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

Concept:

Apply the first law of thermodynamics:

\( \Delta U = Q - W \)

Given:

• Initial volume: \( V_1 = 0.3~m^3 \)
• Final volume: \( V_2 = 0.15~m^3 \)
• Pressure: \( P = 0.105~MPa = 105~kPa \)
• Heat rejected: \( Q = -37.6~kJ \)

Step 1: Work done

\( W = P (V_2 - V_1) = 105 \times (-0.15) = -15.75~kJ \)

Step 2: Change in internal energy

\( \Delta U = -37.6 - (-15.75) = -21.85~kJ \)

Explanation:

Given Data:

• Mass of the object, m = 5 kg
• Vertical distance fallen, h = 20 m
• Final velocity of the object, v = 10 m/s
• Initial velocity of the object, u = 0 m/s (since it falls from rest)
• Gravitational acceleration, g = 9.8 m/s²

Step 1: Calculate the potential energy lost by the object:

The potential energy (P.E.) lost by the object is given by:

P.E. = m × g × h

Substitute the given values:

P.E. = 5 × 9.8 × 20 = 980 J

Thus, the object loses 980 J of potential energy as it falls.

Step 2: Calculate the kinetic energy gained by the object:

The kinetic energy (K.E.) gained by the object is given by:

K.E. = 0.5 × m × v²

Substitute the given values:

K.E. = 0.5 × 5 × (10)² = 0.5 × 5 × 100 = 250 J

Thus, the object gains 250 J of kinetic energy during the fall.

Step 3: Calculate the work done by air resistance:

The work done by air resistance is equal to the difference between the potential energy lost and the kinetic energy gained:

Work done by air resistance = Potential energy lost - Kinetic energy gained

Work done by air resistance = 980 - 250 = 730 J

Since air resistance opposes the motion of the object, the work done by air resistance is negative:

Work done by air resistance = -730 J

Concept:

The collector absorbs solar energy to operate a heat engine. The maximum efficiency of the engine is given by Carnot's principle. The useful power is calculated by multiplying collected solar power with Carnot efficiency.

Calculation:

• Source temperature: \( T_h = 90^\circ C = 363~K \)
• Sink temperature: \( T_c = 20^\circ C = 293~K \)
• Collected solar energy: \( 1880~kJ/m^2 \cdot h = \frac{1880000}{3600} = 522.22~W/m^2 \)

Carnot efficiency:

\( \eta = 1 - \frac{T_c}{T_h} = 1 - \frac{293}{363} = 0.1939 \)

Useful power per m²:

\( 522.22 \times 0.1939 = 101.28~W/m^2 \)

Required power output: 1 kW = 1000 W

\( \text{Area} = \frac{1000}{101.28} \approx 9.87 \approx 10~m^2 \)

Explanation:

The efficiency of the Carnot engine is given by:

\(\eta = \frac{{{T_H} - {T_L}}}{{{T_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}}\)

From the above formula, we know that efficiency is varied by:

• Decreasing the value of Temperature TL
• Keeping constant the value of TH

​

• The efficiency of the Carnot cycle is the function of the source and sink temperature.
• Carnot cycle efficiency depends upon the temperature range of operation.
• The efficiency of the Carnot engine is increased by decreasing the lower temperature and increasing the higher temperature but more effects on the efficiency of the Carnot cycle is by decreasing lower temperatures as compared to an increase in higher temperature.

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: 

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.