Thermodynamics MCQ Quiz - Objective Question with Answer for Thermodynamics - Download Free PDF

Explanation:

Zeroth Law of Thermodynamics

• The Zeroth Law of Thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law provides a fundamental basis for the concept of temperature.

Working Principle: The Zeroth Law essentially implies that temperature is a fundamental and measurable property of matter. If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then system A and system B must be in thermal equilibrium with each other. This logical reasoning allows for the establishment of temperature as a transitive property.

Significance in Measurement of Temperature: The Zeroth Law is crucial because it allows for the creation of temperature scales and the use of thermometers. It enables us to compare temperatures of different systems and ensures that thermometers can provide consistent and reliable temperature readings.

Applications: The principles of the Zeroth Law are applied in various fields, including:

• Thermometry: The development and calibration of thermometers rely on the Zeroth Law, ensuring accurate temperature measurement.
• Engineering: Temperature control and monitoring in processes such as heating, cooling, and chemical reactions are based on the Zeroth Law.
• Scientific Research: Accurate temperature measurement is essential in experiments and studies involving thermal properties of materials and systems.

Explanation:

In thermodynamics, the first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. It is one of the fundamental principles of thermodynamics and states that energy cannot be created or destroyed in an isolated system. The first law is often expressed mathematically as:

ΔU = Q - W

where:

• ΔU is the change in internal energy of the system.
• Q is the heat added to the system.
• W is the work done by the system.

For a closed system undergoing a thermodynamic cycle, the first law of thermodynamics states that the net work done by the system over one complete cycle is equal to the net heat added to the system over the cycle. This is because, over a complete cycle, the system returns to its initial state, meaning the change in internal energy (ΔU) is zero.

Therefore, the correct interpretation of the first law in the context of a thermodynamic cycle is:

Net work done equals net heat transfer.

This is option 1 from the given choices.

Correct Option Analysis:

For a closed system undergoing a thermodynamic cycle, the first law of thermodynamics can be written as:

ΔU = Q - W

During a complete cycle, the system returns to its initial state, which means the change in internal energy (ΔU) is zero:

ΔU = 0

Substituting this into the first law equation gives:

0 = Q - W

Rearranging this equation, we find:

Q = W

This signifies that the net heat added to the system (Q) over the cycle is equal to the net work done by the system (W) over the cycle.

In other words, the energy added to the system as heat is completely converted into work during the cycle, which aligns with the principle of conservation of energy.

Important Information for Analysis of Other Options:

Option 2: Entropy always increases.

This statement is not correct in the context of the first law of thermodynamics. The concept of entropy is related to the second law of thermodynamics, which states that the entropy of an isolated system always increases over time, and processes occur in the direction of increasing entropy. However, this does not directly pertain to the first law, which is concerned with energy conservation.

Option 3: Pressure and temperature are inversely related.

This statement is not a general principle of thermodynamics. The relationship between pressure and temperature depends on the specific thermodynamic process being considered. For example, in an isothermal process, the temperature remains constant, while pressure can change. In an adiabatic process, both pressure and temperature can change. Therefore, pressure and temperature are not universally inversely related.

Option 4: Internal energy remains constant.

While the change in internal energy (ΔU) over a complete cycle is zero for a closed system, it does not mean that internal energy remains constant throughout the cycle. During different stages of the cycle, the internal energy can change due to heat transfer and work interactions. The first law of thermodynamics indicates the net change in internal energy over a cycle is zero, but internal energy can vary within individual processes of the cycle.

In summary, the correct interpretation of the first law of thermodynamics for a closed system undergoing a thermodynamic cycle is that the net work done by the system equals the net heat transfer into the system (option 1). The other options are either related to different thermodynamic principles or are not universally applicable.

Explanation:

The Carnot cycle is a theoretical thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot in 1824. It is considered the most efficient cycle possible for converting heat into work, or extracting work from heat. The efficiency of the Carnot cycle is given by the difference in temperature between the heat source and the heat sink, divided by the temperature of the heat source. Despite its theoretical efficiency, the Carnot cycle is not suitable for practical engines using gaseous working fluids for several reasons.

The correct answer is option 3: it is impossible to achieve perfectly reversible processes. Let's delve into the detailed explanation of why this is the case:

Correct Option Analysis:

Impossibility of Perfectly Reversible Processes:

The Carnot cycle consists of two isothermal processes (one at a high temperature and one at a low temperature) and two adiabatic processes. To achieve the highest efficiency, each of these processes must be perfectly reversible. A reversible process is an idealization and assumes no entropy generation, no friction, no unrestrained expansion, and no heat transfer through a finite temperature difference. In reality, these conditions are impossible to meet:

• Friction: All practical engines experience some form of friction, whether it is in the pistons, the bearings, or other moving parts. This friction generates entropy and makes the process irreversible.
• Heat Transfer: In the Carnot cycle, heat must be transferred isothermally, which requires an infinitely slow process to ensure that the system remains in thermal equilibrium with the heat reservoirs. In practice, heat transfer occurs over a finite temperature difference, which makes the process irreversible.
• Unrestrained Expansion: In real engines, there are always losses associated with unrestrained expansion or compression of gases. This leads to entropy generation and irreversibility.
• Entropy Generation: Any practical process will generate entropy due to various irreversibilities, including friction, rapid expansion or compression, and heat transfer through finite temperature differences. This makes it impossible to achieve the idealized reversible processes assumed in the Carnot cycle.

Due to these inherent practical limitations, it is impossible to achieve the perfectly reversible processes required for the Carnot cycle. As a result, while the Carnot cycle provides a useful benchmark for the maximum possible efficiency, it cannot be realized in a practical engine using a gaseous working fluid.

Analysis of Other Options:

Option 1: The cycle requires very high pressures that are hard to manage:

While high pressures can pose challenges in managing and maintaining engines, it is not the primary reason why the Carnot cycle is not suitable for practical engines. The main issue lies in the impossibility of achieving perfectly reversible processes, not the pressures involved.

Option 2: It is easy to maintain isothermal processes in practice:

This statement is incorrect. In fact, it is quite difficult to maintain isothermal processes in practice, especially in a dynamic system like an engine. Isothermal processes require very slow heat transfer to ensure thermal equilibrium, which is not feasible in practical applications.

Option 4: The work output from the cycle is quite low:

While the work output of any cycle depends on the specific conditions and design, the Carnot cycle is theoretically the most efficient cycle. Therefore, the work output should not be inherently low. The main issue is achieving the theoretical efficiency in practice.

In conclusion, the primary reason the Carnot cycle is not suitable for practical engines using a gaseous working fluid is the impossibility of achieving perfectly reversible processes. This inherent limitation makes it impossible to realize the theoretical efficiency of the Carnot cycle in real-world applications.

Concept:

The process is isothermal: \( PV = \text{constant} = 15 \)

Work done in isothermal process:

\( W = \int_{P_1}^{P_2} \frac{15}{P} dP = 15 \cdot \ln\left(\frac{P_2}{P_1}\right) \)

Calculation:

\( W = 15 \cdot \ln\left(\frac{10}{1}\right) = 15 \cdot 2.3025 = 34.5375 \, \text{bar·m}^3 \)

\( = 34.5375 \times 10^5 = 3.45375 \times 10^6 \, \text{N·m} = 3.453 \, \text{MN·m} \)

Since pressure increases (volume decreases), it's compression.

Explanation:

Heat Engine Cycle

• A heat engine is a device that converts heat energy into mechanical work by operating in a cyclic process. The working fluid in the cycle absorbs heat energy from a high-temperature source, converts part of this energy into work, and then rejects the remaining energy to a low-temperature sink. A heat engine operates on the basis of the first law of thermodynamics, which states that energy cannot be created or destroyed but can only change forms. For a complete cycle of a heat engine, the net heat transfer into the system equals the net work output from the system.

According to the First Law of Thermodynamics for a cycle:

\( \Delta U_{cycle} = 0 \Rightarrow \sum Q_{cycle} = \sum W_{cycle} \)

This means that the **net heat added** to a heat engine over a cycle is always equal to the net work output

Explanation: 

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. 

• The law of conservation of energy states that the total energy of an isolated system is constant. 
• Energy can neither be created nor be destroyed but can be transformed from one form to another.
   

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings

δQ = ΔU + δW

ΔU = δQ - δW

According to the First Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

ΣQ = ΣW.

∴ Option (3) is the Correct Answer.

Explanation:

There are 4 laws to thermodynamics:

Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

First law of thermodynamics – Energy can neither be created nor destroyed. It can only change forms. In any process, the total energy of the universe remains the same.

For a thermodynamic cycle, the net heat supplied to the system equals the net work done by the system.

δQ = ΔU + δW

Second law of thermodynamics – The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

ΔS = ΔQ/T

ΔSTotal = ΔSSystem + ΔSsurrounding

The second law of thermodynamics introduces the concept of entropy.

The second law of thermodynamics deals with the direction taken by natural or spontaneous processes.

Third law of thermodynamics – As the temperature approaches absolute zero, the entropy of a system approaches a constant minimum.

Concept:

\(W = \mathop \smallint \limits_{{v_1}}^{{v_2}} pdV = P\left( {{V_2} - {V_1}} \right)\)

Work Done = Area under P – V diagram

Calculation:

Given:

Work done = Area of Parallelogram ABCDA

Area of parallelogram = 1/2 × (Sum of parallel sides) × height

Work done = Area of Parallelogram ABCDA = 1/2 × (AB + CD) × BC

⇒ 1/2 × (V + 2V) × P = 1.5PV

\(v = {v_f} + x\;\left( {{v_g} - {v_f}} \right)\)

v = V/m = 0.5 m³/kg

\(0.5 = 0.00106 + x\;\left( {0.8908 - 0.00106} \right)\)

\(x = 0.56\)

\(0 \le x \le 1\)

x = 0, Saturated liquid

x = 1, Saturated vapor

As x value lies between 0 and 1, therefore it is a mixture of saturated liquid and saturated vapor.

Explanation:

Since, the properties like internal energy, enthalpy and entropy of a system cannot be directly measured. They are related to change in the energy of the system.

Hence, we can determine Δu, Δh, Δs but not the absolute values of these properties.

⇒ Therefore, It is necessary to choose a reference state to which, these properties are arbitrary assigned some numerical values.  

So for water, the triple point (T = 0.01°C & P = 611 Pa) is selected as reference a state, where the “Internal energy” (u) and “Entropy” (s) of saturated liquid are assigned a zero value.

#Note: h = u + Pv

At triple point, u = 0, but p × ν ≠ 0

Concept:

The change in internal energy is given by, \(Δ U = m{c_v}Δ T\)

Calculation:

Given:

V₂ = 2V; V₁ = V, and P₁ = P₂ = P

\(Δ U = m\frac{R}{{\gamma - 1}}\left( {{T_2} - {T_1}} \right)\)

where \({C_v} = \frac{R}{{\gamma - 1}}\) and ΔT = T₂ – T₁

\(Δ U = \frac{1}{{\gamma - 1}}\left( {mR{T_2} - mR{T_1}} \right)\)

As we know from the ideal gas equation PV = mRT.

\(Δ U = \frac{1}{{\gamma - 1}}\left( {{P_2}{V_2} - {P_1}{V_1}} \right)\)

\(Δ U = \frac{P}{{\gamma - 1}}\left( {{}{2V} - {}{V}} \right)\)

∴ we get, \(Δ U = \frac{{PV}}{{\gamma - 1}}\).

Mistake Points

In the Isobaric process, pressure is constant throughout the process.

ΔW = P₂V₂ - P₁V₁ = mR(T₂ -T₁)

ΔQ = mC_p(T2 -T1)

ΔU = mC_v(T2 -T1)

Hence for isobaric process also, the change in internal energy is given by ΔU = mCv(T2 -T1)

Concept:

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant. Energy can be transformed from one form to another but cannot be created or destroyed.

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings

δQ = ΔU + δW

ΔU = δQ - δW

According to the First Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

ΣQ = ΣW

Net work done = Net heat in the cycle

Calculation:

Given:

Q1 = 20 kJ, Q2 = - 28 kJ, Q3 = - 2 kJ, Q4 = 40 kJ

Net work done in the cycle = Net heat in the cycle

Wnet = Q1 + Q2 + Q3 +  Q4 

= 20 - 28 - 2 + 40

= 30 kJ

The total work for this cycle process is 30 kJ.

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Concept:

Combined equations of the first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}\)

Hence on the T-S diagram, the slope of the constant pressure line = T/Cp

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}\)

Hence on the T-S diagram, the slope of the constant volume line = T/C_v

As, C_p > C_v , T/Cv > T/Cp , Hence the slope of the Isochoric curve will be more than the slope of the Isobaric Curve on TS plane.

Explanation:

The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

This law is the basis for the temperature measurement.

• By replacing the third body with a thermometer, the Zeroth law can be restated as two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact.
• The thermometer is based on the principle of finding the temperature by measuring the thermometric property.