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

Last updated on Jun 13, 2025

Latest Thermodynamics MCQ Objective Questions

Thermodynamics Question 1:

A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa the pressure remaining constant throughout the process. There is a transfer of 37.6 KJ of heat from the gas during the process. The internal energy of gas changes by

  1. 15.75 KJ
  2. -31.25 KJ
  3. -21.85 KJ
  4. -37.6 KJ

Answer (Detailed Solution Below)

Option 3 : -21.85 KJ

Thermodynamics Question 1 Detailed Solution

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

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

 

Thermodynamics Question 2:

An object of mass 5 kg falls from rest through a vertical distance of 20 m and gains a velocity of 10 m/s. The work done by the resistance of the air on the object will be

  1. 550 J
  2. -750 J
  3. -550 J
  4. 750 J

Answer (Detailed Solution Below)

Option 2 : -750 J

Thermodynamics Question 2 Detailed Solution

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

Thermodynamics Question 3:

It is proposed that solar energy be used to warm a large collector plate. The energy would in turn, be transferred as heat to a fluid within a heat engine, and the engine would reject energy as heat to the atmosphere having assumed temperature of 20°C. Experiments indicate that about 1880 KJ/m2h of energy can be collected when the plate is operating at 90°C. The minimum collector area that would be required for plant producing 1 KW of useful shaft power will be approximately

  1. 15 m2
  2. 30 m2
  3. 10 m2
  4. 20 m2

Answer (Detailed Solution Below)

Option 3 : 10 m2

Thermodynamics Question 3 Detailed Solution

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

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

 

Thermodynamics Question 4:

If T1 is the source temperature and T2 is the sink temperature, the more effective way to increase the efficiency of a Carnot engine is

  1. Decrease T2 keeping T1 constant
  2. Decrease T1 keeping T2 constant
  3. Increase T2 keeping T1 constant
  4. Increase T1 keeping T2 constant

Answer (Detailed Solution Below)

Option 1 : Decrease T2 keeping T1 constant

Thermodynamics Question 4 Detailed Solution

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

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  • 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.

Thermodynamics Question 5:

The Zeroth Law of Thermodynamics establishes the basis for which of the following?

  1. Entropy increase in isolated systems
  2. Heat transfer through conduction
  3. Conservation of energy
  4. Measurement of temperature

Answer (Detailed Solution Below)

Option 4 : Measurement of temperature

Thermodynamics Question 5 Detailed Solution

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.

Top Thermodynamics MCQ Objective Questions

Identify the complete and correct statement of the first law of thermodynamics.

  1. During a process, extracting work without supplying heat, while sacrificing the energy of the system.
  2. Difference between the heat and work interactions during a process, which is property of the system.
  3. When a closed system executes a complete cycle, the sum of heat interactions is equal to the sum of work interactions.
  4. When a system undergoes a cycle, the integral of heat is equal to the integral of work.

Answer (Detailed Solution Below)

Option 3 : When a closed system executes a complete cycle, the sum of heat interactions is equal to the sum of work interactions.

Thermodynamics Question 6 Detailed Solution

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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.

The direction of a natural process is dictated by the ______ law of thermodynamics.

  1. zeroth
  2. third
  3. first
  4. second

Answer (Detailed Solution Below)

Option 4 : second

Thermodynamics Question 7 Detailed Solution

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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.

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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.

ΔST = 0K = 0

Figure shows the P-V diagram of an ideal gas. The work done by the gas in the process ABCDA is:

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  1. 0.5PV
  2. 2PV
  3. 1.5PV
  4. 4PV

Answer (Detailed Solution Below)

Option 3 : 1.5PV

Thermodynamics Question 8 Detailed Solution

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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:

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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

A rigid container of volume 0.5 m3 contains 1.0 kg of water at 120°C (vf = 0.00106 m3/kg, vg = 0.8908 m3/kg). The state of water is

  1. Compressed liquid
  2. Saturated liquid
  3. A mixture of saturated liquid and saturated vapor
  4. Superheated vapor

Answer (Detailed Solution Below)

Option 3 : A mixture of saturated liquid and saturated vapor

Thermodynamics Question 9 Detailed Solution

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

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.

At triple point for water, which of the following term is not equal to zero?

  1. Enthalpy
  2. Entropy
  3. Internal energy
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Enthalpy

Thermodynamics Question 10 Detailed Solution

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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

Therefore h ≠ 0 at triple point.

If specific heat ratio for a gas is γ, the change in internal energy of a mass of gas at constant pressure P, when volume changes from V to 2V is,

  1. \(\frac{{PV}}{{\gamma - 1}}\)
  2. \(\frac{R}{{\gamma - 1}}\)
  3. PV
  4. \(\frac{{\gamma PV}}{{\gamma - 1}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{{PV}}{{\gamma - 1}}\)

Thermodynamics Question 11 Detailed Solution

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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 = mCp(T2 -T1)

ΔU = mCv(T2 -T1)

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

The heat transfer in a cyclic process are 20 kJ, -28 kJ, -2 kJ and 40 kJ. Determine the total work for this cycle process.

  1. 30 kJ
  2. 45 kJ
  3. 40 kJ
  4. 60 kJ

Answer (Detailed Solution Below)

Option 1 : 30 kJ

Thermodynamics Question 12 Detailed Solution

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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.

Joule-Thompson coefficient for an ideal gas is

  1. Higher than zero
  2. Less than zero
  3. Zero
  4. 1

Answer (Detailed Solution Below)

Option 3 : Zero

Thermodynamics Question 13 Detailed Solution

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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.

Two processes isobaric and isochoric are represented on T-s diagram. They are starting from the same point. Out of these process which shall have a higher slope?

  1. isobaric
  2. both have same slope
  3. isochoric
  4. depend on final point

Answer (Detailed Solution Below)

Option 3 : isochoric

Thermodynamics Question 14 Detailed Solution

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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

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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/Cv

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

A thermometer works on the principle of

  1. Law of stable equilibrium
  2. Zeroth law of thermodynamics
  3. First law of thermodynamics
  4. Second law of thermodynamics

Answer (Detailed Solution Below)

Option 2 : Zeroth law of thermodynamics

Thermodynamics Question 15 Detailed Solution

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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.

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  • 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.
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