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

Last updated on Feb 25, 2025

Latest Thermodynamics MCQ Objective Questions

Thermodynamics Question 1:

Water of mass m gram is slowly heated to increase the temperature from T1 to T2. The change in entropy of the water, given specific heat of water is 1 Jkg–1K–1, is : 

  1. zero
  2. m (T2–T1
  3. \(\mathrm{m} \ln \left(\frac{\mathrm{~T}_{1}}{\mathrm{~T}_{2}}\right)\)
  4. \(\mathrm{m} \ln \left(\frac{\mathrm{~T}_{2}}{\mathrm{~T}_{1}}\right)\)

Answer (Detailed Solution Below)

Option 4 : \(\mathrm{m} \ln \left(\frac{\mathrm{~T}_{2}}{\mathrm{~T}_{1}}\right)\)

Thermodynamics Question 1 Detailed Solution

Calculation:

dQ = msdT 

\(\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{~T}}=\frac{\mathrm{msdT}}{\mathrm{~T}}\)

\(\Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{~T}}=\mathrm{ms} \ln \frac{\mathrm{~T}_{\mathrm{f}}}{\mathrm{~T}_{\mathrm{i}}}\)

\(\Delta \mathrm{S}=\mathrm{m} \ln \frac{\mathrm{~T}_{2}}{\mathrm{~T}_{1}}\)

Thermodynamics Question 2:

An ideal gas initially at 0ºC temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is 3/2, the change in temperature due to the thermodynamics process is ________ K. 

Answer (Detailed Solution Below) 273

Thermodynamics Question 2 Detailed Solution

Calculation:

\(γ=\frac{3}{2}\)

Tvγ–1 = C

\(273 \mathrm{~V}_{0}^{0.5}=\mathrm{T}\left(\frac{\mathrm{~V}_{0}}{4}\right)^{0.5}\)

T = 273 × 2 = 546

∴ ΔT = 273 k

Thermodynamics Question 3:

Match the List-I with List-II 

List-I

List-II 

A.

Pressure varies inversely with volume of an ideal gas. 

I.

Adiabatic process 

B.

Heat absorbed goes partly to increase internal energy and partly to do work. 

II.

Isochoric process

C.

Heat is neither absorbed nor released by a system 

III.

Isothermal process 

D.

No work is done on or by a gas 

IV.

Isobaric proc

 

Choose the correct answer from the options given below : 

  1. A–I, B–IV, C–II, D–III
  2. A–III, B–I, C–IV, D–II 
  3. A–I, B–III, C–II, D–IV
  4. A–III, B–IV, C–I, D–II 

Answer (Detailed Solution Below)

Option 4 : A–III, B–IV, C–I, D–II 

Thermodynamics Question 3 Detailed Solution

Calculation:

A → P ∝ \(\frac{1}{\mathrm{~V}}\)

⇒ PV = constant 

⇒  nRT = const. T = const. 

Hence Isothermal III 

B → IV

W ≠ 0, ΔU 0, ΔQ 0 [only isobaric] 

C → I ΔQ = 0 Adiabatic 

D → II w = 0 Isochoric 

∴ Correct order is A–III, B–IV, C–I, D–II 

Thermodynamics Question 4:

The internal energy (U) of an ideal gas depends on pressure (P) and volume (V) according to the equation:

P = 2UV-1

Based on this equation, which of the following conclusions can be drawn about the gas?

  1. The molar specific heat of the gas in an isobaric process is R.
  2. The molar specific heat of the gas in an isobaric process is 1.5R.
  3. The molar specific heat of the gas in an isobaric process is 2R.
  4. The molar specific heat of the gas in an isobaric process is 3R.

Answer (Detailed Solution Below)

Option 2 : The molar specific heat of the gas in an isobaric process is 1.5R.

Thermodynamics Question 4 Detailed Solution

Concept:

Internal Energy and Molar Specific Heat:

  • The internal energy (U) of an ideal gas is related to its pressure (P) and volume (V) as per the given equation:

         U = PV/2

We use the first law of thermodynamics and the specific heat relation:

         U = PV/2

Calculation:

U=RT/2

⇒ dU=R/2dT

⇒  dU/dT=R/2 =CV

Using the formula for molar specific heat at constant pressure:

         Cp = R+CV

Substituting values:

         Cp = R+R/2

         Cp = 3/2 R

         Cp = 1.5R

The molar specific heat of the gas in an isobaric process is 1.5R.

Thermodynamics Question 5:

What is the change in entropy when 10 g of ice at 0°C is converted into water at the same temperature? The latent heat of ice is 80 cal/g.

  1. 0 cal/K
  2. 2.93 cal/K
  3. 29.3 cal/K
  4. 293 cal/K 

Answer (Detailed Solution Below)

Option 2 : 2.93 cal/K

Thermodynamics Question 5 Detailed Solution

Calculation:

To find the change in entropy \(\Delta S\)  when ice is converted into water at the same temperature, we use the formula:

\(\Delta S = \frac{Q}{T}\)

where:

Q is the heat absorbed or released during the phase change.

T  is the absolute temperature (in Kelvin) at which the phase change occurs.

Given:

\( \text{Mass of ice, } m=10g\\ \text{Latent heat of fusion of ice, } L=80\text{cal/g}\\ \text{Temperature, } T=0^\circ\text{C}=273\text{K}\)

First, calculate Q, the total heat absorbed:

\(Q = m \cdot L = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal}. \)

Now, calculate \(\Delta S\):

\(\Delta S = \frac{Q}{T} = \frac{800 \, \text{cal}}{273 \, \text{K}} \approx 2.93 \, \text{cal/K}.\)

The change in entropy is \(2.93 \, \text{cal/K}\).

Thus, option '2' is correct.

Top Thermodynamics MCQ Objective Questions

_________  of thermodynamics is used to understand the concept of energy conservation.

  1. Zeroth law
  2. First law
  3. Second law
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : First law

Thermodynamics Question 6 Detailed Solution

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Concept:

  • 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.

  • The first law of thermodynamics states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.

 

The first law of thermodynamics is a restatement of the law of conservation of energy

i.e., According to the first law of Thermodynamics:

ΔQ = ΔW + ΔU

  • Now the First Law of Thermodynamics helped us in understanding the principle of conservation of energy, whereas according to the Second Law of thermodynamics for natural system heat always flows in one direction (higher temperature to lower temperature body) unless it aided by an external factor.

 

And to measure the direction of force we use term entropy which can be expressed as

\({\rm{\Delta }}S = \;\smallint \frac{{dQ}}{T}\)

ΔQ = heat exchange

ΔW = work done due to expansion

ΔU = internal energy of the system

ΔS = change in entropy

T = temperature

Explanation:

As explained above according to the first law of thermodynamics energy cannot be created or destroyed in an isolated system, energy can only be transferred or changed from one form to another.

This is the ideal statement which is used in thermodynamics for explaining the concept of energy conservation among systems and surrounding.

Hence option 2 is correct among all

Tricks to remember:

This is the conclusive point for all three laws of thermodynamics.

Zeroth law – Concept of temperature

First law – Concept of internal energy/ energy conservation

Second law – Concept of entropy/ heat flow

If the temperature of the source is increased, the efficiency of the Carnot engine

  1. Increases
  2. Decreases
  3. Remains Constant
  4. First increases and then becomes constant

Answer (Detailed Solution Below)

Option 1 : Increases

Thermodynamics Question 7 Detailed Solution

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CONCEPT:

  • Carnot engine: The theoretical engine which works on the Carnot cycle is called a Carnot engine.
    • It gives the maximum possible efficiency among all types of heat engines.
    • The part of the Carnot engine which provides heat to the engine is called a heat source.
    • The temperature of the source is maximum among all the parts.
    • The part of the Carnot engine in which an extra amount of heat is rejected by the engine is called as a heat sink.
    • The amount of work which is done by the engine is called as work done.

The efficiency (η) of a Carnot engine is given by:

\(η = 1 - \frac{{{T_C}}}{{{T_H}}} = \;\frac{{Work\;done\left( W \right)}}{{{Q_{in}}}} = \;\frac{{{Q_{in}} - \;{Q_R}}}{{{Q_{in}}}}\)

Where Tis the temperature of the sink, Tis the temperature of the source, W is work done by the engine, Qin is the heat given to the engine/heat input and QRis heat rejected.

EXPLANATION:

The efficiency (η) of the Carnot engine is given by:

η = 1 - TC/TH

  • Here if TH increases, the value of TC/TH decreases, and hence the value of (1 - TC/TH) increases.
  • If the temperature of the source (TH) is increased then the efficiency of the Carnot engine increases. So option 1 is correct.

A Carnot engine works between the temperature 227° C and 127° C. If the work output of the engine is 104 J, then the amount of heat rejected to the sink will be:

  1. 1 × 104 J
  2. 2 × 104 J
  3. 4 × 104 J
  4. 5 × 104 J

Answer (Detailed Solution Below)

Option 3 : 4 × 104 J

Thermodynamics Question 8 Detailed Solution

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CONCEPT:

  • Carnot engine an ideal reversible engine that operates between two temperatures T(Source), and T(Sink).
  • Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
  • The steps of the Carnot cycle are
    1. Isothermal expansion
    2. Adiabatic expansion
    3. Isothermal compression
    4. Adiabatic compression
  • The efficiency of the Carnot engine is defined as the ratio of net work done per cycle by the engine to heat absorbed per cycle by the working substance from the source. 
  • The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

SOLUTION:

Given - T1 = 227° C = 500 K, T2 = 127° C = 400 K and W = 104 J
  • The efficiency is given by

\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{{{400}}}{{{500}}}=\frac{1}{5}\)

  • The amount of heat absorbed by the Carnot engine can be calculated is 

\(⇒ Q_1= \frac{W}{\eta}=\frac{10^4}{\frac{1}{5}}=5×10^4\, J\)

  • The amount of heat rejected to the sink will be:

⇒ Q2 = Q1 - W

⇒ Q2 = 5 × 104 - 1 × 104 = 4 × 104 J

In which Thermodynamic process is there no flow of heat between the system and the surroundings?

  1. Isobaric
  2. Isochoric
  3. Adiabatic
  4. Isothermal

Answer (Detailed Solution Below)

Option 3 : Adiabatic

Thermodynamics Question 9 Detailed Solution

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CONCEPT:

Isobaric process Isochoric process Adiabatic process Isothermal process
It allows to set up the relationship between the changes in volume and temperature under constant pressure.

The process in which the volume of the gas remains constant is called the isochoric process. 

For example: A gas is filled in a closed container then the volume of the gas will remain constant.

The thermodynamic process in a system, during which no heat transfer occurs between thermodynamic systems and surrounding is called an adiabatic process. It allows to set up the relationship between the changes in pressure and volume under constant temperature:

V1/T= V2/T2 so ∝ T  

Where [Vand V2 are volume and T1 and T2 are different temperatures]

 

\(\frac{P_1}{T_1} = \frac{P_2}{T_2} = Constant\)

PVγ  = Constant

Where γ is ratio of specific heat

P1V1 = P2V2 so P V = Constant    

Where [P1 and P2 are the pressure of gases and Vand V2 are volume]

EXPLANATION:

  • In an adiabatic process, no heat flows between the system and surroundings. So option 3 is correct.

An ideal gas heat engine operates in Carnot's cycle between 227° C and 127° C It absorbs 6 × 104 J at high temperature. The amount of heat converted into work is ____

  1. 4.8 × 104 × J
  2. 3.5 × 104 × J
  3. 1.6 × 104 × J
  4. 1.2 × 104 × J

Answer (Detailed Solution Below)

Option 4 : 1.2 × 104 × J

Thermodynamics Question 10 Detailed Solution

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CONCEPT:

  • Carnot engine an ideal reversible engine that operates between two temperatures T(Source), and T(Sink).
  • Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
  • The steps of the Carnot cycle are
    1. Isothermal expansion
    2. Adiabatic expansion
    3. Isothermal compression
    4. Adiabatic compression
  • The efficiency of the Carnot engine is defined as the ratio of network done per cycle by the engine to heat absorbed per cycle by the working substance from the source. 
  • The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

CALCULATION :

Given - T1 = 227+273 = 500 K, T2 = 127 +273 = 400 K, Heat absorbed by the engine is Q1 = 6 × 104J

  • The efficiency of the heat engine is given by

\(\Rightarrow \eta = \frac{W}{Q_1} = 1-\frac{T_2}{T_1}\)

\(\Rightarrow \frac{W}{6\times 10^{4}} = 1-\frac{400}{500}\)

\(\Rightarrow \frac{W}{6\times 10^{4}} = 1-\frac{4}{5} \)

\(\Rightarrow \frac{W}{6\times10^{4}} = \frac{1}{5} \)

\(\Rightarrow {W} = \frac{6\times 10^{4}}{5} = 1.2 \times 10^{4} J\)

  • Hence, option 4 is the answer

In thermodynamics ___________ is not a state variable.

  1. density
  2. internal energy
  3. enthalpy
  4. Heat

Answer (Detailed Solution Below)

Option 4 : Heat

Thermodynamics Question 11 Detailed Solution

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CONCEPT:

  • State variables are defined as the thermodynamical variables which depend only on the initial and final state of a thermodynamical system
    • These variables don't depend on how the thermodynamical system changed itself from the initial to the final state.
    • Temperature, Pressure, Internal energy, and Density are the examples of state variables.
    • State variables are also known as state functions.
  • Path variables are defined as the thermodynamical variables which depend on the way in which the thermodynamical system achieved the initial and final states.
    • Heat, Work is examples of Path variables 

F1 P.Y 26.8.20 Pallavi D2

 

EXPLANATION:

  • Internal energy, pressure, density, and enthalpy are examples of state variables. Since they depend only on the final and initial states of the thermodynamical system.
  • Heat is a measure of the amount of energy present in a thermodynamical system. As the amount of energy changes the heat present in the system changes. Hence heat is the path variable. Therefore option 4 is correct answer.

Which variable is held constant in Charles's Law?

  1. Temperature
  2. Volume
  3. Heat
  4. Pressure

Answer (Detailed Solution Below)

Option 4 : Pressure

Thermodynamics Question 12 Detailed Solution

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CONCEPT:

Charles law:

F1 P.Y Madhu 23.04.20 D5

  • If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

i.e. V ∝ T
or V/T = constant
\( \Rightarrow \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\)

EXPLANATION:

  • From above it is clear that in Charles's Law, pressure remains constant. Therefore option 4 is correct.

What is the source temperature of the Carnot engine in K required to get 70% efficiency?

Given sink temperature = 27 °C

  1. 1000 K
  2. 90 K
  3. 270 K
  4. 727 K

Answer (Detailed Solution Below)

Option 1 : 1000 K

Thermodynamics Question 13 Detailed Solution

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CONCEPT:

  • Carnot engine: A theoretical thermodynamic cycle proposed by Leonard Carnot. It gives the estimate of the maximum possible efficiency that a heat engine during the conversion process of heat into work and conversely, working between two reservoirs, can possess.
    • So practically and theoretically there can not be any engine with more efficiency than Carnot engine.

The efficiency of the Carnot's Heat engine is given by:

 \(η=1-\frac{T_c}{T_h}\)

where Tc is the temperature of the cold reservoir/sink and Th is the temperature of the hot reservoir/source.

  • The efficiency of this type of engine is independent of the nature of the working substance and is only dependent on the temperature of the hot and cold reservoirs.

CALCULATION:

Given that sink temperature Tc = 27°C = 300K

η = 70% = 0.7

\(η=1-\frac{T_c}{T_h}\)

\(0.7=1-\frac{300}{T_h}\)

Th = 300/0.3 = 1000 K 

So the correct answer is option 1.

The heat given to an ideal gas in isothermal conditions is used to:

  1. increase temperature
  2. do external work
  3. increase temperature and in doing external work
  4. increase internal energy

Answer (Detailed Solution Below)

Option 2 : do external work

Thermodynamics Question 14 Detailed Solution

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CONCEPT:

First Law of Thermodynamics:

  • It is a statement of conservation of energy in the thermodynamical process.
  • According to it heat given to a system (ΔQ) is equal to the sum of the increase in its internal energy (ΔU) and the work done (ΔW) by the system against the surroundings.

i.e ΔQ = ΔU + ΔW          [∴ ΔW = p ΔV]

  • It makes no distinction between work and heat as according to it the internal energy (and hence temperature) of a system may be increased either by adding heat to it or doing work on it or both.

EXPLANATION:

  • When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as an isothermal process.
  • As we know that, the internal energy of the system is a function of temperature alone, so in the isothermal process, the change in internal energy is zero.

⇒ ΔQ = 0 + ΔW =  ΔW

  • Therefore, the heat given to an ideal gas in isothermal conditions is used to do external work. Hence option 2 is correct.

110 joule of heat is added to a gaseous system, whose internal energy is 40 J. Then the amount of external work done is 

  1. 150 J
  2. 70 J
  3. 110 J
  4. 40 J

Answer (Detailed Solution Below)

Option 2 : 70 J

Thermodynamics Question 15 Detailed Solution

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Concept:

  • The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.
  • When heat energy is supplied to a thermodynamic system or any machine. 
  • Two things may occur:
    • The internal energy of the System or machine may change.
    • The system may do some external work.

According to the first law of Thermodynamics:

ΔQ = ΔW + ΔU

Where ΔQ = Heat supplied to the system or heat exchange, ΔW = work done by the system, ΔU = change in internal energy of the system 

Explanation:

Given that, ΔQ = 110 J, ΔU = 40 J

According to the  first law of thermodynamics:

ΔQ = ΔW + ΔU

ΔU = ΔQ - ΔW

40 J = 110 J - ΔW

ΔW = 110 - 40 = 70 J

Then the amount of external work done is 70 J.

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