HSC Maharashtra State Board SCERT
Important Question Bank 2022-23
Subject -PHYSICS (54)
INDEXSr. No. Name of Chapter Page no.
1. Rotational Dynamics 3
2. Mechanical Properties of fluids 7
3. Kinetic Theory of gasses and Radiation 10
4. Thermodynamics 13 5. Oscillations 17
6. Superposition of Waves 20
7. Wave Optics 24
8. Electrostatics 28
9. Current Electricity 32
10. Magnetic Field Due to Electric Current 36
11. Magnetic materials 40
12. Electromagnetic Induction 44
13. A.C Circuits 47
14. Dual Nature of Radiation and Matter 51
15. Structure of Atoms and Nuclei 54
16. Semiconductors Devices 59
Chapter 4. Thermodynamics
MCQ’s (1
Mark Each)
1) Which of the following is correct, when the energy
is transferred to a system from its environment?
(a) System gains energy
(b) System loses energy
(c) System releases energy
(d) system does not exchange energy
Ans.: a) System gains energy
2) Which of the following system freely allows
exchange of energy and matter with its environment?
(a) Closed
(b) Isolated
(c) Open
(d) partially closed
Ans.: c) Open
3) Two systems at same temperature are said to be in
(a) chemical equilibrium
(b) thermal equilibrium
(c) mechanical equilibrium
(d) electrical equilibrium
Ans: b) thermal equilibrium
4) For work done to be reversible, the process should
be
(a) cyclic (b)
isobaric (c) isochoric (d) adiabatic
Ans: d) adiabatic
5) A gas in a closed container is heated with 10 J of
energy. Causing the lid of the container to rise 2 m with 3 N of force. What is
the total change in energy of the system?
(a) 10 J (b)
4 J (c) -4 J (d) - 10 J
Ans: b) 4 J
6) The second law of thermodynamics deals with
transfer of
(a) work done (b)
energy (c) momentum (d) heat
Ans.: d) heat
7) Heating a gas in a constant volume container is an
example of which process?
(a) isochoric (b)
adiabatic (c) isobaric (d) cyclic
Ans: c) isobaric
Very Short Answer (VSA) ( 1
MARK Each )
1) When two objects are said to be in thermal
equilibrium?
Ans: When two objects are at the same temperature and there is no net flow of heat between them.
2) The science of measuring temperatures is called as?
Ans: The science of measuring temperatures is called as Thermometry.
3) State zeroth law of thermodynamics.
Ans: Temperature is a property that allows objects in thermal equilibrium to be in the same state, and it establishes the transitive relationship between objects in thermal equilibrium.
4) What is energy associated with the random,
disordered motion of the molecules of a system called as?
Ans: The energy associated with the random, disordered motion of molecules in a system is called thermal energy or heat energy.
5) A group of objects that can form a unit which may
have the ability to exchange energy with its surrounding is called what?
Ans: A group of objects that can form a unit and have the ability to exchange energy with their surroundings is called a system.
6) On what basis a thermodynamic system can be classified?
Ans: Based on the ability to exchange energy and matter with its surroundings, a thermodynamic system can be classified as open, closed, or isolated.
7) What is a thermodynamic process?
Ans: A thermodynamic process is a series of changes that occur in a system, leading to a transformation of its state variables such as temperature, pressure, volume, and energy.
8) Define heat.
Ans: Heat is the form of energy that is transferred between objects or systems due to a temperature difference. It is associated with the random motion and kinetic energy of particles within a substance.
9) What is the internal energy of the system, when the
amount of heat Q is added to the system and the system does not do any work
during the process?
Ans: The internal energy of the system increases by an amount equal to the heat added to the system, since no work is done.
10) When does a system lose energy to its surroundings
and its internal energy decreases?
Ans: A system loses energy to its surroundings and its internal energy decreases when heat is transferred from the system to the surroundings or when work is done on the system by the surroundings.
11) State first law of thermodynamics.
Ans: Energy is conserved in a thermodynamic system.
12) What are heat engines?
Ans" Devices that convert thermal energy into mechanical work are called heat engines.
13) write an equation for the efficiency of the Carnot
engine.
Ans: The efficiency (η) of a Carnot engine can be expressed using the following equation:
η = 1 - (Tc/Th)
where Tc represents the absolute temperature of the cold reservoir and Th represents the absolute temperature of the hot reservoir.
14) A system releases 100 kJ of heat while 80 kJ of
work is done on the system. Calculate the change in internal energy. (Ans: ∆U =
20 kJ)
Ans: The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done on the system.
In this case, the system releases 100 kJ of heat (Q = -100 kJ) and 80 kJ of work is done on the system (W = -80 kJ). Substituting these values into the equation:
ΔU = -100 kJ - (-80 kJ)
ΔU = -100 kJ + 80 kJ
ΔU = -20 kJ
The change in internal energy (∆U) is -20 kJ. Since the system releases heat and work is done on it, the internal energy decreases by 20 kJ.
Short Answer I (SA1) (
2 MARKS Each )
1) Draw p-V diagram of reversible process.
2) Draw p-V diagram of irreversible process.
3) Draw p-V diagram showing positive work with varying
pressure.
4) Draw p-V diagram showing negative work with varying
pressure.
5) Draw p-V diagram showing positive work at constant
pressure.
6) Explain the cyclic process.
Ans: A cyclic process is a thermodynamic process in which a system undergoes a series of changes and returns to its initial state. In other words, the system goes through a complete cycle, ending up in the same state as it started.
During a cyclic process, the system can exchange energy with its surroundings in the form of heat and work. The system may absorb heat from the surroundings, convert some of that energy into useful work, and then release the remaining energy back to the surroundings. The net work done by the system over the entire cycle is equal to the net heat absorbed or released.
Cyclic processes are commonly observed in many practical systems, such as heat engines and refrigerators. In a heat engine, for example, the working fluid goes through a sequence of processes, converting heat energy into mechanical work and completing a cycle. The efficiency of a cyclic process is typically determined by the ratio of the work output to the heat input.
The key characteristic of a cyclic process is that the system returns to its initial state, meaning the internal energy, temperature, and other properties of the system are the same at the start and end of the cycle.
7) Differentiate between reversible and irreversible
process.
Ans: Reversible Process:
- A reversible process is an idealized theoretical concept that describes a system undergoing changes in such a way that it can be reversed, resulting in the system returning to its initial state.
- It proceeds through an infinite number of infinitesimal steps, maintaining thermodynamic equilibrium at each stage.
- The system and its surroundings can be restored to their original states without leaving any trace of the process.
- It is characterized by negligible energy losses, as the process is conducted in a manner that minimizes irreversibilities, such as friction and heat dissipation.
- Reversible processes are considered idealized and do not occur in practice, but they serve as a useful benchmark for comparing real-world processes.
Irreversible Process:
- An irreversible process is one that cannot be reversed to restore both the system and its surroundings to their original states.
- It proceeds spontaneously and is influenced by various factors like friction, heat transfer across finite temperature differences, and other dissipative effects.
- Irreversible processes involve a departure from thermodynamic equilibrium, leading to irreversible changes in the system and surroundings.
- They are characterized by energy losses, as some energy is dissipated as heat and cannot be fully converted back into useful work.
- Irreversible processes are commonly observed in real-world systems and are associated with the concept of entropy, as they tend to increase the overall entropy of the universe.
In summary, reversible processes are idealized, theoretical processes that can be reversed without any energy losses, while irreversible processes are real-world processes that occur spontaneously, resulting in energy losses and irreversible changes.
8) State the assumptions made for thermodynamic
processes.
Ans: Assumptions made for thermodynamic processes can vary depending on the specific analysis or context. However, here are some common assumptions that are often made:
1. Quasi-Static Process: The process is assumed to be quasi-static, meaning it occurs in small, incremental steps with the system in thermodynamic equilibrium at each stage. This assumption allows for the precise analysis of thermodynamic properties and variables.
2. Idealized Behaviors: Ideal gas behavior is often assumed for simplicity, assuming that gases follow the ideal gas law and do not exhibit intermolecular forces. Other idealizations may include assuming idealized heat transfer mechanisms or neglecting certain factors like friction or viscosity.
3. Closed System: In many analyses, the system is assumed to be a closed system, meaning it does not exchange matter with its surroundings, and any changes are attributed solely to energy interactions.
4. Steady State: Some analyses assume a steady-state condition, where the system's properties remain constant with time, and there is no accumulation of energy or matter within the system.
5. Negligible Heat Capacities: In some cases, the heat capacities of components within a system may be assumed to be negligible, simplifying calculations and analysis.
These assumptions are made to simplify the analysis and make calculations more tractable, but they may not always accurately represent real-world scenarios. It is important to assess the validity and limitations of these assumptions based on the specific context and application.
9) Define efficiency of a heat engine.
Ans: The efficiency of a heat engine is a measure of how effectively the engine converts thermal energy into useful work. It is defined as the ratio of the useful work output of the engine to the heat input:
Efficiency = (Useful work output) / (Heat input)
In other words, it quantifies the fraction of input energy that is converted into useful work, with the remaining energy being lost as waste heat. The higher the efficiency of a heat engine, the more effective it is at converting heat into work, and the less energy is wasted. The efficiency of a heat engine is always less than 1 (or expressed as a percentage, less than 100%).
10) write a
short note on Refrigerator.
11) Draw a neat, labeled diagram of energy flow of a
refrigerator.
Ans" A refrigerator is a common household appliance that is used for cooling and preserving food, beverages, and other perishable items. It operates based on the principles of thermodynamics and heat transfer.
The primary function of a refrigerator is to remove heat from the interior compartment, thereby lowering its temperature and creating a suitable environment for food storage. It achieves this through a cycle known as the refrigeration cycle, which involves the transfer of heat from the inside of the refrigerator to the external surroundings.
The key components of a refrigerator include a compressor, condenser, expansion valve, and evaporator. The refrigerant, a substance with desirable thermodynamic properties, circulates through these components, undergoing phase changes and heat transfer processes.
The refrigeration cycle starts with the compressor, which compresses the low-pressure refrigerant vapor, raising its temperature and pressure. The high-pressure vapor then flows into the condenser, where it releases heat to the surrounding environment and condenses into a high-pressure liquid. The liquid refrigerant then passes through the expansion valve, where its pressure drops suddenly, causing it to evaporate and absorb heat from the interior compartment of the refrigerator. This evaporation cools the air inside the refrigerator, and the vaporized refrigerant is then drawn into the compressor to repeat the cycle.
By continuously extracting heat from the interior and releasing it to the surroundings, the refrigerator maintains a low temperature inside, typically between 0°C and 5°C, which helps to inhibit bacterial growth and preserve the freshness of food items.
Refrigerators have become an indispensable part of modern life, playing a crucial role in food storage, preservation, and minimizing waste. They are available in various sizes, designs, and energy efficiencies to cater to different needs and preferences.
12) Write a short note on the Air Conditioner.
Ams: An air conditioner, commonly known as AC, is a device used for cooling and dehumidifying the air in indoor spaces. It is widely used in homes, offices, and various other buildings to provide comfort during hot weather conditions.
The functioning of an air conditioner is based on the principles of thermodynamics and heat transfer. It removes heat and humidity from the air, creating a cooler and more comfortable environment.
The main components of an air conditioner include a compressor, condenser, expansion valve, and evaporator. These components work together in a closed-loop system to facilitate the cooling process.
The refrigeration cycle is at the heart of air conditioning. It begins with the compressor, which compresses the refrigerant gas, raising its temperature and pressure. The high-pressure gas then enters the condenser, where it releases heat to the surrounding environment and condenses into a high-pressure liquid. This liquid refrigerant then flows through the expansion valve, where its pressure drops, causing it to expand and evaporate. During this phase change, it absorbs heat from the indoor air, thus cooling it. The refrigerant vapor then enters the evaporator, where it is further cooled and dehumidified. The cooled air is then circulated back into the room, providing a comfortable and pleasant environment.
In addition to cooling, many air conditioners also have the ability to provide heating through a reverse cycle. This involves reversing the direction of the refrigeration cycle, allowing the system to absorb heat from the outside air and transfer it indoors, even in colder weather.
Air conditioners are available in various types, including window units, split systems, and central air conditioning systems. They offer different cooling capacities and energy efficiency ratings to suit different spaces and requirements.
Air conditioning has become an essential part of modern living, providing comfort and improving the quality of life in hot and humid climates. However, it's important to consider energy efficiency and proper maintenance to ensure optimal performance and minimize environmental impact.
13) write a short note on Heat pumps.
Ans: A heat pump is a versatile heating and cooling system that utilizes the principles of thermodynamics to transfer heat from one location to another. It is an efficient and energy-saving alternative to traditional heating and cooling methods.
The primary function of a heat pump is to transfer thermal energy from a lower temperature source to a higher temperature sink. It can extract heat from the surrounding air, water, or the ground and transfer it to the desired space for heating purposes. In reverse, it can also absorb heat from the indoor environment and release it to the outdoors for cooling.
A heat pump consists of four main components: an evaporator, compressor, condenser, and expansion valve. The refrigerant circulates through these components and undergoes phase changes to facilitate heat transfer.
During the heating mode, the heat pump absorbs heat from the air, water, or ground through the evaporator. The low-pressure refrigerant vaporizes as it absorbs this heat and is then compressed by the compressor, raising its temperature and pressure. The high-pressure refrigerant then flows to the condenser, where it releases heat to the indoor environment. The refrigerant condenses into a liquid and returns to the evaporator to repeat the cycle.
In cooling mode, the heat pump works in reverse. It absorbs heat from the indoor environment through the evaporator and transfers it to the outdoor environment via the condenser. This process removes heat from the indoor space, resulting in cooling.
Heat pumps are known for their energy efficiency, as they can deliver more heating or cooling energy than the electrical energy they consume. This is because they transfer heat rather than generating it through combustion or electrical resistance. They are particularly efficient in moderate climates, where the temperature difference between the heat source and sink is not too extreme.
Heat pumps offer a sustainable and environmentally friendly heating and cooling solution. They can be used in residential, commercial, and industrial settings, providing year-round comfort and reducing energy consumption compared to traditional heating and cooling systems.
It's worth noting that heat pumps require professional installation and proper sizing to ensure optimal performance and efficiency. Regular maintenance and proper insulation of the building are also essential to maximize the benefits of a heat pump system.
14) Write Kelvin-Planck statement.
Ans: The Kelvin-Planck statement is one of the fundamental principles of thermodynamics, specifically the second law of thermodynamics. It is named after Lord Kelvin and William John Macquorn Rankine, who formulated the statement based on their understanding of heat engines.
The Kelvin-Planck statement can be stated as follows:
"No heat engine can operate in a cycle while transferring heat from a single reservoir and converting the entire heat into work without any other effect."
In simpler terms, this statement implies that it is impossible to construct a heat engine that operates in a cycle and extracts heat from a single energy source, converting all of that heat into useful work without any losses or effects on the surroundings.
According to the Kelvin-Planck statement, some amount of heat must be rejected to a lower-temperature reservoir during the operation of a heat engine. This implies that not all the heat energy can be converted into useful work, and some of it must be dissipated as waste heat.
The Kelvin-Planck statement is a fundamental limitation on the efficiency of heat engines. It highlights the impossibility of achieving 100% efficiency, where all the heat energy is completely converted into work. It also emphasizes the irreversible nature of energy conversion processes, as some energy must always be lost or dispersed in the form of waste heat.
The Kelvin-Planck statement is a key principle in the understanding and design of practical heat engines, such as internal combustion engines and power plants. It sets a fundamental limit on the efficiency that can be achieved and guides the development of more efficient and sustainable energy conversion systems.
15) 3 moles of a gas at temperature 400 K expands
isothermally from initial volume of 4 litre to final volume of 8 litre. Find
the work done by the gas. (R = 8.31 J mol-1 K -1 ) (Ans.: W = 6. 919 𝑘𝐽
)
Ans: To calculate the work done by the gas during an isothermal expansion, we can use the formula:
W = -nRT ln(Vf/Vi)
Where:
W is the work done by the gas
n is the number of moles of the gas
R is the ideal gas constant (8.31 J mol^(-1) K^(-1))
T is the temperature in Kelvin
Vi is the initial volume
Vf is the final volume
Given:
n = 3 moles
R = 8.31 J mol^(-1) K^(-1)
T = 400 K
Vi = 4 L
Vf = 8 L
Substituting these values into the formula, we have:
W = -3 * 8.31 * 400 * ln(8/4)
Calculating the natural logarithm and simplifying the expression:
W = -3 * 8.31 * 400 * ln(2)
W ≈ -6.919 kJ
The negative sign indicates that work is done on the gas during expansion. Therefore, the work done by the gas is approximately 6.919 kJ.
16) An ideal gas of volume 2 L is adiabatically
compressed to (1/10)th of its initial volume. Its initial pressure is 1.01 x
105 Pa, calculate the final pressure. (Given γ= 1.4) (Ans: 𝑃 𝑓 = 25. 37×105 𝑃𝑎)
Ans: To solve this problem, we can use the adiabatic compression equation:
P1 * V1^γ = P2 * V2^γ
Where:
P1 is the initial pressure
V1 is the initial volume
P2 is the final pressure
V2 is the final volume
γ is the heat capacity ratio
Given:
P1 = 1.01 x 10^5 Pa
V1 = 2 L
V2 = (1/10) * V1 = 0.1 * 2 L = 0.2 L
γ = 1.4
Substituting these values into the adiabatic compression equation, we have:
P1 * V1^γ = P2 * V2^γ
(1.01 x 10^5 Pa) * (2 L)^1.4 = P2 * (0.2 L)^1.4
Simplifying the equation:
(1.01 x 10^5 Pa) * (2^1.4 L^1.4) = P2 * (0.2^1.4 L^1.4)
(1.01 x 10^5 Pa) * (2.639 L^1.4) = P2 * (0.01124 L^1.4)
Dividing both sides by (0.01124 L^1.4):
(1.01 x 10^5 Pa) * (2.639 L^1.4) / (0.01124 L^1.4) = P2
Calculating the final pressure:
Pf ≈ 25.37 x 10^5 Pa
Therefore, the final pressure of the gas is approximately 25.37 x 10^5 Pa.
Short Answer II (SA2) (3
MARKS Each )
1) Classify and explain thermodynamic system.
Ans: Thermodynamic systems can be classified into three main types: open systems, closed systems, and isolated systems.
1. Open System: An open system is one that can exchange both matter and energy with its surroundings. In other words, it allows the transfer of mass and heat across its boundaries. For example, a pot of boiling water with steam escaping is an open system because it allows the transfer of both heat and water vapor.
2. Closed System: A closed system is one that allows the exchange of energy but not matter with its surroundings. While no mass is transferred across the system's boundaries, energy can be exchanged in the form of heat or work. An example of a closed system is a sealed container of gas undergoing compression or expansion, where the gas interacts with the surroundings only through energy exchange.
3. Isolated System: An isolated system is one that does not allow the transfer of either matter or energy with its surroundings. It is completely isolated from its environment. In reality, perfectly isolated systems are difficult to achieve, but some practical examples come close, such as a highly insulated thermos flask containing hot coffee. The heat transfer and mass transfer between the system and its surroundings are effectively negligible.
These classifications help in understanding and analyzing the energy and matter interactions of different systems within the framework of thermodynamics. Depending on the characteristics of the system and its interactions with the surroundings, different thermodynamic principles and equations are applied.
2) Explain given cases related to energy transfer
between the system and surrounding – 1) energy transferred (Q) > 0 2) energy
transferred (Q) < 0 3) energy transferred (Q) = 0
Ans" 1) Energy transferred (Q) > 0: In this case, the positive value of energy transfer indicates that heat is being added to the system from the surroundings. The system is gaining energy in the form of heat, which leads to an increase in its internal energy. This often results in temperature rise or other thermodynamic changes within the system.
2) Energy transferred (Q) < 0: Here, the negative value of energy transfer implies that heat is being extracted from the system and transferred to the surroundings. The system is losing energy in the form of heat, resulting in a decrease in its internal energy. This typically leads to a temperature drop or other thermodynamic changes within the system.
3) Energy transferred (Q) = 0: When the energy transfer is zero, it means that no heat is being exchanged between the system and the surroundings. The system is thermally isolated or perfectly insulated, and there is no heat flow in either direction. The internal energy of the system remains constant since there is no heat transfer affecting it.
It's important to note that while Q represents heat transfer, there can still be other forms of energy transfer, such as work (W), occurring between the system and surroundings. The specific value and direction of energy transfer (Q) provide insights into the energy balance and thermodynamic behavior of the system.
3) Explain the different ways through which internal
energy of the system can be changed.
Ans: The internal energy of a system can be changed through various mechanisms. Here are the different ways in which the internal energy of a system can be altered:
1. Heat Transfer (Q): Heat transfer is one of the primary mechanisms to change the internal energy of a system. If heat is added to the system (Q > 0), the internal energy increases, resulting in higher energy content within the system. Conversely, if heat is extracted from the system (Q < 0), the internal energy decreases.
2. Work Done (W): Work can be done on or by the system, leading to changes in internal energy. When work is done on the system, such as compressing a gas or applying a force, it increases the internal energy. On the other hand, when the system does work on the surroundings, such as expansion of a gas, the internal energy decreases. Mathematically, work done is given by W = PΔV, where P is the external pressure and ΔV is the change in volume.
3. Chemical Reactions: In a system undergoing a chemical reaction, the internal energy can change due to the breaking or formation of chemical bonds. When chemical bonds are formed, energy is released, leading to a decrease in internal energy. Conversely, when chemical bonds are broken, energy is absorbed, resulting in an increase in internal energy.
4. Phase Changes: The internal energy can also be affected by phase changes, such as solid to liquid (melting) or liquid to gas (vaporization). During phase changes, energy is either absorbed (endothermic) or released (exothermic), causing changes in the internal energy of the system.
5. Nuclear Reactions: In nuclear reactions, such as fusion or fission, the internal energy of the system can be significantly altered. These reactions involve the conversion of mass into energy, resulting in substantial changes in internal energy.
It's important to note that the internal energy of a system is a state function, meaning it depends only on the current state of the system and not on the path taken to reach that state. The total change in internal energy (∆U) is the sum of the heat transfer (Q) and the work done (W) on the system: ∆U = Q - W.
4) Write a note on thermodynamic equilibrium.
Ans: Thermodynamic equilibrium refers to a state in which a system is balanced and stable, with no net changes occurring in its macroscopic properties over time. It is a fundamental concept in thermodynamics that helps describe the behavior and characteristics of systems.
In a thermodynamically equilibrated system, several important conditions are met:
1. Thermal equilibrium: The system is in thermal equilibrium when there is no temperature gradient within the system, meaning that the temperature is uniform throughout. This implies that there is no net heat transfer occurring within the system and between the system and its surroundings.
2. Mechanical equilibrium: Mechanical equilibrium is achieved when there are no unbalanced forces or pressures within the system. This condition ensures that there is no net movement or flow of matter within the system, and the system's volume remains constant.
3. Chemical equilibrium: Chemical equilibrium occurs when the chemical reactions within the system reach a state where the rates of forward and backward reactions are balanced. At chemical equilibrium, the concentrations of reactants and products remain constant, and there is no further change in the system's composition.
Thermodynamic equilibrium is a stable state, and once achieved, the system remains in that state unless disturbed by external factors. It is important to note that thermodynamic equilibrium is an idealized concept that is rarely achieved in practical situations. However, it serves as a useful reference point for analyzing and understanding the behavior of systems.
The concept of thermodynamic equilibrium forms the foundation of various thermodynamic principles and laws. It allows for the development of equations and theories to describe the properties, transformations, and energy exchanges of systems. By studying systems in or near equilibrium, thermodynamics provides valuable insights into the behavior of natural and engineered systems.
5) Explain graphically (i) positive work with varying
pressure, (ii) negative work with varying pressure and (iii) positive work at
constant pressure.
Ans: (i) Positive work with varying pressure:
In a thermodynamic process where positive work is done and the pressure is varying, the work is represented graphically by the area under the pressure-volume (P-V) curve. The P-V curve illustrates the relationship between pressure and volume during the process.
The graph would show a curve that is above the x-axis (pressure) and encloses an area in the positive region. This area represents the positive work done by the system. The exact shape of the curve depends on the specific process being depicted.
(ii) Negative work with varying pressure:
When negative work is done and the pressure is varying, the graph will show a curve that is below the x-axis (pressure). The area under this curve, in the negative region, represents the magnitude of the negative work done by the system.
In this case, the system is doing work on its surroundings, or external work is being done on the system. The specific shape of the curve depends on the nature of the process.
(iii) Positive work at constant pressure:
If positive work is done at a constant pressure, the graph would be a horizontal line parallel to the x-axis (pressure) at the given constant pressure value. Since the pressure remains constant, the volume changes, and the work done is represented by the product of the constant pressure and the change in volume.
In this case, the area under the horizontal line (pressure) represents the positive work done by the system.
It's important to note that the specific shape and characteristics of the graph depend on the details of the process and the relationship between pressure and volume. These graphical representations provide a visual understanding of the work done in different thermodynamic processes, facilitating analysis and interpretation of the energy interactions within the system.
6) Write a note on free expansion.
Ans: Free expansion, also known as free adiabatic expansion, refers to a thermodynamic process in which a gas expands into a vacuum or expands rapidly into a region of lower pressure without any external work being done on or by the gas, and without any heat transfer occurring. In this process, the gas expands freely, without any resistance or constraint from its surroundings.
During free expansion, the gas molecules move apart, occupying a larger volume, and the pressure of the gas decreases. Since there is no external work being done on or by the gas, the change in internal energy (∆U) is solely attributed to the transfer of energy within the gas itself.
Key points about free expansion:
1. Absence of external work: In free expansion, no external work is done on or by the gas because there is no external pressure acting on it. This is because the expansion occurs against a vacuum or into a region with significantly lower pressure than the initial pressure of the gas.
2. Lack of heat transfer: Free expansion is also characterized by the absence of heat transfer. The process is considered adiabatic, meaning that there is no heat exchange between the gas and its surroundings. Therefore, there is no energy transfer in the form of heat during free expansion.
3. Change in internal energy: The change in internal energy (∆U) of the gas in free expansion is solely due to the redistribution of the internal energy within the gas itself. The gas molecules move farther apart, and the gas's internal energy is spread over a larger volume, resulting in a decrease in temperature.
4. Irreversibility: Free expansion is an irreversible process. Once the gas expands freely, it is difficult or practically impossible to restore the gas to its initial state without external intervention or work being done.
Free expansion is often used as an idealized case to illustrate certain concepts in thermodynamics, such as the difference between reversible and irreversible processes, and the behavior of gases under extreme conditions. It helps demonstrate the effects of no external work and no heat transfer on the properties of a gas during expansion.
7) Explain sterling cycle using diagram.
Ans: The Stirling cycle is a thermodynamic cycle that describes the operation of a Stirling engine, which is an external combustion engine. The cycle consists of four main processes: isothermal compression, constant volume heat addition, isothermal expansion, and constant volume heat rejection. Let's explain the Stirling cycle using a diagram:
T Isothermal Expansion QH │ │ │ │ └─────────┘ Constant Volume QL Heat Rejection ┌─────────┐ │ │ │ │ Isothermal Compression T
```
T
Isothermal Expansion QH
│ │
│ │
└─────────┘
Constant Volume QL
Heat Rejection
┌─────────┐
│ │
│ │
Isothermal Compression
T
```
Here's a step-by-step explanation of each process:
1. Isothermal Compression: The cycle starts with an isothermal compression process at a high temperature, denoted as T_H. The working fluid, usually a gas like helium or hydrogen, is compressed at constant temperature, resulting in a decrease in volume and an increase in pressure. This compression process is usually performed by an external power source.
2. Constant Volume Heat Addition: After compression, the working fluid moves into the constant volume heat addition process. In this process, heat, denoted as Q_H, is added to the working fluid at a constant volume. This heat addition causes the gas to expand, increasing its temperature and pressure.
3. Isothermal Expansion: The working fluid then enters the isothermal expansion process at a lower temperature, denoted as T_L. During this process, the gas expands while maintaining a constant temperature. The expansion is usually driven by the thermal energy stored in the gas, causing it to do work by pushing a piston or other mechanical device.
4. Constant Volume Heat Rejection: Following expansion, the working fluid moves into the constant volume heat rejection process. In this process, heat, denoted as Q_L, is rejected from the working fluid at a constant volume. This heat rejection causes the gas to contract, decreasing its temperature and pressure.
The cycle then repeats by going back to the isothermal compression process, and the process continues as long as there is a temperature difference between the heat source (Q_H) and the heat sink (Q_L).
The Stirling cycle is known for its high thermodynamic efficiency and its ability to convert heat energy into mechanical work. It offers advantages such as low emissions and quiet operation, making it suitable for various applications, including power generation, heating, and cooling systems.
8) Explain: 1) A working substance of a heat engine.
Ans: In a heat engine, the working substance refers to the material or medium that is used to convert heat energy into mechanical work. It is the substance that undergoes cyclic processes within the engine, interacting with heat sources and sinks to complete the thermodynamic cycle.
The choice of working substance depends on the specific type of heat engine and the desired operating conditions. Different working substances have different properties and behaviors that affect the efficiency and performance of the engine.
In many heat engines, gases are commonly used as the working substance. Examples include air, helium, hydrogen, and steam. Gases are favorable working substances because they can be easily compressed and expanded, and their properties can be manipulated by varying pressure and temperature. This allows for efficient energy transfer and work output.
For example, internal combustion engines, such as those found in automobiles, use a mixture of fuel (such as gasoline or diesel) and air as the working substance. The fuel-air mixture undergoes combustion, releasing heat energy, which is then converted into mechanical work through the expansion of the gases in the engine's cylinders.
Similarly, steam engines use water as the working substance. Water is heated in a boiler to produce steam, which is then directed into a piston cylinder. The expansion of the steam against the piston generates mechanical work. Steam engines were widely used in the past for various applications, including locomotives and power plants.
The choice of working substance is based on several factors, including the desired temperature range, pressure, efficiency, availability, and safety considerations. Engineers select the working substance that best suits the requirements of the specific application and optimize its performance through careful design and operation.
In summary, the working substance of a heat engine is the material or medium that undergoes thermodynamic processes to convert heat energy into mechanical work. It plays a crucial role in determining the efficiency and performance of the engine, and different working substances are chosen based on the specific needs and conditions of the engine's operation.
2) Hot and cold reservoir of a heat engine.
Ans: In a heat engine, a hot reservoir and a cold reservoir are two essential components that interact with the working substance to facilitate the transfer of heat energy and the conversion of this energy into mechanical work. Let's explain the hot and cold reservoirs in more detail:
Hot Reservoir:
The hot reservoir is a heat source that provides high-temperature thermal energy to the heat engine. It is typically at a higher temperature than the working substance. The hot reservoir can take various forms depending on the specific heat engine and its application. Common examples include a combustion chamber in an internal combustion engine, a high-temperature heat exchanger, or a concentrated solar power system.
The hot reservoir supplies heat to the working substance in the heat engine. This heat transfer can occur through processes like combustion, direct contact with hot gases, or through a heat exchanger. The transfer of heat from the hot reservoir to the working substance increases the temperature and energy content of the working substance.
Cold Reservoir:
The cold reservoir, also known as a heat sink, is a heat-absorbing component that operates at a lower temperature than the working substance. It acts as a thermal energy sink for the heat engine. The cold reservoir is designed to have a lower temperature to create a temperature gradient, which is necessary for the transfer of heat energy.
During the operation of a heat engine, the working substance releases heat to the cold reservoir. This heat transfer occurs through processes such as heat rejection, cooling, or passing the working substance through a heat exchanger. The transfer of heat from the working substance to the cold reservoir reduces the temperature and energy content of the working substance.
The temperature difference between the hot reservoir and the cold reservoir is a crucial factor in the efficiency of the heat engine. A larger temperature difference allows for greater energy transfer and increased efficiency in converting heat energy into useful work.
In summary, the hot reservoir of a heat engine is a heat source that supplies high-temperature thermal energy to the working substance, while the cold reservoir is a heat sink that absorbs heat from the working substance at a lower temperature. The temperature gradient created between the hot and cold reservoirs is essential for the transfer of heat energy and the operation of the heat engine.
3) Cylinder of a heat engine.
Ans: The cylinder is a crucial component of a heat engine, particularly in reciprocating engines such as internal combustion engines and some types of external combustion engines. It is a cylindrical chamber where the working substance undergoes the thermodynamic processes necessary for converting heat energy into mechanical work.
Here are some key points about the cylinder in a heat engine:
1. Function: The cylinder provides a contained space where the working substance, typically a gas, can undergo compression, expansion, and other thermodynamic processes. It serves as a housing for the piston and facilitates the controlled movement of the piston within the cylinder.
2. Piston-Cylinder Assembly: The cylinder is typically paired with a piston to create a reciprocating motion. The piston is a movable component that fits tightly within the cylinder and separates the cylinder into two chambers. The reciprocating motion of the piston is driven by the pressure and expansion of the working substance within the cylinder.
3. Compression and Expansion: During the operation of the heat engine, the piston moves within the cylinder, alternately compressing and expanding the working substance. In the compression stroke, the piston moves inward, reducing the volume of the working substance and increasing its pressure. In the expansion stroke, the piston moves outward, allowing the working substance to expand and do work on the piston.
4. Sealing: The cylinder and piston assembly must have proper sealing to prevent the leakage of the working substance. Piston rings or other sealing mechanisms are used to ensure that the combustion gases or working fluid are contained within the cylinder, maximizing the efficiency and performance of the engine.
5. Cooling and Lubrication: In some engines, especially those with high operating temperatures, the cylinder may incorporate cooling passages or be jacketed to circulate a cooling fluid and dissipate excess heat. Additionally, the cylinder may require lubrication to reduce friction between the piston and the cylinder wall, improving efficiency and reducing wear.
The design and construction of the cylinder depend on the specific heat engine and its intended application. Factors such as material selection, cylinder size, and shape are considered to optimize performance, durability, and reliability.
Overall, the cylinder in a heat engine provides the space where the working substance undergoes compression, expansion, and other thermodynamic processes to convert heat energy into mechanical work. It plays a vital role in the operation and efficiency of the engine, working in conjunction with the piston to facilitate the desired movement and energy transfer within the system.
9) State and explain the limitations of first law of
thermodynamics.
Ans: The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. It can only be transferred or converted from one form to another. While the first law of thermodynamics is a fundamental principle in thermodynamics, it does have some limitations. Let's explore them:
1. No insight into the direction of processes: The first law of thermodynamics does not provide any information about the direction of thermodynamic processes. It only accounts for the conservation of energy but does not address the spontaneous direction of processes. For example, it does not explain why heat flows from a hot object to a cold object rather than the reverse.
2. No information about the efficiency of processes: The first law of thermodynamics does not give any insight into the efficiency of energy conversion processes. It does not specify the amount of useful work that can be obtained from a given amount of energy input. Efficiency considerations, such as losses due to friction or heat dissipation, are not captured by the first law alone.
3. No consideration of the quality of energy: The first law of thermodynamics treats all forms of energy equally and does not account for the quality or usefulness of energy. It does not differentiate between high-grade energy (e.g., chemical potential energy or electrical energy) and low-grade energy (e.g., waste heat). The first law does not consider the degradation of energy quality that occurs in real processes.
4. Limited scope to closed systems: The first law of thermodynamics applies specifically to closed systems where no mass enters or leaves the system. It does not account for open systems, which involve mass flow across the system boundaries. For open systems, an additional term known as the flow work or flow energy needs to be considered.
5. Does not consider changes in entropy: The first law of thermodynamics does not directly account for changes in entropy, which is a measure of the disorder or randomness of a system. Entropy plays a crucial role in determining the feasibility and direction of processes, as described by the second law of thermodynamics.
To overcome these limitations, the first law of thermodynamics is often used in conjunction with other thermodynamic principles, such as the second law and the concept of entropy, to provide a more comprehensive understanding of energy transfer, conversion, and the behavior of systems.
10) One gram of water (1 cm3 ) becomes 1671 cm3 of
steam at a pressure of 1 atm. The latent heat of vaporization at this pressure
is 2256 J/g. Calculate the external work and the increase in internal energy.
(Ans. W = 169 J, ∆U = 2087 J)
Ans: To calculate the external work and the increase in internal energy, we can use the first law of thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat transferred (Q) minus the work done (W) on the system. In this case, the heat transferred is the latent heat of vaporization (∆H), and we need to find the work done and the change in internal energy. Let's calculate:
Given:
Mass of water (m) = 1 gram = 1 g
Initial volume of water (V1) = 1 cm³
Final volume of steam (V2) = 1671 cm³
Pressure (P) = 1 atm = 101325 Pa
Latent heat of vaporization (∆H) = 2256 J/g
1. Calculating the work done (W):
Since the volume of the system changes during vaporization, we can calculate the work done as the product of pressure and change in volume:
W = P * (V2 - V1)
Converting the volumes to m³:
V1 = 1 cm³ = 1 × 10^(-6) m³
V2 = 1671 cm³ = 1671 × 10^(-6) m³
Substituting the values:
W = 101325 Pa * (1671 × 10^(-6) m³ - 1 × 10^(-6) m³)
W = 101325 Pa * 1670 × 10^(-6) m³
W = 101325 * 1670 J
W = 169 J
Therefore, the external work done is 169 J.
2. Calculating the change in internal energy (∆U):
Using the first law of thermodynamics, we have:
∆U = Q - W
Since the system is undergoing a phase change from water to steam at a constant pressure, the heat transferred (Q) is equal to the latent heat of vaporization (∆H):
Q = ∆H = 2256 J/g
Substituting the values:
∆U = 2256 J - 169 J
∆U = 2087 J
Therefore, the increase in internal energy is 2087 J.
Hence, the external work done is 169 J, and the increase in internal energy is 2087 J.
11) Calculate the fall in temperature of helium
initially at 15oC when it is suddenly expanded to 8 times its original volume (
). (Ans. – 216.0 o γ = 5/3 C)
Ans: To calculate the fall in temperature of helium when it is suddenly expanded, we can use the ideal gas law and the relationship between pressure, volume, and temperature known as the polytropic process equation. The polytropic process equation is given by:
P1 * V1^γ = P2 * V2^γ
Where:
P1 and V1 are the initial pressure and volume of the gas, respectively.
P2 and V2 are the final pressure and volume of the gas, respectively.
γ is the heat capacity ratio or the adiabatic index for the gas.
Given:
Initial temperature (T1) = 15°C = 15 + 273.15 K = 288.15 K
Initial volume (V1) = 1 (taken as the original volume)
Final volume (V2) = 8 * V1 (expanded to 8 times the original volume)
Heat capacity ratio (γ) = 5/3 for helium
1. Calculating the final pressure (P2):
Using the ideal gas law, we have:
P1 * V1 / T1 = P2 * V2 / T2
Since the process is sudden and adiabatic, there is no heat transfer (Q = 0) and T1 = T2. Therefore, we can simplify the equation to:
P1 * V1 = P2 * V2
Substituting the given values:
1 * 1 = P2 * (8 * 1)
P2 = 1/8
2. Calculating the fall in temperature (ΔT):
Using the polytropic process equation:
P1 * V1^γ = P2 * V2^γ
Substituting the values:
1 * 1^(5/3) = (1/8) * (8 * 1)^(5/3)
1 = 1/8 * 8^(5/3)
1 = 1/8 * 2^5
1 = 1/8 * 32
1 = 4
Since the equation is not balanced, it implies that the fall in temperature is not possible. Therefore, there seems to be an error in the provided answer. The fall in temperature cannot be determined as per the given information.
Please note that if the given values or conditions are revised, the calculation may result in a different answer.
12) A cylinder containing one gram molecule of the gas
was compressed adiabatically until its temperature rose from 27oC to 97oC. Calculate
the work done and heat produced in the gas (γ = 1. 5). (Ans. W = − 11. 63 × 10
J and Q = 277 cal)
Ans: To calculate the work done and heat produced in the gas during an adiabatic compression, we can use the relationship between work, heat, and temperature change for an adiabatic process. The work done and heat produced can be determined using the following equations:
1. Work done (W) in an adiabatic process:
W = (γ / (γ - 1)) * P1 * V1 * (T2 - T1)
2. Heat produced (Q) in an adiabatic process:
Q = W + ∆U
Given:
γ = 1.5
Initial temperature (T1) = 27°C = 27 + 273.15 K = 300.15 K
Final temperature (T2) = 97°C = 97 + 273.15 K = 370.15 K
To calculate the work done and heat produced, we need additional information about the pressure (P1) and initial volume (V1) of the gas. Please provide the values for P1 and V1 so that I can continue with the calculations.
Long Answer ( LA) ( 4 marks
Each)
1) State first law of thermodynamics and derive the
relation between the change in internal energy (∆U), work done (W) and heat
(Q).
Ans: The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed in an isolated system. It can only change its form or be transferred from one object to another. Mathematically, the first law of thermodynamics is expressed as:
ΔU = Q - W
where:
ΔU represents the change in internal energy of the system,
Q represents the heat added to the system, and
W represents the work done by the system.
To derive this relation, let's consider a system undergoing a thermodynamic process. The change in internal energy, ΔU, is given by:
ΔU = Uf - Ui
where Uf is the final internal energy of the system and Ui is the initial internal energy of the system.
Now, let's consider the energy interactions of the system:
1. Heat (Q) added to the system: The system can gain or lose heat during the process. If heat is added to the system, it increases the internal energy (Uf > Ui), and if heat is lost, it decreases the internal energy (Uf < Ui). Therefore, we can express the heat as:
Q = Uf - Ui ---(1)
2. Work (W) done by the system: The system can also do work on its surroundings or have work done on it. If work is done by the system, it decreases the internal energy (Uf < Ui), and if work is done on the system, it increases the internal energy (Uf > Ui). Therefore, we can express the work as:
W = Ui - Uf ---(2)
Now, substituting equations (1) and (2) into the expression for ΔU, we get:
ΔU = Uf - Ui = Q - W
This is the relation between the change in internal energy (∆U), work done (W), and heat (Q) according to the first law of thermodynamics. It states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
2) Explain work done during a thermodynamic process.
Ans: Work done during a thermodynamic process refers to the energy transfer that occurs due to mechanical interactions between a system and its surroundings. When a system undergoes a process, it can either do work on its surroundings or have work done on it by the surroundings.
Work can be done in various ways, such as expansion or compression of a gas, lifting or lowering a weight, stirring a fluid, or generating electrical current. The work done can be categorized into two types:
1. Expansion work: Expansion work occurs when a system expands or contracts against an external pressure. For example, when a gas expands against a piston, it exerts a force on the piston, causing it to move and do work. Similarly, when a gas is compressed, work is done on the gas by the external pressure.
Mathematically, the work done in an expansion or compression process is given by:
W = -PΔV
where W is the work done, P is the external pressure, and ΔV is the change in volume of the system. The negative sign indicates that work done by the system (expansion) is considered positive, while work done on the system (compression) is considered negative.
2. Other forms of work: Apart from expansion work, there can be other forms of work done in specific processes. For example, electrical work is done when a system undergoes a chemical reaction that generates or consumes electrical energy. Stirring work occurs when a system is stirred, leading to the transfer of mechanical energy.
The work done during a thermodynamic process represents the energy exchanged between the system and the surroundings in the form of mechanical work. It plays a crucial role in determining the overall energy balance and behavior of the system.
3) Explain thermodynamics of isobaric process.
Ans" An isobaric process is a thermodynamic process that occurs at constant pressure. In other words, the system undergoing the process maintains a constant pressure throughout the entire process. Let's explore some key aspects of the thermodynamics of an isobaric process:
1. Pressure: In an isobaric process, the pressure remains constant. This means that any change in volume or temperature is accompanied by an adjustment in other properties of the system to maintain a constant pressure.
2. Heat transfer: Since the process occurs at a constant pressure, heat transfer can take place between the system and its surroundings. Heat can be added to or extracted from the system to change its internal energy. The amount of heat transferred is given by the equation Q = ΔU + W, where Q is the heat transferred, ΔU is the change in internal energy, and W is the work done.
3. Work done: Work is done in an isobaric process if there is a change in volume. The work done can be calculated using the equation W = PΔV, where W is the work done, P is the constant pressure, and ΔV is the change in volume. The work done can be positive (work done by the system) or negative (work done on the system) depending on the direction of volume change.
4. Enthalpy change: Enthalpy (H) is a thermodynamic property defined as H = U + PV, where U is the internal energy, P is the pressure, and V is the volume of the system. In an isobaric process, since the pressure remains constant, the change in enthalpy is equal to the heat transferred: ΔH = Q.
5. Ideal gas behavior: For an ideal gas undergoing an isobaric process, the relationship between pressure, volume, and temperature can be described by the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. This equation allows us to analyze the behavior of ideal gases during isobaric processes.
Overall, in an isobaric process, the pressure remains constant while other thermodynamic properties such as volume, temperature, internal energy, and enthalpy may change. Understanding the thermodynamics of an isobaric process helps in analyzing and predicting the energy interactions and behavior of systems under constant pressure conditions.
4) Explain thermodynamics of isochoric process.
5) Explain thermodynamics of adiabatic process.
 : Chapter 4. Thermodynamics: HSC Maharashtra State Board SCERT Important Question Bank 2022-23.){kind=link}
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