands or compresses isothermally. If P₁, V₁ are initial pressure and volume and P₂, V₂ are final pressure and final volume then

P₁V₁ = P₂V₂

Since, the internal energy of an ideal gas depends upon the temperature therefore, during isothermal process change in internal energy will be zero i.e. ΔU = 0. By 1ˢᵗ law of thermodynamics

Q = ΔU + W
Q = 0 + W
Or Q = W

Hence all the heat energy supplied during isothermal process is converted into work. Thus when gas expand and do external work 'W' then amount of heat 'Q' has to be supplied to the gas in order to produce an isothermal change. Since Transfer of heat from one place to another require time hence to keep the temperature of the gas constant, the expansion or compression must take place slowly.

Graphical Representation of isothermal process:

Graphically the process on PV-plane is represented by a curved line known as isotherm, this is shown in Fig.

ii. Adiabatic Process:

"The process, in which the system does not exchange heat energy i.e. no heat enters or leaves the system, is known as adiabatic process".

In this case Q = 0. By applying 1ˢᵗ law of thermodynamics we can write

Q = ΔU + W
Or ΔU = - W

Thus, if the gas expands and does external work, it is done at the cost of internal energy of its molecules. Therefore, the temperature of the gas falls. On the other hand, if the gas is compressed adiabatically the temperature of the gas will rise. The adiabatic change occur when gas expands or is compressed rapidly so that heat should not get any time to leave or enter into the system.

Examples

i. The rapid escape of air from a burst tyre.
ii. The rapid expansion and compression of air through which a sound wave is passing.
iii. Cloud formation in the atmosphere.

During the adiabatic processes there is change in internal energy, therefore temperature of the system will not remain same. in this case:

PVγ =constant

Where γ = Cp / Cv

Graphical representation

The adiabatic process on the PV plane is represented by the curve which is steeper than the corresponding "isotherm". The curve is known as "adiabat," and is shown in Fig.

11.Describe reversible and irreversible processes with examples.

Reversible Process

The process which can be retraced by reversing the controlling factors without producing any change in the surrounding is known as reversible process.

Explanation In reversible process, the working substance passes through the same stages as in the direct proves but thermal and mechanical effects at each stage are exactly reversed. It means that during a reversible process, if heat is absorbed in the direct process, it will be given out in reverse process. Similarly, if work is done by the working substance in direct process, an equal amount of work will be done on working substance in reverse process. Hence thermal and mechanical effects are exactly reversed.

Examples i. Slow expansion and compression of the gas.
ii. Liquefaction and evaporation.

Irreversible process

A process which cannot be retraced in the backward direction by reversing the controlling factors is known as irreversible process.

Explanation All sudden changes which involve frictional effects or dissipation of energy are called irreversible. The energy dissipation may be the result of conduction, convection or radiation.

Example (i) Work done against friction is an irreversible process.
(ii) A chemical explosion.
(iii) All engines in practical life.

12.What is a heat engine? Explain its principle.

Heat Engine "The device which converts heat energy into mechanical work is known as a heat engine".

Usually the heat comes from burning of fuel. The first heat engine was a steam engine. It was developed on the fact that when water is boiled in a vessel covered with a lid the steam inside tried to push the lid showing ability to do work.

Principle

A heat engine consists of a hot reservoir which is a source of heat and a cold reservoir which is also known as sink. The working substance is needed which absorbs heat Q₁ from source, converts some of it into mechanical work W and rest of the heat energy Q₂ is rejected to cold reservoir or sink. For a continuous supply of work the heat engine is made cyclic i.e., the working substance is brought back to its initial state repeatedly.

13.State and explain second law of thermodynamics.

The first Law of thermodynamics tells us that heat energy can be converted into equivalent amount of work, but it tells nothing about the conditions under which this conversion takes place. Second law is concerned with the circumstances in which heat can be converted into work and direction of flow of heat.

Explanation The engine or the system is represented by a block diagram as shown in Fig. It absorbs a quantity of heat Q₁ from the source at temperature T₁ after doing work W expels heat Q₂ to low temperature reservoir at temperature T₂. The working substance undergoes a cyclic process which finally bring back the system to its initial state. Therefore, the change in internal energy ΔU is zero. Using first law of thermodynamics i.e.
ΔQ = ΔU + W

Where ΔQ = Q₁ − Q₂
∴ W = Q₁ − Q₂

In a real heat engine of a motor car convert a part of the energy obtained from burning of fuel into work and rest of the energy is rejected to the atmosphere. The petrol engine converts 25% and diesel engine converts 35 to 40% of total energy into mechanical work.

Lord Kelvin's Statement According to Lord Kelvin, it is impossible to devise a process which may convert heat taken from a single reservoir entirely into work without leaving any change in the working system. This is illustrated in Fig.

14.What is a Carnot engine? Describe the construction, principle and working of Carnot engine. Derive the expression for the efficiency of Carnot engine. Also state Carnot theorem.

Carnot Engine

In 1840 Sadi Carnot proposed an ideal heat engine which involves isothermal and adiabatic processes. He showed that efficiency of such engine in maximum when it works in reversible cyclic process between two heat reservoir at different temperature. Such heat engine is free from all sorts of frictional losses and losses of heat due to conduction.

The Carnot's cycle using an ideal gas as a working substance is shown on PV diagram as shown in Fig. The Carnot's engine consists of a cylinder with non-conducting walls, piston and conducting base. An ideal gas is enclosed in the cylinder as a working substance.

Principle Like other cyclic heat engines Carnot's engine also take heat energy from the hot body and convert into work while the remaining part of energy is rejected to the sink or cold body.

Working of Carnot Cycle The cyclic process in which the engine operates is known as Carnot's cycle. It consists of following four processes (i) Isothermal expansion (ii) Adiabatic expansion (iii) Isothermal compression (iv) Adiabatic compression. This is shown in Fig.

i. Isothermal expansion:
This process is realized by placing a Carnot's engine on a hot reservoir at temperature T₁ and gas is allowed to expand. During the expansion temperature of the gas falls and engine absorb Q₁ amount of heat from the high temperature reservoir. In this way temperature of the gas will remain constant. This process is represented by curve AB in the Fig.

ii. Adiabatic expansion:
Carnot's engine is placed on an insulator and gas is allowed to expand further adiabatically. The temperature of the gas falls from T₁ to T₂. This process is represented by the curve BC as shown is Fig.

iii. Isothermal compression:
The engine is now placed at cold reservoir. Gas is compressed slowly. During the compression the temperature of the gas increases. In order to keep the temperature constant engine rejects heat energy Q₂ to the cold reservoir. This process is represented by the curve CD as shown is Fig.

iv. Adiabatic compression:
The engine is finally placed on an insulator and gas is compressed slowly. The temperature of the gas rises from T₂ to and the system returns to its initial state. This is represented by the curve DA as shown in Fig.

Expression for efficiency In Carnot's cycle the system finally returns to its initial state therefore there is no change in its internal energy i.e., ΔU = 0.
The network done 'W' in the cyclic process is equal to the area enclosed by the curve ABCDA on PV diagram. The net heat absorb Q in one cycle is given by
Q = Q₁ − Q₂

Using first law of thermodynamics.
Q = ΔU + W
Q₁ − Q₂ = 0 + W
Or W = Q₁ − Q₂ -------- (1)

The efficiency η of the heat engine is defined as
η = Output/Input
η = W/Q₁ ----------(2)

Substituting value of W from equation (1), we get
η = (Q₁ − Q₂)/Q₁ = Q₁/Q₁ − Q₂/Q₁
η = (1 − Q₂/Q₁) -------- (3)

The energy transfer in isothermal expansion or compression comes out to be proportional to Kelvin's temperature of source and sink.
i.e., Q₁ ∝ T₁
and Q₂ ∝ T₂

Where T₁ and T₂ are the Kelvin's temperature of HTR and LTR respectively
∴ Q₂/Q₁ = T₂/T₁ ----------(4)

Using equation (4) and equation (3), we get.
η = (1 − T₂/T₁) ---------- (5)

The efficiency is usually taken in percentage, therefore,
Percentage Efficiency = (1 − T₂/T₁) × 100

Results From the above discussion it becomes evident:
i. Efficiency of a Carnot engine is always less than 100%.
ii. The efficiency of Carnot engine depends upon the temperature difference of H.T.R and L.T.R.
iii. The efficiency of the engine is independent of the nature of working substance.

Carnot Theorem No heat engine can be more efficient than a Carnot engine operating between the same two temperatures.

OR

"All reversible engines operating between the same two temperatures have the same efficiency, irrespective of the nature of working substance."

In most practical cases the sink is at environment temperature so the efficiency can only be increases by raising the temperature of hot reservoir. All the real heat engines are less efficient than Carnot engine due to friction and other heat losses.

15.What is a refrigerator? Describe its principle and working. Also find the co-efficient of performance of refrigerator.

Refrigerator

Refrigerator is a device which maintains the temperature of a body below that of its surrounding. It principle operates in a cyclic process but in reverse as that of the heat engine as shown in Fig.

Working of refrigerator

A refrigerator absorbs heat from a cold reservoir and gives it off to a hot reservoir. This shows that in a refrigerator, the work is done on the system while in a heat engine work is done by the system.

A refrigerator works on the basis of Clausius statement of second law of thermodynamics, i.e., a heat engine is operating in reverse. Heat Qc is drawn from Low Temperature Reservoir (L.T.R) by compressor and is thrown into High Temperature Reservoir (H.T.R) with the help of external work done. The heat rejected to H.T.R (QH) is given by

Qc + W = QH or W = QH − Qc

The main purpose of refrigerator is to extract as much heat Qc as possible from L.T.R with the expenditure of as little work W as possible.

Co-efficient of Performance of Refrigerator

The ratio of heat removed from L.T.R (Qc) to the work done (W) is called co-efficient of performance of a refrigerator.

A better refrigerator will remove a greater amount of heat from inside the refrigerator for the expenditure of a smaller mechanical work or electrical energy. The co-efficient of performance of a refrigerator can be given by
E = Qc/W = Qc/(QH − Qc)

Co-efficient of performance in terms of temperature, where Q ∝ T, is
E = T₁/(T₂ − T₁)

16.What is entropy? Explain in detail.

Entropy

Entropy is a state function in thermodynamics and it was introduced by Rudolf Clausius in 1856 to give quantitative meanings to the second law of thermodynamics. It is another variable like pressure, volume, temperature and internal energy. Like other state functions it is the change in entropy of the system and not its absolute value, which is important, when a system undergoes a reversible process.

Expression Let a system undergoes a reversible process in which it absorb ΔQ amount of heat at absolute temperature T. the change in entropy "ΔS" is given by
ΔS = ΔQ/T

Where ΔS is positive when heat is absorbed by the system ΔS is negative when heat is removed from the system. SI unit for entropy is JK⁻¹. Like internal energy the change in entropy depends upon the initial and final states of the system.

Change in entropy during irreversible process:
Consider an irreversible process in which heat is transferred from hot to the cold body through a metallic rod. Let the temperature of hot and cold bodies are T₁ and T₂ respectively.

The amount of heat removed from the hot body is totally transferred to the cold body. The change in entropy of the hot body ΔS₁ is given by:
ΔS₂ = −Q/T₁

Change in entropy of the cold body ΔS₂ is:
ΔS₂ = Q/T₂

Net change in entropy ΔS
ΔS = ΔS₁ + ΔS₂
ΔS = −Q/T₁ + Q/T₂
ΔS = Q/T₂ − Q/T₁

As T₂ is less then T₁ i.e. T₂ < T₁
∴ Q/T₂ > Q/T₁
⇒ ΔS > 0

Therefore, the entropy increases during all natural processes where heat flows from one system to another.

Example Every time entropy increases, the opportunity to convert some heat into work is lost. For example, there is an increase in entropy when hot and cold waters are mixed. Finally,the warm water cannot be separated into a hot layer and a cold layer. There has been no loss of energy but some of the energy is no longer available for conversion into work. Therefore, increase in entropy means degradation of energy from a higher level where more work can be extracted to a lower level at which less or no useful work can be done. The energy in a sense is degraded, going from more orderly form to less orderly form, eventually ending up as thermal energy.

In all real processes where heat transfer occurs, the energy available for doing useful work decreases. In other words, the entropy increases. Even if the temperature of some system decreases, thereby decreasing the entropy, it is at the expense of net increase in entropy for some other system. When all the systems are taken together as the universe, the entropy of the universe always increases.

17.Increase in entropy means degradation of energy. Discuss.

Whenever the entropy of the system increases then available heat energy which can be converted into work is lost. For example, there is an increase in entropy when hot and cold water are mixed. The warm water which results cannot be separated into hot layer and cold layer. Although, there is no loss of energy but some of the energy is no longer available for conversion into work. Therefore, as a result of increase in entropy, the energy is degraded from a higher level where more work, can be extracted to lower level at which less or no useful work can be done.

18.Explain second law of thermodynamics in terms of entropy.

The second law of thermodynamics in terms of entropy can also be defined as:
"When a system undergoes a natural process, it will always proceed in a direction that cases the entropy of system and environment to increase."

Explanation It is observed that natural processes proceed toward the state of greater disorder, thus there in relation between entropy and molecular disorder. For example, in irreversible process heat flow from bot to cold body results in increase in disorder. It is because molecules are initially sorted on hot and cold regions. This order is lost when the system comes to thermal equilibrium. Addition of heat to the system increases disorder because of increase in average molecular speed. Similarly, in fine expansion of gas increases disorder because the molecules have greater randomness of position after expansion. In both examples the entropy of the system increases.