Test: Second Law of Thermodynamics, Entropy - 2 - Mechanical Engineering MCQ

To determine the dimensions of entropy, we start by recalling the definition of entropy in terms of heat transfer and temperature:

dS = δQ/T

Where dS is the differential entropy, δQ is the heat transferred, and T is the temperature.

The dimensional formula for heat (δQ) can be derived from its relation to energy. Energy (E) has dimensions of ML²T^-2 (from the basic formula for kinetic energy ½mv² or potential energy mgh).

Temperature (T) is assigned the dimension θ.

Using these, we can express the dimensions of dS as follows:

[dS] = [δQ/T] = ML²T^-2/θ = M L² T^-2 θ^-1

Thus, the dimensional formula for entropy is M L² T^-2 θ^-1.

In thermodynamics, when a system reaches a state of equilibrium, particularly thermal equilibrium, the property that assumes its maximum value is entropy. This concept is derived from the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. It reaches a maximum value when the system achieves equilibrium, meaning no further changes occur in the state of the system without external influences.

• Availability refers to the maximum useful work that can be extracted from a system as it reaches equilibrium with its surroundings. It is not maximized at equilibrium; rather, it decreases because less work can be extracted as the system approaches equilibrium.
• Gibbs function (Gibbs free energy) and Helmholtz function (Helmholtz free energy) both describe the amount of reversible work obtainable from a system at constant temperature and pressure (Gibbs) or at constant temperature and volume (Helmholtz). These functions are minimized in the state of equilibrium under their respective conditions (constant T and P for Gibbs, constant T and V for Helmholtz).

Therefore, the correct answer is: B: Entropy

Assertion (A):

"If a graph is plotted for absolute temperature as a function of entropy, the area under the curve would give the amount of heat supplied."

• Explanation: In thermodynamics, the relationship between heat $Q$ supplied to a system and the change in entropy $\Delta S$ is given by: $$\Delta Q = T \, \Delta S$$ So, the area under a $T$ vs. $S$ curve represents the amount of heat supplied, because heat supplied is $\Delta Q = \int T \, dS$, which is the area under the curve.

Thus, Assertion (A) is true.

Reason (R):

"Entropy represents the maximum fraction of work obtainable from heat per degree drop in temperature."

• Explanation: This statement is not entirely correct. Entropy is a measure of the dispersal of energy in a system, and its change corresponds to the heat transfer divided by the temperature. However, the phrase "maximum fraction of work obtainable from heat per degree drop in temperature" is misleading. Entropy does not directly represent the fraction of work obtainable per temperature drop. The maximum work that can be obtained is related to the change in free energy, and not just entropy or temperature change alone.

Thus, Reason (R) is false.

Conclusion:

The correct answer is: c) A is true but R is false.

All the spontaneous processes are irreversible in nature. The irreversibility is caused by finite potential gradient like temperature gradient etc. or by any dissipative effect like friction. There are two types of irreversibility
i.) Internal irreversibility
ii.) External irreversibility
The external irreversibility occurs due to the temperature difference between the source and a working fluid at heat supply and the temperature difference between the sink and the working fluid at heat rejection. If the hypothetical heat source and sink is considered then the process becomes reversible.

• Statement 1: In an irreversible process, entropy of the universe always increases. This is a fundamental concept of the second law of thermodynamics.
• Statement 2: The sum of the entropy of all bodies (the entire system and surroundings) in an irreversible process always increases, reflecting the increase in disorder.
• Statement 3: Once created, entropy cannot be destroyed. Entropy can be transferred, but the total entropy of a closed system cannot decrease.

Hence, all three statements are correct. Option C is the right answer.

Entropy increase in process = 100 (0.4 – 0.3) = 10 kJ/kg
Entropy change of surroundings = 5 kJ/K
Thus net entropy increases and the process is irreversible

For an ideal gas mixture, the entropy can be calculated based on the properties of the individual gases making up the mixture. The key factor here is how the entropy of each constituent gas is evaluated in the context of the mixture's overall properties.

Here's a breakdown of how entropy typically depends on state variables for an ideal gas:

1. Entropy as a function of temperature and pressure: This is a common way to express the entropy, but it may not be directly additive for a mixture due to the pressure term varying for each gas due to its partial pressure.
2. Entropy as a function of temperature and partial pressure: While individual gas entropies could be calculated this way, summing them does not straightforwardly account for the interactions in the mixture's volume or the total pressure.
3. Entropy as a function of temperature and volume: For a mixture, if each gas were to occupy the total volume V of the mixture at the mixture's temperature T, the entropy of each gas can be calculated separately and then summed. This approach utilizes the fact that ideal gases do not interact chemically, and each gas expands or contracts to fill the entire volume independently of others. The entropy calculation then considers each gas's partial pressure implicitly through its mole fraction in the mixture and the total volume.
4. Entropy as a function of pressure and volume: This would require additional information about the temperature to compute the entropy since temperature is a key variable affecting entropy changes and values.

Thus, for calculating the entropy of a mixture of ideal gases where the entropies of the constituents are additive, the most practical and typical approach uses the mixture's temperature and the total volume that the mixture occupies. Therefore, the correct answer is:

C: Temperature and volume of the mixture

It is isobaric compression.

Entropy:

• Entropy is a measure of the energy of a system that is unavailable for doing useful work. Increase in entropy of a system leads to increase in randomness and degradation of energy.
• No process between two equilibrium states is possible if it would result in a decrease in the total entropy of the system and surroundings.
• Since entropy is a point function that depends only on states i.e. independent of the path followed..
• So the change of entropy will be zero when the process undergoes a complete cycle (where initial state and final state is the same)

Principle of Increase of Entropy:

• The entropy of the isolated system is the measure of the irreversibility undergone by the system.
• More is the irreversibility more increase is the entropy of the system or dgradation of energy. As such the reversible process is an ideal process and it never really occurs. This means the certain amount of the irreversibility is always there in the system, this also means that the entropy of the isolated system always goes on increasing, it never reduces. Here let us keep in mind that isolated system can always be formed by including any system and surroundings within a single boundary.

That so why we are using p–v or T–s diagram.

In a Temperature-Entropy (T-S) diagram, such as the one described, the area under the curve between two points on the graph represents the total heat absorbed or rejected by the system, depending on the nature of the process. The formula that relates this area to heat transfer is:

Q = ∫ T dS

This equation comes from the first law of thermodynamics for reversible processes in a closed system, where dQ = T dS.

Options Analysis:

• A: Total work done during the process - This is not directly represented by the area under the T-S curve. Work interactions are related to pressure-volume (P-V) diagrams.
• B: Total heat absorbed during the process - This is exactly what is represented by the area under the curve in a T-S diagram for the reasons described above.
• C: Total heat rejected during the process - This could also be represented by an area under the curve in a T-S diagram, but typically it would be represented by a curve moving in the opposite direction (decreasing entropy).
• D: Degree of irreversibility - Does not directly relate to the area under the curve. While entropy change itself can indicate irreversibility (if entropy increases), the T-S diagram primarily quantifies heat transfer.

Thus, the correct answer, matching your description and consistent with your given answer, is: B: Total heat absorbed during the process

  does not necessarily means reversible process. If dQ = 0 .

Throttling is a constant enthalpy expansion process.

When two masses of water at different temperatures T₁ and T₂ with T₁ > T₂ are mixed under isobaric (constant pressure) and adiabatic (no heat exchange with the surroundings) conditions, the entropy change can be analyzed as follows:

Key Considerations:

• Adiabatic Process: No heat is transferred to or from the environment, so changes in entropy are confined to the system itself (the two masses of water).
• Isobaric Mixing: The pressure remains constant during the mixing, with changes in entropy driven by the temperature differences.
• Temperature Equalization: Heat flows from the hotter to the cooler water until thermal equilibrium is reached, increasing the system's randomness or disorder and hence its entropy.

Analysis of Entropy Change:

• Entropy Change in the System: The entropy of the hotter water decreases as it loses heat, while the entropy of the cooler water increases as it gains heat. The increase in entropy of the cooler water is greater than the decrease in the warmer water due to the logarithmic relationship of entropy with temperature, leading to a net increase in the system's entropy.
• Entropy Change of the Universe: As the process is adiabatic, the entropy change of the surroundings is zero. Therefore, the total entropy change of the universe is the entropy change of the system, which is positive.

Conclusion: The entropy change of the universe during this mixing process is:

A: Necessarily positive

This reflects the second law of thermodynamics, which states that the entropy of an isolated system never decreases, confirming the increase in entropy due to spontaneous mixing and temperature equalization.

If a closed system is undergoing an irreversible process, the change in entropy of the system is given by

dS > 0, or dS = 0, or dS < 0

In an irreversible process in which heat is removed from the system then its entropy can decrease. 

The entropy of a closed system is given by

When the process is irreversible then entropy generation in the system (δs)gen is always positive, the heat transfer will decide whether the entropy will increase or decrease.

When heat is added to the system

When Heat is removed from the system

When the process is adiabatic, dQ = 0,

∴ when a closed system is undergoing an irreversible process the entropy may increase, decrease or remain constant.

The absolute thermodynamic temperature scale is also known as the Kelvin temperature scale. As the heat is absorbed and rejected respectively by the Carnot engine operating between two reservoirs of different temperatures T1 and T2, these two temperatures on the Kelvin scale feature the same relationship with each other.
The heat absorbed and the heat rejected can be measured during two reversible isothermal and two reversible adiabatic processes in a Carnot engine. The triple point of water is used in defining the Kelvin temperature scale.

The entropy change during heat addition

An isentropic process is also known as a reversible adiabatic process.

If the process is reversible and adiabatic: dQ_rev = 0 ⇒ dS = 0

Important Point:
Entropy analysis for a closed system:

Case 1: For an internally reversible process (δs)_gen = 0

• When heat is added to the system ⇒ (δQ) = +ve ⇒ ds = +ve
• When heat is rejected from the system ⇒ (δQ) = -ve ⇒ ds = -ve
• For adiabatic process ⇒ (δQ) = 0 ⇒ ds = 0 (ISENTROPIC)

Case 2: For internally irreversible process (δs)gen ≠ 0 ⇒ (δs)gen = +ve

• When heat is added to the system ⇒ (δQ) = +ve ⇒ ds = +ve
• When heat is rejected from the system ⇒ (δQ) = -ve ⇒ ds = -ve or +ve or zero
  • ​Since (δs)_gen = +ve and (δQ) = -ve so change in entropy, ds can be zero is (δQ/T) = δs)gen. In this case, the process will be ISENTROPIC
• For adiabatic process ⇒ (δQ) = 0 ⇒ ds = +ve

Thus for a process to be isentropic there are two cases:

1. Reversible adiabatic process
2. The irreversible process where (δQ/T) = δs)gen

Note: Irreversible adiabatic process is not an isentropic process.

AB constant pressure heat addition.

To accurately evaluate the statement and its reasoning, let's break down the assertion (A) and the reason (R) individually:

Assertion (A): The second law of thermodynamics is called the law of degradation of energy. This assertion is true. The second law states that energy quality degrades as it is transformed or transferred, typically from a more useful form to a less useful form (like from mechanical energy to heat). This degradation reflects an increase in entropy, indicating a decrease in the energy's ability to do work.

Reason (R): Energy does not degrade each time it flows through a finite temperature difference. This statement is false. Energy does degrade each time it is transferred through a finite temperature difference, particularly in the form of heat. When energy flows from a hotter to a cooler body, the amount of useful energy (exergy) that can be converted into work decreases, which is a manifestation of energy degradation.

Given this evaluation:

• A is indeed true as the second law is often described in terms of energy degradation.
• R is false because it contradicts the fundamental concept of energy degradation when energy flows through a temperature difference.

Therefore, the correct answer is:

C: A is true but R is false

All spontaneous processes are irreversible. Statement-1 and statement-4 heat is transferred with a finite temperature difference they are irreversible.

We know Carnot efficiency

But inventor claims 60% efficiency (means 40% heat rejection). It is then impossible.

An inventor claims that his new engine, operating between temperatures of 377°C and 27°C, will reject 50% of the heat absorbed. To verify this claim, we convert these temperatures to Kelvin and compare the efficiency to the theoretical maximum provided by the Carnot cycle.

Temperature Conversion

• Source temperature, T1 = 377°C = 650.15 K
• Sink temperature, T2 = 27°C = 300.15 K

Carnot Cycle Efficiency

The efficiency of a Carnot cycle, which is the theoretical limit for any heat engine operating between two temperatures, is calculated as follows:

η_Carnot = 1 - (T2 / T1)

η_Carnot = 1 - (300.15 / 650.15) ≈ 53.8%

This means the Carnot cycle would reject about 46.2% of the heat absorbed, indicating the theoretical maximum efficiency.

Inventor's Engine Efficiency

The inventor's engine has a claimed efficiency:

η_{Inventor's Engine} = 1 - 0.50 = 50%

This efficiency is less than the Carnot efficiency but still within realistic bounds, as no engine can surpass the Carnot cycle's efficiency for the same operating temperatures.

Assessment of Engine Type

• A: Carnot Cycle - Not exactly since this engine has lower efficiency.
• B: Stirling Cycle - A possibility, but specifics are unknown.
• C: Impossible Cycle - Not applicable since the efficiency does not exceed Carnot.
• D: Possible Cycle - Most appropriate choice given the efficiency is realistic and does not violate thermodynamic laws.

Conclusion: The correct assessment of the inventor's engine is D: Possible Cycle.

Concept:

For reversible heat engine efficiency

Calculation: