Class 11 Physics (Physics Part-II) Chapter 11 Thermodynamics

Key Concept: Heat Transfer

a) Absorbed from the reservoir

[Solution Description] In the isothermal expansion phase of the Carnot cycle, the system absorbs heat from the high-temperature reservoir. The temperature remains constant at $T_1$, and the volume of the gas expands from $V_1$ to $V_2$, with the absorbed heat equal to the work done by the gas: $$Q_1 = W_{1 \to 2}$$.

Key Concept: Extensive vs. Intensive Variables

c) Density

[Solution Description] Intensive properties are those that do not depend on the size or extent of the system. Temperature is an example of an intensive property because it remains constant irrespective of the quantity of matter.

Key Concept: Thermodynamics Laws, Internal Energy

c) 0 J

[Solution Description] For a complete cycle, the system returns to its original state, so the change in internal energy $\Delta U = 0$ according to the first law of thermodynamics, which states that $\Delta U = Q – W$. Since it’s a cyclic process,

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

However, since it’s a full cycle returning to original state:

$$\Delta U = 0 \, \text{J}$$

Key Concept: Internal Energy Change

c) Internal Energy

[Solution Description] During an isochoric process, the heat absorbed by the gas does not go into doing work (since no work is done), instead it increases the internal energy of the gas.

Key Concept: Zeroth Law Application

c) System X and Z are in thermal equilibrium.

[Solution Description] According to the Zeroth Law of Thermodynamics, if two systems are each in equilibrium with a third system, they must be in equilibrium with each other. Therefore, systems X and Z must be in thermal equilibrium.

Key Concept: Process Analysis, Mathematical Application

c) $W = RT \ln\left(\frac{P_1}{P_2}\right)$

[Solution Description] Since it is an isothermal process for an ideal gas, we use the formula for work done:

$$W = RT \ln\left(\frac{V_2}{V_1}\right)$$

For an isothermal process, using Boyle’s Law:

$$P_1 V_1 = P_2 V_2$$

Substituting $$V_2 = \frac{P_1 V_1}{P_2}$$ in the equation for $W$:

$$W = RT \ln\left(\frac{P_1 V_1 / P_2}{V_1}\right) = RT \ln\left(\frac{P_1}{P_2}\right)$$

Hence, the work done is related to the pressures as given by:

$$W = RT \ln\left(\frac{P_1}{P_2}\right)$$

Key Concept: Basic Definition

b) Clausius statement

[Solution Description] The Clausius statement of the Second Law of Thermodynamics directly states that it is impossible for heat to flow from a colder object to a hotter object spontaneously or without external work being done.

Key Concept: Energy Conversion

c) Conversion of work to heat

[Solution Description] Rumford’s experiment demonstrated the conversion of mechanical work into heat energy, thereby providing evidence against the caloric theory and supporting the understanding of heat as a form of energy.

Key Concept: Real vs Ideal Gases

d) Real gases deviate from $PV = nRT$ at high pressures

[Solution Description] For an ideal gas, the equation of state is given by $PV = nRT$, where $P$ is pressure, $V$ is volume, $T$ is temperature, $n$ is moles of gas, and $R$ is the gas constant. Real gases deviate from this behavior under high pressure and low temperature conditions because intermolecular forces and the finite size of molecules become significant. Thus, real gases do not always adhere strictly to this relationship.

Key Concept: Basic Concept

c) Mass of the substance

[Solution Description] While specific heat capacity depends on the nature of the substance and sometimes slightly on the temperature, it is independent of the mass of the substance itself.

Key Concept: Definition Focus

d) No engine operating between two temperatures can have efficiency greater than that of the Carnot engine

[Solution Description] According to Carnot’s theorem, no engine operating between two temperatures can have an efficiency greater than that of a Carnot engine.

Key Concept: Comparison with Non-Equilibrium

d) Quasi-static processes have well-defined state variables

[Solution Description] Quasi-static processes are characterized by well-defined state variables since they maintain thermal and mechanical equilibrium throughout the process. In contrast, non-equilibrium processes do not have such clearly defined variables because the system is not in equilibrium during these processes.

Key Concept: Basic Definition

b) A reversible engine

[Solution Description] A Carnot engine is defined as a reversible heat engine that operates between two temperatures.

Key Concept: Equation of State

c) Final pressure is 2.5 atm

[Solution Description] Using the ideal gas equation for initial and final states:

For the initial state:

$$P_i V_i = nRT$$

Since $nRT$ is constant for isothermal conditions, we have:

$$P_i V_i = P_f V_f$$

Substituting $$P_i = 5 \, \text{atm}$$, $$V_i = 10 \, \text{L}$$, and $$V_f = 20 \, \text{L}$$:

$$5 \times 10 = P_f \times 20$$

Solving for $P_f$:

$$P_f = \frac{50}{20} = 2.5 \, \text{atm}$$

So, the final pressure is 2.5 atm.

Key Concept: Basic Definition

b) No process can completely convert absorbed heat into work

[Solution Description] The Kelvin-Planck statement implies that it is impossible for any device to operate on a cycle and produce no effect other than the transfer of heat from a single reservoir, converting all heat into work.

Key Concept: Conceptual Understanding

b) Nature of the substance

[Solution Description] The specific heat capacity of a substance is primarily dependent on the nature of the substance itself. Different substances have different atomic structures and bonding which affect how they absorb heat. Mass, volume, and shape do not directly influence the value of specific heat capacity; it is an intrinsic property.

Key Concept: First Law Application

b) 60 J

[Solution Description] According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$, where $\Delta Q$ is the heat supplied, $\Delta W$ is the work done by the system, and $\Delta U$ is the change in internal energy. Here, $\Delta Q = 100$ J and $\Delta W = 40$ J. We have:

$$100 = \Delta U + 40$$

Subtracting 40 from both sides gives:

$$\Delta U = 100 – 40 = 60$$

Therefore, the change in internal energy is 60 J.

Key Concept: Basic Definition

a) Heat flow from a cold object to a hot object without external work

[Solution Description] The Clausius statement asserts that it is impossible for heat to be transferred from a colder object to a hotter object without performing work. This implies that option A describes an impossible process.

Key Concept: Basic Concept

b) Entropy is a measure of disorder.

[Solution Description] Entropy is a measure reflecting the amount of disorder or randomness in a system. It quantifies how much energy in a system is unavailable for doing work.

Key Concept: Equilibrium Concepts

d) Constant macroscopic variables

[Solution Description] In thermodynamics, equilibrium refers to a state where all macroscopic variables remain constant over time, indicating no net change within the system.

Key Concept: Work in Isothermal Process

c) 2740 J

[Solution Description] For an isothermal process, the work done by an ideal gas is given by:

$$W = \mu RT \ln\left(\frac{V2}{V1}\right)$$

where $\mu$ is the number of moles, $R$ is the universal gas constant $(8.314\: J/mol\cdot K)$, $T$ is the temperature, and $V_1$ and $V_2$ are initial and final volumes respectively.

Plugging in the given values:

$$\mu = 1 \: \text{mol}$$

$$R = 8.314 \: J/mol\cdot K$$

$$T = 300 \: K$$

$$V1 = 5 \: L = 0.005 \: m^3 \: (\text{conversion from L to } m^3)$$

$$V2 = 15 \: L = 0.015 \: m^3 \: (\text{conversion from L to } m^3)$$

Work done, $W$:

$$W = (1)(8.314)(300) \ln\left(\frac{0.015}{0.005}\right)$$

$$= 2494.2 \times \ln(3)$$

$$\approx 2494.2 \times 1.0986$$

$$\approx 2740.65 \: J$$

Key Concept: Basic Concept

c) It can be reversed with no change in the universe

[Solution Description] A reversible process is defined as one that can be reversed without changing the surroundings or the system permanently. It involves quasi-static changes, meaning it happens infinitely slowly to maintain thermodynamic equilibrium, and it does not include any dissipative effects.

Key Concept: Temperature Scale

c) By establishing temperature as a measurable quantity

[Solution Description] The Zeroth Law of Thermodynamics implies that if two systems are both in thermal equilibrium with a third one, they are in equilibrium with each other, establishing a consistent measurement for temperature. This consistency is crucial for defining temperature as a measurable quantity on a scale.

Key Concept: Intermediate Application

c) X and Y will maintain their initial states without further heat exchange

[Solution Description] As per the Zeroth Law of Thermodynamics, if system X is in thermal equilibrium with Z, and Y is in thermal equilibrium with Z, then X and Y are in thermal equilibrium with each other. If the diathermic wall allowing energy flow (and thus maintaining equilibrium) is replaced by an adiabatic wall that prevents energy flow, then X and Y will not be able to exchange heat and remain isolated from each other. However, since they were already at equilibrium before the change, their temperatures will remain equal as no new energy exchanges occur after the wall replacement.

Key Concept: Reversible Process

c) Both a and b

[Solution Description] The Carnot cycle consists of four processes: two isothermal and two adiabatic. All processes within this cycle are thermodynamically reversible. Therefore, both isothermal expansion and adiabatic expansion are reversible processes in a Carnot cycle.

Key Concept: Zeroth Law of Thermodynamics

c) Are in thermal equilibrium with each other

[Solution Description] The Zeroth Law states that if two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.

Key Concept: Thermal Equilibrium Conditions

c) They have the same temperature.

[Solution Description] Two systems are in thermal equilibrium when their macroscopic variables do not change over time, meaning they have equal temperatures. The presence of a diathermic wall ensures energy flow until this equilibrium is reached.

Key Concept: Work Calculation

a) $W = \frac{P_1V_1 – P_2V_2}{\gamma – 1}$

[Solution Description] To calculate the work done in an adiabatic process, we use the formula:

$$W = \frac{P_1V_1 – P_2V_2}{\gamma – 1}$$

This formula represents the work done by or on the gas as it moves from one state to another without exchanging heat with its surroundings.

Key Concept: Internal Energy Dependence

d) Internal energy depends on the state described by pressure, volume, and temperature.

[Solution Description] Internal energy is a thermodynamic state variable that depends on the state of the system, which includes pressure, volume, and temperature. It does not depend on the path taken to reach this state.

Key Concept: First Law Application

c) 70 J

[Solution Description] The First Law of Thermodynamics is expressed as: $\Delta Q = \Delta U + \Delta W$. Here, $\Delta U = 20\ \text{J}$ (internal energy increase), and $\Delta W = 50\ \text{J}$ (work done by the system). We need to find $\Delta Q$ (heat added):

$$\Delta Q = \Delta U + \Delta W = 20\ \text{J} + 50\ \text{J} = 70\ \text{J}$$

Therefore, the heat added to the system is 70 J.

Key Concept: Wall Types

d), Diathermic wall

[Solution Description]

In thermodynamic systems, diathermic walls allow energy flow between two systems. Adiabatic walls do not allow energy flow.

Key Concept: Extensive vs Intensive

c) Pressure

[Solution Description] Intensive variables do not depend on the size or extent of the system. Pressure is an intensive variable because it remains unchanged irrespective of the system’s size.

Key Concept: Reversible vs Irreversible Processes

c) It can be reversed without any change in the surroundings.

[Solution Description] A reversible process is one that can be reversed without leaving any change in both the system and the surroundings. This means it is carried out infinitely slowly so the system remains in equilibrium at all steps, allowing every state to be retraced backward.

Key Concept: Work Done Formula

b) $W = nRT \ln\left(\frac{V_2}{V_1}\right)$

[Solution Description] The work done by an ideal gas during an isothermal process is calculated using the formula $$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$, where $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the absolute temperature, $V_1$ is the initial volume, and $V_2$ is the final volume.

Key Concept: Zeroth Law Basics

b) System $A$ is in thermal equilibrium with system $C$.

[Solution Description] The Zeroth Law of Thermodynamics implies that if system $A$ is in thermal equilibrium with system $B$, and system $B$ is in thermal equilibrium with system $C$, then system $A$ is also in thermal equilibrium with system $C$. This is a foundational concept for defining temperature scales.

Key Concept: Ideal Gas Law

a) $PV = nRT$

[Solution Description] The ideal gas law is a fundamental equation in thermodynamics that relates pressure $(P)$, volume $(V)$, and temperature $(T)$ for an ideal gas. It is expressed as $PV = nRT$, where $n$ is the number of moles and $R$ is the universal gas constant.

Key Concept: Adiabatic vs Conducting Walls

b) Conducting wall

[Solution Description] Conducting walls, also known as diathermic walls, allow heat to transfer between systems, enabling them to reach thermal equilibrium. In contrast, adiabatic walls do not permit heat flow and thus do not allow systems to reach thermal equilibrium through direct heat exchange.

Key Concept: Isothermal vs Adiabatic

d) The process is isothermal, and $PV = \text{constant}$

[Solution Description] This process is an isothermal process where the temperature remains constant, and according to Boyle’s Law, $PV = \text{constant}$. In contrast, in an adiabatic process, there is no heat exchange, and $PV^\gamma = \text{constant}$ with $\gamma = \frac{C_p}{C_v}$.

Key Concept: Caloric Theory

c) Caloric flows from the hot body to the cold body

[Solution Description] The caloric theory posited that heat was a fluid known as caloric that flowed from hotter objects to cooler ones until thermal equilibrium was reached. Therefore, according to this outdated theory, caloric flows from the hot body to the cold body upon contact.

Key Concept: Equation Application

b) 300 J

[Solution Description] According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$. In this scenario, we have $\Delta Q = 500$ J and $\Delta W = 200$ J. Substituting these values into the equation gives:

$$500 = \Delta U + 200$$

Solving for $\Delta U$, we get:

$$\Delta U = 500 – 200 = 300 \, \text{J}$$

Thus, the change in internal energy is 300 J.

Key Concept: Thermal Equilibrium Conditions

d) Same temperature

[Solution Description] For two bodies to achieve thermal equilibrium, they must reach the same temperature. This means that initially, they may exchange heat until this condition is met. Other factors like mass or volume do not directly affect the achievement of thermal equilibrium.

Key Concept: Adiabatic Process Relations

d) $V_2/V_3 = (T_1/T_2)^{1/(\gamma-1)}$

[Solution Description] During the adiabatic process in a Carnot cycle, the volume and temperature are related by the expression:

$$\frac{V_2}{V_3} = \left(\frac{T_1}{T_2}\right)^{1/(\gamma-1)}$$

where $\gamma$ is the heat capacity ratio.

Key Concept: First Law Basic

c) Change in internal energy equals heat supplied minus work done.

[Solution Description] The First Law of Thermodynamics states that the change in internal energy $(\Delta U)$ of a system is equal to the heat supplied to the system $(\Delta Q)$ minus the work done by the system $(\Delta W)$, expressed as $\Delta Q = \Delta U + \Delta W$.

Key Concept: First Law Application

b) 250 J

[Solution Description] According to the First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$, where $\Delta Q$ is the heat added, $\Delta U$ is the change in internal energy, and $\Delta W$ is the work done by the system.

Given: $$\Delta Q = 200$$ J, $$\Delta W = -50$$ J (work done on the system).

Rearrange the equation for $\Delta U$:

$$\Delta U = \Delta Q – \Delta W$$

Substitute the values:

$$\Delta U = 200 – (-50) = 200 + 50 = 250 \text{ J}$$

Therefore, the change in internal energy is 250 J.

Key Concept: Conceptual Understanding

c) Amount of substance

[Solution Description] Molar specific heat capacity is not affected by the amount of substance because it is already defined per mole. It depends on other factors like the nature of the substance, temperature, and conditions under which heat is supplied but not the quantity.

Key Concept: Process Dependency

c) Volume

[Solution Description] Specific heat capacity can depend on factors such as temperature and pressure, but it does not inherently depend on volume. Therefore, volume is not a determining factor for the specific heat capacity of water.

Key Concept: Comparative Analysis

d) Radiation does not need a medium while conduction does

[Solution Description] Conduction requires direct contact and primarily occurs in solids, whereas radiation does not require a medium and involves electromagnetic waves.