Starting with the equation and using the appropriate Maxwell's relation along with the expression for heat capacity (see useful information), the derivative for a substance can be expressed in terms of its specific heat , density , coefficient of volume expansion , and temperature . For ice, , , and . If the value of at 270 K is , then the value of is:
Official Solution
Correct Option:
(1)
Step 1: Use Maxwell's relations. From thermodynamics, Maxwell's relations give the following expression for :
Step 2: Substituting known values. Using the values , , and , we can calculate:
02
PYQ 2017
medium
physicsID: iit-jam-
Consider a Carnot engine operating between temperatures of 600 K and 400 K. The engine performs 1000 J of work per cycle. The heat (in Joules) extracted per cycle from the high temperature reservoir is:
Official Solution
Correct Option:
(1)
Step 1: Understanding the Carnot engine. The efficiency of a Carnot engine is given by:
where is the cold reservoir temperature and is the hot reservoir temperature. The work done per cycle . The heat extracted from the hot reservoir is , and the efficiency is also given by:
Thus, we can solve for :
Step 2: Conclusion. Thus, the heat extracted per cycle from the high temperature reservoir is 3000 J.
03
PYQ 2017
medium
physicsID: iit-jam-
An isolated box is divided into two equal compartments by a partition (see figure). One compartment contains a van der Waals gas while the other compartment is empty. The partition between the two compartments is now removed. After the gas has filled the entire box and equilibrium has been achieved, which of the following statement(s) is (are) correct?
1
Internal energy of the gas has not changed.
2
Internal energy of the gas has decreased.
3
Temperature of the gas has increased.
4
Temperature of the gas has decreased.
Official Solution
Correct Option:
(1)
Step 1: Understanding the effect of partition removal.
When the partition is removed, the gas expands to fill the entire volume. This expansion is an irreversible process, and in most cases, it will result in a decrease in the temperature of the gas, assuming no heat exchange with the surroundings. However, the internal energy of the gas does not change in an isolated system because the expansion work is done without heat exchange. Step 2: Conclusion.
Thus, the correct answer is options (A) and (D).
04
PYQ 2017
medium
physicsID: iit-jam-
Consider two identical, finite, isolated systems of constant heat capacity at temperatures and ( ). An engine works between them until their temperatures become equal. Taking into account that the work performed by the engine will be maximum ( ) if the process is reversible (equivalently, the entropy change of the entire system is zero), the value of is:
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Understanding the maximum work.
The maximum work done in a thermodynamic process where the entropy change of the system is zero occurs when the temperatures become equal. The work done is given by , based on the first and second laws of thermodynamics. Step 2: Conclusion.
Therefore, the correct answer is option (D).
05
PYQ 2018
medium
physicsID: iit-jam-
Which one of the figures correctly represents the T–S diagram of a Carnot engine?
1
(A)
2
(B)
3
(C)
4
(D)
Official Solution
Correct Option:
(2)
Step 1: Recall the Carnot cycle in the T–S plane. A Carnot cycle consists of two isothermal processes and two adiabatic processes. In a T–S diagram: - Isothermals appear as horizontal lines (constant temperature). - Adiabatic processes appear as vertical lines (since entropy remains constant).
Step 2: Identify the correct rectangular shape. A Carnot engine forms a perfect rectangle in the T–S plane: - Top and bottom edges: isothermal expansion and compression. - Left and right edges: adiabatic compression and expansion.
Step 3: Compare with the given diagrams. Only diagram (B) shows a proper rectangle with horizontal and vertical sides, matching the structure of a Carnot cycle.
Step 4: Conclusion. Thus the correct T–S diagram for a Carnot engine is option (B).
06
PYQ 2018
medium
physicsID: iit-jam-
A monoatomic ideal gas undergoes a closed cycle shown in the P–V diagram. The ratio is ........... (Specify your answer up to two digits after the decimal point.)
Official Solution
Correct Option:
(1)
Step 1: Note the adiabatic relation.
For a monoatomic ideal gas, , and .
Step 2: Apply between points and .
.
Step 3: Solve for pressure ratio.
.
Step 4: But is the upper point at same , so ratio is inverted.
Thus final ratio becomes .
07
PYQ 2018
medium
physicsID: iit-jam-
Which of the following relations is (are) true for thermodynamic variables?
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Use Maxwell relations. Thermodynamic identities: (not +P dV) → (C) is false. → (D) correct.
Step 2: Entropy differential identities. Using exact relations for : → (A) correct. → (B) is incorrect sign.
Step 3: Conclusion. Correct statements are (A) and (D).
08
PYQ 2018
medium
physicsID: iit-jam-
The equation of state for one mole of a non-ideal gas is given by , where the coefficients and are temperature dependent. If the volume changes from to in an isothermal process, the work done by the gas is
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Start from the definition of work. For expansion from to : .
Step 2: Substitute the equation of state. . Thus, .
Step 3: Integrate term-by-term. , .
Step 4: Final expression. .
Step 5: Conclusion. This matches option (C).
09
PYQ 2018
medium
physicsID: iit-jam-
Consider an ensemble of thermodynamic systems, each of which is characterized by the same number of particles, pressure and temperature. The thermodynamic function describing the ensemble is
1
Enthalpy
2
Helmholtz free energy
3
Gibbs free energy
4
Entropy
Official Solution
Correct Option:
(3)
Step 1: Identify ensemble constraints. The ensemble has fixed: • Number of particles, • Pressure, • Temperature (N, P, T).
Step 2: Connect constraints to thermodynamic potentials. Each ensemble is described by a thermodynamic potential that is minimised under its natural variables: • Helmholtz free energy → natural variables , • Gibbs free energy → natural variables , • Enthalpy → natural variables .
Step 3: Compare with given ensemble. Since the ensemble is at fixed , the appropriate potential is Gibbs free energy .
Step 4: Conclusion. The thermodynamic function describing such an ensemble is the Gibbs free energy.
10
PYQ 2019
medium
physicsID: iit-jam-
A diatomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is .................
Official Solution
Correct Option:
(1)
Step 1: Use the formula for work done in an adiabatic process. For an adiabatic expansion, the work done by the gas is given by:
where is the number of moles, is the molar heat capacity at constant volume for a diatomic gas, and is the change in temperature. For a diatomic gas, , where is the universal gas constant. Step 2: Calculate the work done. The temperature change is:
Now, substitute into the work equation:
Solving for :
Step 3: Conclusion. Thus, the number of moles required to obtain 2300 J of work from the expansion is 1.00 mol.
11
PYQ 2019
medium
physicsID: iit-jam-
A thermodynamic system is described by the coordinates. Choose the valid expression(s) for the system.
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understanding thermodynamic identities. The relations between partial derivatives of state variables in thermodynamics are derived from Maxwell’s relations, which are based on the thermodynamic potentials. For a system with the variables, the correct relations come from the use of the Helmholtz free energy or other thermodynamic potentials.
Step 2: Analyzing the options. (A) : Correct. This is a valid thermodynamic identity derived from Maxwell’s relations. (B) : Incorrect. This does not hold as a valid identity. (C) : Incorrect. This is not a valid thermodynamic identity. (D) : Incorrect. This does not hold as a valid identity.
Step 3: Conclusion. The correct answer is (A) as it is a valid thermodynamic identity based on Maxwell’s relations.
12
PYQ 2019
medium
physicsID: iit-jam-
During free expansion of an ideal gas under adiabatic condition, the internal energy of the gas
1
Decreases
2
Initially decreases and then increases
3
Increases
4
Remains constant
Official Solution
Correct Option:
(4)
Step 1: Understanding free expansion of an ideal gas. In the free expansion of an ideal gas, the gas expands into a vacuum without doing any work. Since the process is adiabatic (no heat exchange with the surroundings), the internal energy of the gas decreases. This is because the gas expands and does not gain any energy from external sources.
Step 2: Analyzing the options. (A) Decreases: Correct. The internal energy decreases as the gas expands adiabatically, following the first law of thermodynamics. (B) Initially decreases and then increases: Incorrect. The internal energy continuously decreases during the free expansion. (C) Increases: Incorrect. The internal energy does not increase in free expansion. (D) Remains constant: Incorrect. The internal energy decreases during the expansion, so it does not remain constant.
Step 3: Conclusion. The correct answer is (A) Decreases because the gas loses internal energy as it expands without heat exchange.
13
PYQ 2019
medium
physicsID: iit-jam-
In a heat engine based on the Carnot cycle, heat is added to the working substance at constant
1
Entropy
2
Pressure
3
Temperature
4
Volume
Official Solution
Correct Option:
(3)
Step 1: Understanding the Carnot cycle. The Carnot cycle is a theoretical thermodynamic cycle that defines the maximum possible efficiency of a heat engine. It consists of two isothermal processes (constant temperature) and two adiabatic processes (no heat exchange). Step 2: Heat addition in the Carnot cycle. In the Carnot cycle, heat is added to the working substance during an isothermal expansion, meaning the temperature remains constant. Thus, the heat is added at a constant temperature. Step 3: Conclusion. Therefore, the correct answer is (C), as heat is added at constant temperature in the Carnot cycle.
14
PYQ 2019
medium
physicsID: iit-jam-
Isothermal compressibility is given by
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Understanding isothermal compressibility. Isothermal compressibility describes how the volume of a substance changes with pressure at constant temperature. It is defined as:
where is the isothermal compressibility, is the volume, is the pressure, and is the temperature. Step 2: Identifying the correct option. The correct form of isothermal compressibility is , which matches option (C).
15
PYQ 2020
medium
physicsID: iit-jam-
A body at a temperature is brought into contact with a reservoir at temperature . Thermal equilibrium is established at constant pressure. The heat capacity of the body at constant pressure is . The total change in entropy of the body and the reservoir in units of is ............ (Round off to 2 decimal places)
Official Solution
Correct Option:
(1)
Step 1: Entropy change of the body. The body is heated from temperature to at constant pressure.
Step 2: Entropy change of the reservoir. The heat lost by the reservoir is equal to the heat gained by the body. The reservoir temperature remains constant at , so its entropy change is:
Step 3: Total entropy change.
Step 4: Conclusion. Hence, the total change in entropy of the system and reservoir in units of is .
16
PYQ 2020
medium
physicsID: iit-jam-
One mole of an ideal monatomic gas at pressure , volume and temperature is expanded isothermally to volume . Thereafter, the gas is heated isochorically (at constant volume) till its pressure becomes . If is the universal gas constant, the total heat transfer in the process, in units of , is ................... (Round off to 2 decimal places)
Official Solution
Correct Option:
(1)
Step 1: For the isothermal expansion (A → B).
Step 2: For the isochoric heating (B → C). Since volume is constant, . For an ideal gas: At the end of isothermal expansion, . After heating at constant volume, final pressure . For a monatomic gas, .
Step 3: Total heat transfer.
Step 4: Conclusion. Hence, the total heat transfer in the process in units of is .
17
PYQ 2020
medium
physicsID: iit-jam-
One mole of an ideal gas having specific heat ratio ( ) of 1.6 is mixed with one mole of another ideal gas having specific heat ratio of 1.4. If and are the molar specific heat capacities of the gas mixture at constant volume and pressure, respectively, which of the following is/are correct? ( denotes the universal gas constant.)
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Recall relationships
For an ideal gas: $ \gamma_1 = 1.6 \gamma_2 = 1.4 C_P = 1.48 C_V$
Answer: (A) and (C) are correct
18
PYQ 2020
medium
physicsID: iit-jam-
Consider two spherical perfect blackbodies with radii and at temperatures and , respectively. They both emit radiation of power 1 kW. The ratio of their radii, , is given by ...............
Official Solution
Correct Option:
(1)
Physics Principle
The power ( ) radiated by a perfect blackbody is given by the Stefan-Boltzmann Law:
$ P \sigma \sigma \approx 5.67 \times 10^{-8}\ W\cdot m^{-2}\cdot K^{-4} A m^2 T K A A = 4\pi R^2 R R_1 R_2 T T_1 = 1000\ K T_2 = 2000\ K P P_1 = 1\ kW P_2 = 1\ kW P_1 = P_2 4\pi \sigma R_1/R_2 R_1/R_2$, is 4.
19
PYQ 2020
medium
physicsID: iit-jam-
An ideal gas undergoes an isothermal expansion along path AB, adiabatic expansion along BC, isobaric compression along CD, isothermal compression along DE, and adiabatic compression along EA, as shown in the figure. The work done by the gas along the process BC is 10 J. The change in the internal energy along process EA is 16 J. The absolute value of the change in the internal energy along the process CD is ............ J.
Official Solution
Correct Option:
(1)
Step 1: Recall the first law of thermodynamics. The internal energy change over a complete cycle is zero: The work done over one complete cycle equals the total heat absorbed:
Step 2: Analyze given processes. - For isothermal processes (AB and DE): . - For adiabatic processes (BC and EA): , hence . Given ⇒ . Also, .
Step 3: Apply energy balance over full cycle. Hence, the absolute value of internal energy change:
Step 4: Conclusion. The absolute value of the internal energy change during process CD is .
20
PYQ 2020
medium
physicsID: iit-jam-
For a system undergoing a first-order phase transition at a temperature , which one of the following graphs best describes the variation of entropy (S) as a function of temperature (T)?
1
(A)
2
(B)
3
(C)
4
(D)
Official Solution
Correct Option:
(4)
Step 1: Understanding the concept. In a first-order phase transition, the entropy changes discontinuously at the transition temperature , corresponding to the latent heat of the transformation.
Step 2: Interpreting the graphs. Among the given graphs, only Graph (D) shows a sudden jump in entropy at , indicating a discontinuous change.
Step 3: Conclusion. Therefore, Graph (D) represents the correct variation of entropy during a first-order phase transition.
21
PYQ 2021
medium
physicsID: iit-jam-
A Carnot engine operates between two temperatures, and . Each cycle of the engine lasts for 0.5 seconds during which the power delivered is . Let be the heat absorbed by the engine and be the heat lost. Identify the correct statement(s).
1
2
3
The change in entropy of the engine and the hot bath in a cycle is
4
The change in entropy of the engine in 0.5 seconds is zero
Official Solution
Correct Option:
(1)
Step 1: Determine work done per cycle. Given power and each cycle takes : Step 2: Efficiency of Carnot engine. Step 3: Calculate heat absorbed from the hot reservoir. Step 4: Calculate . Step 5: Entropy change in a Carnot cycle. For a reversible Carnot cycle,
because the entropy gained and lost per cycle are equal in magnitude and opposite in sign. Step 6: Final Answer. Correct options are (A) and (D).
22
PYQ 2021
medium
physicsID: iit-jam-
A gaseous system, enclosed in an adiabatic container, is in equilibrium at pressure and volume . Work is done on the system in a quasi-static manner due to which the pressure and volume change to and , respectively, in the final equilibrium state. At every instant, the pressure and volume obey the condition , where and is a constant. If the work done is zero, then identify the correct statement(s).
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Given adiabatic condition. The process follows . Normally, work is done in adiabatic processes, but here it is stated that the net work done is zero. Step 2: Work done in adiabatic process. For an adiabatic process, If , then Step 3: Final Answer. Therefore, the correct statement is
23
PYQ 2021
medium
physicsID: iit-jam-
An isolated ideal gas is kept at a pressure and volume . The gas undergoes free expansion and attains a pressure and volume . Identify the correct statement(s).
1
This is an adiabatic process
2
3
4
This is an isobaric process
Official Solution
Correct Option:
(1)
Step 1: Nature of free expansion. In a free expansion, the gas expands into a vacuum without doing any work ( ) and without heat exchange ( ), since the system is isolated. Step 2: Apply the first law of thermodynamics. Since and , we have This implies that the internal energy and hence the temperature remain constant. Step 3: Process type. A process in which no heat is exchanged with the surroundings is an adiabatic process. Step 4: Final Answer. Therefore, the process is adiabatic (Option A).
24
PYQ 2021
medium
physicsID: iit-jam-
A system undergoes a thermodynamic transformation from state to via two different paths 1 and 2. The heat absorbed and work done along path 1 are 50 J and 30 J, respectively. If the heat absorbed along path 2 is 30 J, the work done along path 2 is:
1
Zero
2
10 J
3
20 J
4
30 J
Official Solution
Correct Option:
(2)
Step 1: Apply the first law of thermodynamics. Step 2: For path 1, Since internal energy change depends only on initial and final states, remains the same for both paths. Step 3: For path 2, Step 4: Final Answer. The work done along path 2 is 10 J.
25
PYQ 2022
hard
physicsID: iit-jam-
For a certain thermodynamic system, the internal energy U = PV and P is proportional to T2. The entropy of the system is proportional to
1
UV
2
3
4
Official Solution
Correct Option:
(4)
To determine the entropy of the thermodynamic system, we start by interpreting the given information:
The internal energy is given as , suggesting the relationship between pressure ( ), volume ( ), and internal energy.
It is also provided that is proportional to , i.e., , where is a proportionality constant.
Hence, we can express the internal energy as:
This indicates that is proportional to .
We need to find the entropy of the system, for which we consider the relationship involving temperature ( ) and internal energy ( ). The basic thermodynamic identity in terms of internal energy, entropy, volume, and temperature is:
Given , we substitute in the above equation:
Simplifying the expression and considering small changes, it can be derived that entropy that follows these conditions should depend on . However, we notice the problem requests an expression for , which leads to exploring the functional relationship further with given options.
Given the options and upon derivations involving standard thermodynamic identities or common proportionality in thermodynamics, the most suitable entropy expression in terms of and is:
Therefore, the entropy of the system is proportional to .
26
PYQ 2022
easy
physicsID: iit-jam-
For an ideal gas, AB and CD are two isothermals at temperatures T1 and T2( ), respectively. AD and BC represent two adiabatic paths as shown in figure. Let VA, VB, VC and VD be the volumes of the gas at A, B, C and D respectively. If , then ______.
Official Solution
Correct Option:
(1)
To solve this problem, we need to use the properties of adiabatic and isothermal processes for an ideal gas. For adiabatic processes, the relation holds, where is the adiabatic index. For isothermal processes, applies.
Given: .
Since BC is adiabatic:
AD is also adiabatic:
From conservation using isothermal relation along AB and CD:
Substitute in the isothermal relations:
By comparing both adiabatic conditions:
when ratios align for , this leads to specific constraints where the numerical resolution yields:
. This confirms the required constraint for continuity in adiabatic expansion to isothermal.
Conclusion:
27
PYQ 2022
hard
physicsID: iit-jam-
A hot body with constant heat capacity 800 //K at temperature 925 K is dropped gently into a vessel containing 1 kg of water at temperature 300 K and the combined system is allowed to reach equilibrium. The change in the total entropy is _____________ J/K (Round off to 1 decimal place). [Take the specific heat capacity of water to be 4200 J/kg K. Neglect any loss of heat to the vessel and air and change in the volume of water.]
Official Solution
Correct Option:
(1)
To compute the change in the total entropy of the system, we consider both the hot body and the water. The heat exchanges between them can be calculated using their heat capacities and the principle of energy conservation. The steps are as follows:
1. **Define the initial conditions and variables:** The heat capacity of the hot body, J/K. The initial temperature of the hot body, K. The mass of the water, kg. The specific heat capacity of water, J/kg·K. The initial temperature of the water, K.
2. **Calculate the final equilibrium temperature :** At thermal equilibrium, the heat lost by the hot body equals the heat gained by the water:
Substitute the known values:
Solve this equation for . After simplifying:
Combine terms:
3. **Calculate the entropy change of the hot body :** The entropy change is given by:
4. **Calculate the entropy change of the water :** The entropy change for the water is:
5. **Calculate the total entropy change :**
This is the change in the total entropy. The change in total entropy for the system is 537.5 J/K.
28
PYQ 2022
medium
physicsID: iit-jam-
Consider a classical ideal gas of N molecules in equilibrium at temperature T. Each molecule has two energy levels, -∈ and ∈. The mean energy of the gas is
1
0
2
3
4
Official Solution
Correct Option:
(3)
To find the mean energy of a classical ideal gas with N molecules, each having two energy levels, we use the concept of statistical mechanics. The possible energy levels for each molecule are given as and .
In statistical mechanics, the probability of a system being in a particular energy state is given by the Boltzmann factor. This factor is given by:
where is the partition function, is the Boltzmann constant, and is the temperature. The partition function, , for a single molecule is calculated as follows:
The probability of a molecule being in the state is:
The probability of a molecule being in the state is:
The mean energy of one molecule, , is calculated by averaging over these two states:
Substituting in the expressions for and :
This simplifies to:
Thus, for N molecules, the total mean energy of the gas is:
The correct answer is selected based on the fact that the mean energy of the gas would result in net energy direction considering the dual levels, leading to:
Therefore, the answer is .
29
PYQ 2022
medium
physicsID: iit-jam-
At a temperature T, let and denote the volume expansivity and isothermal compressibility of a gas, respectively. Then is equal to
1
2
3
4
Official Solution
Correct Option:
(1)
To solve this problem, we need to understand the physical terms involved: volume expansivity ( ) and isothermal compressibility ( ). These are fundamental thermodynamic properties of materials.
Volume Expansivity ( ):
It is a measure of how much the volume of a substance changes with temperature at constant pressure. Mathematically, it is defined as:
Isothermal Compressibility ( ):
It measures the fractional change in volume of a substance as pressure changes at constant temperature. It is expressed as:
The problem asks for the ratio . Substituting the expressions for and , we get:
After simplifying, the volumes ( ) cancel out, leading to:
Using the thermodynamic identity of Maxwell's relations, this ratio simplifies to:
_V
This matches the first option, , which is the correct answer.
Therefore, the correct answer is:
30
PYQ 2022
medium
physicsID: iit-jam-
An ideal gas in equilibrium at temperature T expands isothermally to twice its initial volume. If ΔS, ΔU and ΔF denote the changes in its entropy, internal energy and Helmholtz free energy respectively, then
1
ΔS < 0, ΔU > 0, ΔF < 0
2
ΔS > 0, ΔU = 0, ΔF < 0
3
ΔS < 0, ΔU = 0, ΔF > 0
4
ΔS > 0, ΔU > 0, ΔF = 0
Official Solution
Correct Option:
(2)
To solve this problem related to the thermodynamics of an ideal gas, let's consider the concepts involved:
Isothermal Process: An isothermal process is one in which the temperature of the system remains constant. For an ideal gas undergoing an isothermal process, the internal energy change is zero because internal energy of an ideal gas depends only on temperature, which is unchanged. Hence, .
Entropy Change ( ): The entropy change for an ideal gas undergoing an isothermal expansion from volume to volume is given by:
Since , this becomes . Since , it follows that .
Helmholtz Free Energy Change ( ): The Helmholtz free energy change is defined as:
With , the equation simplifies to . Since , it follows that .
Therefore, the correct statements are: , , and .
Thus, the correct option is:
, ,
31
PYQ 2023
hard
physicsID: iit-jam-
Temperature (T) dependence of the total specific heat (Cv) for a two dimensional metallic solid at low temperatures is
1
2
3
4
Official Solution
Correct Option:
(1)
To determine the temperature dependence of the total specific heat for a two-dimensional metallic solid at low temperatures, we need to understand how specific heat behaves in such systems.
At low temperatures, the specific heat of a metal is typically described by the expression:
where:
is the linear contribution from the electronic states.
is the square contribution from the lattice vibrations (phonons) in two dimensions.
For a two-dimensional metallic solid, at low temperatures, the electronic contribution dominates according to the linear term . However, this specific context implies that both electronic and phononic contributions are considered.
Given the options available, the correct graph should represent this behavior where specific heat initially increases linearly with temperature and shows curvature due to the phononic term. The correct graph among the provided options should reflect an initial linear rise followed by a slight curve, indicating the increasing influence of phononic contributions as the temperature increases.
Examining the options, the correct graph for the temperature dependence of specific heat at low temperatures in a two-dimensional metallic solid is:
This graph shows a clear linear rise initially (due to the term) with a subsequent rise indicative of a contribution (showing curvature), thus accurately representing the expected behavior of at low temperatures.
32
PYQ 2023
medium
physicsID: iit-jam-
For a thermodynamic system, the coefficient of volume expansion and compressibility , where V, T, and P are respectively the volume, temperature, and pressure. Considering that is a perfect differential, we get
1
2
3
4
Official Solution
Correct Option:
(3)
The question provides us with expressions for the coefficient of volume expansion and compressibility for a thermodynamic system. We are asked to find a relationship based on these expressions.
Given:
The coefficient of volume expansion:
The compressibility:
To find the relationship, we start by considering that is a perfect differential. The total differential change in volume can be expressed using the following differential form:
Dividing this equation by volume gives:
Substitute the expressions for and :
Now, apply the condition that is a perfect differential, which implies that the mixed second derivatives satisfy the following equality:
This leads to:
Therefore, the correct answer is:
This shows that the derivative of the coefficient of volume expansion with respect to pressure at constant temperature is the negative of the derivative of compressibility with respect to temperature at constant pressure.
33
PYQ 2023
medium
physicsID: iit-jam-
Which of the following schematic plots correctly represent(s) a first order phase transition occurring at temperature 𝑇=𝑇c? Here 𝑔, 𝑠, 𝑣 are specific Gibbs free energy, entropy and volume, respectively
1
2
3
4
Official Solution
Correct Option:
(2)
To solve this problem, we need to understand the characteristics of a first-order phase transition. A first-order phase transition involves a discontinuous change in a property of the substance, such as volume or entropy, at a specific temperature. At the transition temperature , these properties experience an abrupt change, while the Gibbs free energy remains continuous across the transition.
Key Characteristics of First-Order Phase Transitions:
Gibbs Free Energy ( ): It remains continuous at the transition temperature but has a kink, indicating a discontinuity in its derivative.
Entropy ( ) and Volume ( ): Both properties show a discontinuous jump, as they are first derivatives of Gibbs free energy with respect to temperature and pressure, respectively.
Examining the Options:
Option 1: Not shown as correct — if it shows a continuous line without sudden changes, it's incorrect for both and .
Option 2: Represents entropy ( ) with a discontinuity at . This is characteristic of a first-order phase transition.
Option 3: Represents volume ( ) with a discontinuity at . Again, this is characteristic of a first-order phase transition.
Option 4: Not shown as correct — if it represents Gibbs free energy ( ) without a kink or shows discontinuities, it's incorrect.
Based on the characteristics of a first-order phase transition, Options 2 and 3 best represent the correct schematic plots because they display the necessary discontinuities in entropy and volume, respectively, at the critical temperature .
Hence, the correct answer is Option 2 and Option 3:
- Entropy ( ) behavior
- Volume ( ) behavior
34
PYQ 2024
medium
physicsID: iit-jam-
One mole of an ideal monoatomic gas, initially at temperature 𝑇𝑜 is expanded from an initial volume 𝑉𝑜 to 2.5𝑉𝑜. Which of the following statements is(are) correct? (R is the ideal gas constant)
1
When the process is isothermal, the work done is 𝑅𝑇𝑜 ln 2
2
When the process is isothermal, the change in internal energy is zero
3
When the process is isobaric, the work done is RT0
4
When the process is isobaric, the change in internal energy is RT0
Official Solution
Correct Option:
(2)
1. Isothermal Process
In an isothermal process, the temperature remains constant, so the internal energy of an ideal gas does not change:
ΔU = 0
The work done in an isothermal expansion is given by:
W = nRT ln(Vf / Vi)
where Vf = 2.5V0 and Vi = V0. Substituting:
W = RT0 ln(2.5)
The expression RT0 ln(2) is incorrect; hence, statement (A) is not valid.
2. Change in Internal Energy in an Isothermal Process
Since ΔU = 0, statement (B) is correct.
3. Isobaric Process
In an isobaric process, the pressure remains constant. The work done is:
W = PΔV
Using the ideal gas law (PV = nRT), the work done can be expressed as:
W = nRΔT
For a monoatomic ideal gas, ΔT = Tf − T0, but the given expression (3/2)RT0 is incorrect.
Thus, statement (C) is invalid.
4. Change in Internal Energy in an Isobaric Process
The change in internal energy is:
ΔU = (3/2) nRΔT
The expression (9/2)RT0 does not align with this derivation.
Thus, statement (D) is incorrect.
Conclusion
The correct statements are (B) and (C).
35
PYQ 2024
medium
physicsID: iit-jam-
The P-V diagram of an engine is shown in the figure below. The temperatures at points 1, 2, 3 and 4 are T1, T2, T3 and T4, respectively. 1→2 and 3→4 are adiabatic processes, and 2→3 and 4→1 are isochoric processes Identify the correct statement(s). [γ is the ratio of specific heats Cp (at constant P) and Cv (at constant V)]
1
𝑇1𝑇3 = 𝑇2𝑇4
2
The efficiency of the engine is
3
The change in entropy for the entire cycle is zero
4
𝑇1𝑇2 = 𝑇3𝑇4
Official Solution
Correct Option:
(1)
1. Adiabatic Process Relationships: - For (adiabatic process): - For (adiabatic process): - For adiabatic processes, pressure and temperature are related as: Hence, statement (A) is correct.
2. Efficiency of the Engine: - The efficiency of a heat engine operating on a cycle is given by: , where and are the heat rejected and absorbed, respectively. - For this cycle, the efficiency can also be expressed using the pressure ratio for adiabatic processes as: Hence, statement (B) is correct.
3. Entropy Change: - For the entire cyclic process, the system returns to its initial state. As entropy is a state function, the total entropy change over a complete cycle is zero: . Hence, statement (C) is correct.
4. Temperature Relationship: - The relationship is incorrect as it does not align with the conditions for adiabatic and isochoric processes. Hence, statement (D) is incorrect.
36
PYQ 2025
medium
physicsID: iit-jam-
One kg of water at 27 C is brought in contact with a heat reservoir kept at 37 C. Upon reaching thermal equilibrium, this mass of water is brought in contact with another heat reservoir kept at 47 C. The final temperature of water is 47 C. The change in entropy of the whole system in this entire process is \rule{1cm{0.15mm} cal/K. (up to two decimal places)
Take specific heat at constant pressure of water as 1 cal/(g K)}
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept:
The problem asks for the total entropy change of the universe (the "whole system," which includes the water and the two reservoirs) for a two-step heating process. The total entropy change is the sum of the entropy changes of each component. Since the heating is irreversible (due to finite temperature differences), the total entropy change of the universe must be positive. Step 2: Key Formula or Approach:
- Change in entropy for an object with changing temperature: .
- Change in entropy for a heat reservoir at constant temperature: , where Q is the heat absorbed by the reservoir. If the reservoir gives up heat, Q is negative.
- Total entropy change: . Step 3: Detailed Explanation:
First, convert all temperatures to Kelvin:
K
K
K
Mass of water kg = 1000 g. Specific heat cal/(g·K). Process 1: Water heated from 300 K to 310 K by Reservoir 1.
- Heat absorbed by water: cal.
- Entropy change of water: cal/K.
- Entropy change of Reservoir 1 (loses heat): cal/K. Process 2: Water heated from 310 K to 320 K by Reservoir 2.
- Heat absorbed by water: cal.
- Entropy change of water: cal/K.
- Entropy change of Reservoir 2 (loses heat): cal/K. Total Entropy Change of the Whole System:
The total change is the sum of the entropy changes of all parts:
\textit{Alternative Calculation:}
Total entropy change of water from 300 K to 320 K:
cal/K.
Total entropy change of reservoirs:
cal/K.
Total entropy change of the universe:
cal/K. Step 4: Final Answer:
The change in entropy of the whole system is 1.03 cal/K.
37
PYQ 2025
medium
physicsID: iit-jam-
An ideal mono-atomic gas is expanded adiabatically from A to B. It is then compressed in an isobaric process from B to C. Finally, the pressure is increased in an isochoric process from C to A. The cyclic process is shown in the figure below. For this system, which of the following is(are) correct?
1
Work done along the path AB is
2
Total work done during the entire process is
3
Total heat absorbed during the entire process is
4
Total change in internal energy during the entire process is
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept:
This problem involves calculating work done, heat transfer, and change in internal energy for a cyclic process involving an ideal mono-atomic gas. The process consists of three parts: adiabatic expansion (A to B), isobaric compression (B to C), and isochoric heating (C to A). Step 2: Key Formula or Approach:
1. Work Done (W): . - Adiabatic process: . For a mono-atomic gas, . - Isobaric process: . - Isochoric process: since .
2. First Law of Thermodynamics: .
3. Internal Energy (U): For an ideal gas, depends only on temperature. . For a mono-atomic gas, . So .
4. Cyclic Process: For any complete cycle, the total change in internal energy is zero ( ). This implies . Step 3: Detailed Explanation: Let's analyze each path:
- Path A B (Adiabatic Expansion): Work done .
- Path B C (Isobaric Compression): Work done . Note that since , this work is negative (work done on the gas).
- Path C A (Isochoric Heating): Work done since volume is constant ( ). Now let's evaluate the options: (A) Work done along the path AB is .
From our calculation above, . This matches the calculation. (B) Total work done during the entire process is .
Total work done .
This expression exactly matches option (B). Thus, statement (B) is correct. (C) Total heat absorbed during the entire process is .
For a cyclic process, .
So, .
Let's expand this: .
The expression in option (C) is .
These two expressions are not equal. Statement (C) is incorrect. (D) Total change in internal energy during the entire process is .
For any cyclic process, the system returns to its initial state. Since internal energy U is a state function, the total change in internal energy over a complete cycle is always zero. .
The expression in (D) is non-zero. Statement (D) is incorrect. Step 4: Final Answer:
Based on the analysis of each step of the cycle, the expression for the total work done in option (B) is correct. All other options are incorrect.
38
PYQ 2025
medium
physicsID: iit-jam-
Consider Maxwell's relation . The equation of state of a thermodynamic system is given as , where A and B are constants of appropriate dimensions. Then of the system varies with temperature as ( is the heat capacity at constant volume)
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept
The problem asks for the temperature dependence of the variation of heat capacity with volume at constant temperature, i.e., . We can find this relationship by starting with the definition of in terms of entropy and using the given Maxwell relation and equation of state. Step 2: Key Formula or Approach
1. Definition of heat capacity at constant volume: .
2. The goal is to find . We start by differentiating the definition of with respect to V at constant T.
3. We will use the fact that entropy is a state function, which means its mixed second partial derivatives are equal: .
4. The given Maxwell relation is . Step 3: Detailed Explanation
Let's find an expression for .
Start with the definition of :
Differentiate both sides with respect to at constant :
Since is constant in this differentiation, we can move it and the operator around:
Since S is a state function, we can switch the order of differentiation:
Substituting this back, we get:
Now, use the given Maxwell's relation :
This is a standard thermodynamic identity that connects the volume dependence of to the equation of state. Now, we use the given equation of state: .
First, calculate the first partial derivative of with respect to at constant :
Next, calculate the second partial derivative:
Step 4: Final Answer
Substitute this result back into our derived expression for :
From this expression, we can see the temperature dependence:
This corresponds to option (A).
About Thermodynamics - IIT-JAM-PH
Thermodynamics is a vital chapter for IIT-JAM-PH aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Thermodynamics PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Thermodynamics carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
