Tuesday 2 August 2022

Ch 6- THERMODYNAMICS

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 6- THERMODYNAMICS

THERMODYNAMLCS

VERY SHORT QUESTIONS ANSWER

Q.1.Distinguish between a system a surroundings?

Ans. System: The specific part of interest under study.

Surroundings: Everything outside the system that can interact with it.

Q.2.What do you understand by the ‘ state of system?

Ans. The specific condition of a system described by its properties like temperature, pressure, volume, and composition.

Q.3.What are state functions or state of a system?

Ans. Properties of a system that depend only on its current state and not on the path taken to reach that state.

Q.4.Define intensive property?

Ans. Properties of a system that depend only on its current state and not on the path taken to reach that state.

Q.5. Define extensive property?

Ans. Extensive property: A property of a system that depends on the size or amount of the system.

Q.6. Define isolated system and give one example?

Ans. Isolated system: A system that does not exchange energy or matter with its surroundings.

Example: A thermally insulated flask containing a reaction taking place inside without any heat or matter exchange with the surroundings.

Q.7. Differentiate between open and closed systems?

Ans. Open system: Can exchange both energy and matter with its surroundings.

Closed system: Can exchange only energy but not matter with its surroundings.

Q.8. Define a thermodynamically reversible process?

Ans. Thermodynamically reversible process: A process that occurs infinitely slowly and with no entropy change.

Q.9. Can the reversible process be realized in practice?

Ans. No, reversible processes are idealized and cannot be realized in practice due to the absence of infinite slowness and perfect conditions.

Q.10. How is a reversible process be realized in practice?

Ans. A truly reversible process cannot be realized in practice, but certain processes can be made close to reversible by carrying them out very slowly to minimize entropy changes.

Q.11.Is spontaneous process reversible in nature?

Ans. No.

Q.12. Give one difference between an isothermal and an adiabatic process?

Ans. Isothermal process: Occurs at constant temperature, leading to no change in internal energy.

Adiabatic process: Occurs with no heat exchange with the surroundings, leading to a change in internal energy.

Q.13.Give the physical significance of enthalpy?

Ans. Enthalpy represents the heat flow at constant pressure during a process.

Q.14.State the first law of thermodynamics?

Ans. The first law of thermodynamics states that energy cannot be created or destroyed, only converted from one form to another.

Q.15.Give the main limitation of first law of thermodynamics?

Ans. The first law of thermodynamics does not provide information about the direction of a process (i.e., it does not predict whether a process will occur spontaneously or not).

Q.16.Give the main limitation of first law of thermodynamics?

Ans. The first law of thermodynamics does not indicate the direction of processes.

Q.17.Name the law which says that it is impossible to construct a perpetual motion meachine?

Ans. The law is called the "second law of thermodynamics."

Q.18.Can absolute value of H be determined?

Ans. No, the absolute value of enthalpy (H) cannot be determined, only changes in enthalpy (ΔH) can be measured.

Q.19.Define heat capacity?

Ans. Amount of heat required to raise the temperature of a substance by one degree Celsius or Kelvin.

Q.20.What is the other name for zeroth law of thermodynamics?

Ans. Thermal equilibrium law.

Q.21.Can thermodynamics predict the feasibility of a process?

Ans. No, thermodynamics cannot predict the feasibility of a process; it only determines whether energy conservation is satisfied.

Q.22.Can thermodynamics tell about some system which is not in equilibrium?

Ans. Yes, thermodynamics can analyze systems that are not in equilibrium through the study of non-equilibrium thermodynamics.

Q.23.Can thermodynamics predict the feasibility of a process?

Ans. Yes.

Q.24.Heat and path are path functions is it correct?

Ans. No, heat is a path function, but path is not a thermodynamic property; it refers to the specific route taken during a process.

Q.25.Define zeroth law of thermodynamics?

Ans. If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.

Q.26.Internal energy is a state function how?

Ans. Internal energy is a state function because it depends only on the current state of the system and not on the path taken to reach that state.

Q.27.How is enthalpy change related to internal energy change?

Ans. Enthalpy change (ΔH) is related to internal energy change (ΔU) by the equation ΔH = ΔU + PΔV, where P is the pressure and ΔV is the change in volume.

Q.28.Define the standard state of a substance?

Ans. Standard state of a substance is its most stable form at a defined pressure, temperature (usually 25°C or 298 K), and concentration (usually 1 mol/L for solutions).

Q.29.Define enthalpy of a system?

Ans. Enthalpy of a system is the total heat energy it contains at constant pressure.

Q.30.Define standard enthalpy of formation?

Ans. Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at a specified temperature and pressure (usually 25°C and 1 atm).

Q.31.Which is the standard form of carbon?

Ans. The standard form of carbon is graphite.

Q.32.What is calorific value?

Ans. Calorific value is the amount of heat energy released per unit mass of a substance when it undergoes complete combustion.

Q.33.Stae Hess’s law of constant heat summation?

Ans. Hess's law states that the total enthalpy change of a reaction is independent of the pathway between the initial and final states.

Q.34.Define enthalpy of reaction?

Ans. Enthalpy of reaction is the heat energy exchange (heat absorbed or released) during a chemical reaction at constant pressure.

Q.35.Define bond energy is bond breaking exothermic or endothermic?

Ans. Bond energy is the energy required to break a chemical bond, and bond breaking is an endothermic process because it requires energy input.

Q.36.What is entropy?

Ans. Entropy is a measure of the degree of randomness or disorder in a system.

Q.37.What is the effect of temperature on entropy?

Ans. As temperature increases, the entropy of a system generally increases, as there is a greater dispersion of energy and more microstates available to the particles.

Q.38.Is entropy of the universe constant?

Ans. No, the entropy of the universe tends to increase over time due to the second law of thermodynamics.

Q.39.The entropy of a solution is higher than that of pyre liquid why?

Ans. The entropy of a solution is higher than that of a pure liquid due to the increased disorder and randomness of the solute particles dispersed in the solvent.

Q.40.What is the importance of free energy concept for chemical reactions?

Ans. The concept of free energy helps predict whether a chemical reaction is spontaneous (favorable) or non-spontaneous (unfavorable) and provides insights into its feasibility and direction.

SHORT QUESTIONS ANSWER

Q.1.Define free energy of a system?

Ans. The free energy of a system, denoted by G, is the energy available to do useful work and is a combination of the system's internal energy (U) and the product of its absolute temperature (T) and entropy (S). It is given by the equation G = U - TS.

Q.2.Define internal energy of a system?

Ans. The internal energy of a system is the total energy contained within the system, including the kinetic energy of its particles and the potential energy arising from their interactions. It is a fundamental property used to describe the system's thermodynamic state.

Q.3.Define closed and isolated system with examples?

Ans. Closed System: A closed system allows the exchange of energy with its surroundings (in the form of heat and work), but it does not allow the transfer of matter across its boundaries. The total mass of the closed system remains constant.

Example: A sealed container of gas. Heat can flow in and out, but the gas particles cannot escape or enter the container.

Isolated System: An isolated system does not allow the exchange of energy or matter with its surroundings. It is entirely self-contained, and the total energy and mass within the system remain constant.

Example: A well-insulated thermos flask containing hot coffee. The heat stays trapped within the flask, and no matter (coffee or heat) can enter or leave the system.

Q.4.What are homogeneous and heterogeneous systems?

Ans. Homogeneous System: A homogeneous system is a system that has uniform composition and properties throughout its entire volume. In other words, all the components are uniformly mixed at a molecular level, and there are no visible boundaries or phases within the system.

 

Example: A solution of salt dissolved in water is a homogeneous system because the salt molecules are evenly distributed throughout the water, and there are no visible separate phases.

Heterogeneous System: A heterogeneous system is a system that has non-uniform composition and properties, with distinct regions or phases that can be distinguished from each other.

Example: A mixture of oil and water is a heterogeneous system because the oil forms separate droplets within the water, creating visible boundaries between the two phases.

Q.5.What is thermodynamic equilibrium?

Ans. Thermodynamic equilibrium is a state in which a system's macroscopic properties, such as temperature, pressure, and composition, remain constant over time, and there is no net exchange of heat, work, or matter between the system and its surroundings. In this state, the system is in perfect balance and has reached its most stable configuration.

Q.6.Define Adiabatic and Isochoric processes?

Ans. Adiabatic Process: An adiabatic process is a thermodynamic process in which there is no heat exchange between the system and its surroundings. In other words, the system is thermally insulated, and no heat is added or removed during the process. As a result, the change in the system's internal energy is solely due to work done on or by the system.

Example: When a gas expands or compresses rapidly in a perfectly insulated container, the process is considered adiabatic as there is no heat transfer.

Isochoric Process: An isochoric process, also known as an isovolumetric process, is a thermodynamic process that occurs at constant volume. In an isochoric process, the system does not perform any work on its surroundings or vice versa, since the volume remains constant throughout the process. Any change in the system's internal energy is solely due to heat exchange.

Example: Heating a gas in a rigid container without allowing it to expand or contract results in an isochoric process because the volume remains constant.

Q.7.Define isobaric process?

Ans. An isobaric process is a thermodynamic process in which the pressure of a system remains constant while other variables, such as volume and temperature, may change. The term "isobaric" is derived from two Greek words: "iso" meaning "equal" and "baros" meaning "pressure."

During an isobaric process, the system exchanges energy with its surroundings in the form of heat, allowing for changes in volume and temperature while keeping the pressure constant. In other words, the system experiences pressure equilibrium with its surroundings throughout the process.

An example of an isobaric process is heating a gas confined in a container with a movable piston. As heat is added, the gas molecules gain kinetic energy, leading to an increase in temperature and expansion of the gas, pushing the piston outward. The pressure remains constant during this expansion because the heat input compensates for the increase in volume, maintaining the pressure at a constant value. Similarly, during cooling, the gas contracts, but the pressure remains unchanged because the heat is removed to balance the reduction in volume.

In an ideal isobaric process, the relationship between the variables can be described by the following equation:

P * V = constant,

where P is the constant pressure, V is the volume, and the product of pressure and volume remains constant throughout the process.

Q.8.What do you understand by state and path functions?

Ans. In thermodynamics, state functions and path functions are two fundamental concepts used to describe the properties and behavior of a system.

 

State functions:

State functions, also known as state variables or state properties, are thermodynamic properties that depend only on the current state of the system and are independent of the path taken to reach that state. In other words, their values are determined solely by the initial and final states of the system, not by the process or the intermediate steps used to get from one state to another. The actual path or history of the system doesn't matter; the state function's value remains the same as long as the initial and final states are the same.

Examples of state functions include:

Temperature (T): The temperature of a system is solely determined by its current state and is independent of how the system arrived at that temperature.

Pressure (P): The pressure of a system depends only on the current state and the molecular interactions within the system, not on how the pressure was changed.

The change in a state function between two states can be represented as ΔX, where X is the state function.

Path functions:

Path functions, also known as process functions, are thermodynamic properties that depend on the specific path taken during a process and not just on the initial and final states of the system. The actual route or sequence of changes from one state to another significantly affects the value of the path function.

Examples of path functions include:

Work (W): The work done on or by the system during a process is a path function because it depends on how the work was performed and the specific changes in the system's state during the process.

Heat (Q): The heat transferred to or from the system during a process is also a path function because the amount of heat exchanged depends on the specific process and the temperature variations at each step.

In summary, state functions depend only on the initial and final states of the system, whereas path functions depend on the path taken between those states. Understanding the distinction between these two types of functions is essential in analyzing and understanding thermodynamic processes and their associated properties.

Q.9.What are thermochemical equations?

Ans. Thermochemical equations are balanced chemical equations that also include information about the heat energy (enthalpy) changes associated with a chemical reaction. They provide a way to represent the chemical reaction and the corresponding heat energy exchange (heat flow) between the reactants and products during the reaction.

Thermochemical equations are typically written in the following format:

Reactants Products ΔH

Where:

Reactants are the substances present at the beginning of the chemical reaction.

Products are the substances formed as a result of the chemical reaction.

ΔH represents the change in enthalpy (heat energy) for the reaction. It can be positive (endothermic, heat is absorbed) or negative (exothermic, heat is released) and is usually expressed in units like joules (J) or kilojoules (kJ) per mole of the balanced reaction.

Here's an example of a thermochemical equation for the combustion of methane (CH4):

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ΔH = -802 kJ/mol

In this example, methane (CH4) and oxygen (O2) are the reactants, and carbon dioxide (CO2) and water (H2O) are the products. The value of ΔH is -802 kJ/mol, indicating that the reaction is exothermic, and 802 kilojoules of heat are released per mole of methane reacted.

Thermochemical equations are essential in understanding the energy changes associated with chemical reactions and are commonly used in applications related to chemical processes, engineering, and heat calculations. The enthalpy values provided in thermochemical equations can be used to determine the heat exchanged in a reaction under various conditions, such as different temperatures or reactant quantities.

Q.10.How does an increase in internal energy of a system manifest itself?

Ans. An increase in the internal energy of a system manifests itself through changes in the system's thermodynamic properties, such as temperature, pressure, volume, and the state of matter. Internal energy (U) is a fundamental concept in thermodynamics that represents the total energy stored within a system due to the microscopic motion and interactions of its particles (atoms or molecules).

When the internal energy of a system increases, it means that energy has been added to the system, either in the form of heat transfer (Q) or work done (W) on the system, or both. Here's how an increase in internal energy can manifest itself in different ways:

Temperature Increase: One of the most common manifestations of an increase in internal energy is an increase in the system's temperature. When energy is added to a substance, its particles gain kinetic energy, and their average speed increases. This leads to a rise in temperature, and the substance may change from a solid to a liquid or a liquid to a gas if the conditions are appropriate.

Pressure Change: If the volume of the system is held constant, an increase in internal energy can lead to an increase in pressure. This is particularly evident in gases, where an increase in internal energy leads to higher molecular velocities and more frequent collisions with the container walls, resulting in higher pressure.

Volume Expansion: If the system's pressure is constant, an increase in internal energy can cause the system to expand its volume. For instance, when a gas is heated at constant pressure, its internal energy increases, and the gas molecules move with greater speed, leading to an expansion and an increase in volume.

Phase Changes: In some cases, an increase in internal energy can cause a substance to undergo a phase change, such as melting, vaporization, or sublimation. During these phase transitions, the temperature remains constant, but the internal energy is still increasing due to energy being used to break intermolecular forces and rearrange the particles into a different phase.

It's essential to note that the manifestation of an increase in internal energy depends on the specific conditions of the system, such as whether the process occurs at constant pressure, constant volume, or under different constraints. Additionally, the specific properties and behavior of the substance being considered also play a role in how internal energy changes are observed.

Q.11.What is enthalpy of neutralization?

Ans. The enthalpy of neutralization is the heat energy change associated with the complete neutralization of an acid and a base to form a salt and water. This chemical process is an exothermic reaction, meaning heat is released into the surroundings. The enthalpy of neutralization is typically measured under constant pressure conditions, such as at atmospheric pressure.

The general chemical equation for the neutralization of a strong acid (HA) with a strong base (BOH) is:

HA (aq) + BOH (aq) B(A) (aq) + H2O (l)

Where:

HA is the strong acid,

BOH is the strong base,

B(A) represents the salt formed, and

H2O is water.

During the neutralization process, hydrogen ions (H+) from the acid react with hydroxide ions (OH-) from the base to form water. The remaining ions combine to form a salt. The overall reaction releases energy in the form of heat, which is the enthalpy of neutralization.

The enthalpy of neutralization is typically expressed in units of joules per mole (J/mol) or kilojoules per mole (kJ/mol) of the acid or base reacting. The value of the enthalpy of neutralization is usually negative, indicating that heat is released during the reaction. The specific value of the enthalpy of neutralization depends on the specific acid and base involved in the reaction.

For example, the neutralization of hydrochloric acid (HCl) with sodium hydroxide (NaOH) can be represented by the equation:

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

The enthalpy change for this reaction is typically around -57 kJ/mol, meaning that 57 kilojoules of heat are released for every mole of HCl neutralized by NaOH. However, it's essential to note that the exact enthalpy values can vary depending on the specific experimental conditions and concentrations of the reactants.

Q.12.What is most important use of Hess’s law?

Ans. The most important use of Hess's law in thermodynamics is to calculate the enthalpy change for a chemical reaction indirectly. Hess's law allows us to determine the overall enthalpy change for a reaction by combining the enthalpy changes of one or more intermediate reactions.

Hess's law is based on the principle that the enthalpy change of a chemical reaction is independent of the pathway taken between the initial and final states of the reactants and products. In other words, the change in enthalpy is a state function, and it depends only on the initial and final states of the system, not on the specific steps or intermediates involved in the reaction.

 

This is particularly useful in cases where it is challenging or even impossible to directly measure the enthalpy change of a specific reaction. By using Hess's law, we can determine the enthalpy change indirectly through a series of intermediate reactions whose enthalpy changes are known or can be measured experimentally.

The steps to use Hess's law are as follows:

Identify and write down a series of intermediate reactions that, when combined, will yield the target reaction.

Write the balanced chemical equations for these intermediate reactions, including the known or measured enthalpy changes (∆H) for each reaction.

Apply the law of conservation of energy and algebraically manipulate the intermediate reactions to cancel out any common reactants or products and obtain the target reaction.

The overall enthalpy change for the target reaction will be the sum of the enthalpy changes for the intermediate reactions, taking into account the stoichiometric coefficients.

Hess's law is particularly useful in situations where direct measurements of enthalpy changes are difficult or impractical, and it has applications in various fields, including chemical engineering, thermodynamic calculations, and the study of combustion processes. It allows scientists and engineers to understand and predict the energetics of complex reactions by using simpler, well-understood reactions.

Q.13.Define enthalpy of transition?

Ans. The enthalpy of transition, also known as the heat (enthalpy) of phase transition or heat of transformation, refers to the heat energy change associated with a physical change in a substance from one phase to another at a constant temperature and pressure. Phase transitions involve the conversion of matter between different states, such as solid, liquid, and gas, without any change in chemical composition.

 

The three primary types of phase transitions are:

Melting (fusion): The transition from a solid to a liquid state, characterized by the absorption of heat energy. The enthalpy of transition for melting is also known as the heat of fusion.

Freezing: The transition from a liquid to a solid state, characterized by the release of heat energy. The enthalpy of transition for freezing is equal in magnitude but opposite in sign to the heat of fusion.

Vaporization (boiling or evaporation): The transition from a liquid to a gas state, characterized by the absorption of heat energy. The enthalpy of transition for vaporization is also known as the heat of vaporization.

The enthalpy of transition is quantified as the amount of heat energy required or released per unit mass (usually per mole) of the substance undergoing the phase change. It is typically expressed in units of joules per mole (J/mol) or kilojoules per mole (kJ/mol).

For example, let's consider the process of water boiling and transitioning from a liquid to a gas at its boiling point of 100°C (at atmospheric pressure). The heat of vaporization for water is approximately 40.79 kJ/mol. This means that for each mole of water vaporized at its boiling point, approximately 40.79 kJ of heat energy is absorbed from the surroundings.

Similarly, when water freezes at 0°C (at atmospheric pressure), it releases the same amount of heat energy (40.79 kJ/mol) per mole in the form of heat of fusion.

The enthalpy of transition is an important concept in thermodynamics and is used in various applications, including the design of cooling and heating systems and the study of phase equilibria in materials.

Q.14.Hess’s law is a corollary of the first law of thermodynamics comment?

Ans. Hess's law can be considered a corollary of the first law of thermodynamics, also known as the law of conservation of energy. The first law of thermodynamics states that energy cannot be created or destroyed in a closed system; it can only change forms. In other words, the total energy of a system remains constant.

Hess's law is based on this fundamental principle of energy conservation. It states that the overall enthalpy change (∆H) for a chemical reaction is the same, regardless of the pathway taken between the initial and final states of the reactants and products. This means that the change in enthalpy is a state function and depends only on the initial and final states of the system, not on the specific steps or intermediates involved in the reaction.

To put it another way, if a chemical reaction can occur via multiple intermediate steps, the sum of the enthalpy changes for these steps will be equal to the enthalpy change for the overall reaction. This is in line with the first law of thermodynamics, which dictates that the total energy change in a system is independent of the process's specific details.

Mathematically, Hess's law can be expressed as follows:

∆H (overall) = Σ∆H (intermediate reactions)

Where:

∆H (overall) is the enthalpy change for the overall reaction.

Σ∆H (intermediate reactions) is the sum of the enthalpy changes for the intermediate reactions.

By applying Hess's law, chemists and thermodynamicists can determine the enthalpy changes for complex reactions indirectly through a series of simpler, well-understood reactions. This is extremely valuable when direct measurement of a particular reaction's enthalpy change is challenging or not feasible.

In summary, Hess's law is a corollary of the first law of thermodynamics because it is based on the principle of energy conservation and highlights the relationship between energy changes and the initial and final states of a system, regardless of the specific path taken during a chemical reaction.

Q.15.Prove that the decrease in free energy in a system during a process is equal to the useful work done?

Ans. To prove that the decrease in free energy (∆G) in a system during a process is equal to the useful work done (w), we can start with the definition of free energy change and then relate it to the work done.

The change in Gibbs free energy (∆G) for a closed system is given by the equation:

∆G = ∆H - T∆S

Where:

∆G = Change in Gibbs free energy

∆H = Change in enthalpy (heat energy change)

T = Temperature of the system in Kelvin

∆S = Change in entropy (disorder) of the system

Now, let's consider a system undergoing a process, and during this process, it performs work on its surroundings. According to the first law of thermodynamics, the energy of the system (U) changes due to heat transfer (Q) and work done (w):

ΔU = Q - w

Since the system is closed (no matter transfer), the change in internal energy (∆U) is equal to the change in enthalpy (∆H):

∆H = Q - w

Now, let's rearrange the equation to isolate Q:

Q = ∆H + w

Next, we can use the definition of entropy change (∆S) as follows:

∆S = q/T

Where q is the heat transfer, and T is the absolute temperature.

Now, we'll substitute q with ∆H, as the enthalpy change (∆H) represents the heat transfer at constant pressure:

∆S = ∆H/T

Now, let's modify the expression for ∆G:

∆G = ∆H - T∆S

Substitute the value of ∆S:

∆G = ∆H - (∆H/T) * T

∆G = ∆H - ∆H

∆G = 0

Hence, we find that ∆G = 0.

At equilibrium, the free energy change (∆G) is zero. This means that the system has reached a stable state with minimal free energy, and no useful work is being done.

However, during a non-equilibrium process, the decrease in free energy (∆G) is equal to the useful work done (w):

∆G = -w

Therefore, the decrease in free energy during a process is indeed equal to the useful work done by the system.