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

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:

Apply the Steady Flow Energy Equation (SFEE) for a turbine:

\( h_1 + q = h_2 + w \Rightarrow q = (h_2 + w) - h_1 \)

Calculation:

Given:

Inlet enthalpy, \( h_1 = 3000~\text{kJ/kg} \)

Outlet enthalpy, \( h_2 = 2700~\text{kJ/kg} \)

Work output, \( w = 250~\text{kJ/kg} \)

Now,

\( q = (2700 + 250) - 3000 = 2950 - 3000 = -50~\text{kJ/kg} \)

Negative sign indicates heat loss from the turbine to the surroundings.

Power is defined as \(P=Fv\)

where F is force and v is velocity.

\( P= Fv = 4500 \times 2 = 9000 \, \text{W} = 9 \, \text{kW} \)

Solution:

For an adiabatic process involving a gas, the relation between temperature and volume is given by:
T × V^{(γ - 1)} = constant

Initially, temperature T₀ and volume V change to T and V/5 after adiabatic compression. Here, γ is calculated as:
γ = 1 + (2/5) = 7/5

Thus, we have:
T₀ × V^{(γ - 1)} = T × (V/5)^{(γ - 1)}

Therefore, the final temperature (T) is:
T = 373 × (5)^(2/5)

The average rotational kinetic energy per molecule is:
(K.E.)_rotational = k × T
= 1.38 × 10^-23 × 373 × (5)^(2/5)
= 98× 10^-22 J (approximately)

Hence, the value of n is 98.

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.

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

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:

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.

Concept:

According to the first law of thermodynamics

δQ = dU + δW

Isobaric work done

δW = PdV = P (Vfinal - Vinitial) 

Calculation:

Initial condition ⇒ P₁ = 1 MPa, V₁ = y m³

Final condition ⇒ P₂ = 1 MPa, V₂ = 0.2 m³

Heat Transfer = -40 kJ (from the gas)

Change in Internal energy (u₂ – u₁) = -20 kJ (drop)

According to first law of thermodynamics

δQ = dU + δW

-40 = -20 + δW

⇒ δW = -20 kJ          ---(I)     

Since, the process is isobaric (as pressure remains same)

So, isobaric work done δW = PdV = P (V_final - V_initial) 

δW = P (V_final - V_initial) = -20 kJ

1000 kPa × (0.2 – y) m³ = -20 kJ

\(0.2 - y = \frac{{ - 20}}{{1000}} = - 0.02 \Rightarrow y = 0.22\)

Concept:

Apply the first law of thermodynamics

dQ = dU + dW

Calculation:

Given:

V₁ = 1 m³, V₂ = 2 m³, dQ = 2000 kJ, P = 1000 V kPa ⇒ 1000 × 10³ V Pa.

Heat is added to the system, so it is positive

We know work done

\(\begin{array}{l} dW = \int\limits_{\mathop V\nolimits_1 }^{\mathop V\nolimits_2 } {PdV} \\ dW= \int\limits_1^2 {1000 × \mathop {10}\nolimits^3 VdV} \\ dW= 1000 × \mathop {10}\nolimits^3 \left[ {\mathop {\left( {\frac{{\mathop V\nolimits^2 }}{2}} \right)}\nolimits_1^2 } \right]\\dW = 1000 × \mathop {10}\nolimits^3 × \left[ {\frac{{4 - 1}}{2}} \right]\\dW = 1500~kJ \end{array}\)

We know that

dU = dQ - dW

dU = 2000 - 1500

dU = 500 kJ.

Concept:

Pump:

• A pump is a mechanical device used to force a fluid (a liquid or a gas) to move forward inside a pipeline or hose.
• They are also used to produce pressure by the creation of a suction (partial vacuum), which causes the fluid to rise to a higher altitude.

The efficiency of the pump, \(\eta =\frac{Useful ~power}{Power~rating}\) 

Calculation:

Given:

Power rating = 400 W, Mass of water lifted = 500 kg, H = 30 m, time = 10 minutes = 60 × 10 = 600 seconds, g = 10 m/s²

Force, F = m × g = 500 × 10 = 5000 N

Work = Force × displacement (H) = 5000 × 30 = 150 kJ

\(Power=\frac{Work}{Time}\)

\(Power=\frac{150}{600}=0.25 ~kW=250~W\)

⇒ Useful work = 250 W

\(\eta=\frac{Useful ~power}{Power~rating}\)

\(\eta=\frac{250}{400}=0.625\) 

⇒ η = 62.5%

Concept:

When Q joule heat is added to a body whose mass is m, the temperature rises from T₁ to T₂.

It is given by Q = mc(T₂ - T₁) where c = specific heat of the body.

Calculation:

Given:

m = 2 kg, Q = 500 kJ, T₂ = 200 °C, T₁ = 100 °C   

∵ Q = mc(T₂ – T₁)

⇒ 500 = 2 × c × (200 – 100)

⇒ c = 2.5 kJ/kg°K

Important Points

When difference of temperature is needed do not convert °C into °K.

Concept:

If the balloon containing the ideal gas is initially kept in an evacuated and insulated room. Then if the balloon ruptures and the gas fills up the entire room, the process is known as free or unrestrained expansion.

Now if apply the first law of thermodynamics between the initial and final states.

\(Q=(u_2-u_1)+W\)

In this process, no work is done on or by the fluid, since the boundary of the system does not move. No heat flows to or from the fluid since the system is well insulated.

\(u_2-u_1=0\Rightarrow u_2=u_1\)

Enthalpy is given as 

h = u + Pv

For ideal gases, as we know, internal energy and enthalpy are a function of temperature only, so if internal energy U remains constant, temperature T also remains constant which means enthalpy also remains constant.

So, during the free expansion of an ideal gas, both internal energy and enthalpy remain constant.