A piston of mass $M$ $M$ is hung from a massless spring whose restoring force law goes as $F = -kx^3$ $F = -kx^3$, where $k$ $k$ is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with ' $n$ $n$' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature $T$ $T$) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $L_0$ $L_0$ to $L_1$ $L_1$, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
A schematic of a piston of mass M connected to a spring inside a vertical chamber, separating gas at the bottom from vacuum/atmosphere at the top.

A. $3nRTln left(fracL_1L_0right) + 2Mg(L_1 - L_0) + frack3 (L_1^3 - L_0^3)$
B. $nRTln left(fracL_1^2L_0^2right) + fracMg2 (L_1 - L_0) + frack4 (L_1^4 - L_0^4)$
C. $nRTln left(fracL_1L_0right) + Mg(L_1 - L_0) + frack4 (L_1^4 - L_0^4)$
D. $nRTln left(fracL_1L_0right) + Mg(L_1 - L_0) + frac3k4 (L_1^4 - L_0^4)$

Solution & Explanation

### Related Formula First Law of Thermodynamics:

Delta Q = Delta U + W_textby gas

Work done by an ideal gas during isothermal expansion:

W_textgas = nRTlnleft(fracV_1V_0right) = nRTlnleft(fracL_1L_0 ight)

Conservation of Energy (Work-Energy Theorem): Total energy delivered by the heating filament (

W_textfilament

) must equal the total work needed to lift the piston against gravity and compress the non-linear spring. ### Core Logic Since the process is isothermal, the change in internal energy of the ideal gas is zero (

Delta U = 0

). Hence:

Q = W_textgas

By the Work-Energy Theorem for the piston:

W_textgas + W_textfilament = Delta U_textgravity + Delta U_textspring

Let's evaluate each term: - Increase in gravitational potential energy:

Delta U_textgravity = Mg(L_1 - L_0)

- Increase in spring potential energy:

U_textspring = -int_L_0^L_1 F_textrestoring dx = int_L_0^L_1 kx^3 dx = frack4(L_1^4 - L_0^4)

### Step 1: Finding Total Energy Delivered Isolating

W_textfilament

(the net external energy delivered to the gas system):

W_textfilament = W_textgas + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)

Since

W_textgas = nRTlnleft(fracL_1L_0 ight)

W_textfilament = nRTlnleft(fracL_1L_0 ight) + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)

### Pattern Recognition Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity (

Mgh

), and the potential energy of the spring (integrated from

kx^3

). Keeping this total energy ledger in mind prevents tedious mathematical tangents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics: First Law Class 11 Physics: Work, Energy and Power: Variable Force Integration

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 4

Q12 2025 Adiabatic Compression

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of

800~mathrmcm^3

and temperature

27^circmathrmC

. The change in temperature when the gas is adiabatically compressed to

200~mathrmcm^3

is: (Take

gamma = 1.5

)

A. $327mathrm~K$ $327\mathrm{~K}$
B. $600mathrm~K$ $600\mathrm{~K}$
C. $522mathrm~K$ $522\mathrm{~K}$
D. $300mathrm~K$ $300\mathrm{~K}$

Solution

### Related Formula For an adiabatic process:

T V^gamma - 1 = textconstant

where,

T

= absolute temperature in Kelvin,

V

= volume of the gas,

gamma

= adiabatic exponent. ### Core Logic Given values: - Initial volume,

V_1 = 800mathrm~cm^3

- Final volume,

V_2 = 200mathrm~cm^3

- Initial temperature,

T_1 = 27^circmathrmC = 27 + 273 = 300mathrm~K

- Adiabatic exponent,

gamma = 1.5 implies gamma - 1 = 0.5

### Step 1: Calculating Final Temperature Apply the adiabatic relation:

T_1 V_1^gamma - 1 = T_2 V_2^gamma - 1

T_2 = T_1 left(fracV_1V_2right)^gamma - 1

Substitute the values:

T_2 = 300 left(frac800200right)^0.5 = 300 times (4)^0.5

T_2 = 300 times 2 = 600mathrm~K

### Step 2: Calculating Change in Temperature Now compute the change in temperature (

Delta T

Delta T = T_2 - T_1 = 600mathrm~K - 300mathrm~K = 300mathrm~K

### Pattern Recognition Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing

600mathrm~K

(the final temperature) instead of the difference

300mathrm~K

! Stay sharp. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Q17 2025 Thermodynamic Processes

Match List-I with List-II:

List-I	List-II
(A) Isobaric	(I) $Delta Q=Delta W$ $\Delta Q=\Delta W$
(B) Isochoric	(II) $Delta Q=Delta U$ $\Delta Q=\Delta U$
(C) Adiabatic	(III) $Delta Q=textzero$ $\Delta Q=\text{zero}$
(D) Isothermal	(IV) $Delta Q=Delta U+PDelta V$ $\Delta Q=\Delta U+P\Delta V$

Choose the correct answer from the options given below:

A. \text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}
B. \text{(A)-(IV), (B)-(I), (C)-(III), (D)-(II)}
C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
D. \text{(A)-(II), (B)-(IV), (C)-(III), (D)-(I)}

Solution

### Related Formula First Law of Thermodynamics:

Delta Q = Delta U + Delta W

### Core Logic - **Isobaric**: Pressure is constant, work done

Delta W = PDelta V

. Thus,

Delta Q = Delta U + PDelta V

(Matches IV). - **Isochoric**: Volume is constant,

Delta V = 0 implies Delta W = 0

. Thus,

Delta Q = Delta U

(Matches II). - **Adiabatic**: No heat transfer,

Delta Q = 0

(Matches III). - **Isothermal**: Temperature is constant, internal energy change

Delta U = 0

for an ideal gas. Thus,

Delta Q = Delta W

(Matches I). ### Step 1: Match Compilation Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). ### Pattern Recognition Adiabatic definition is always heat-isolated (

Q=0

), isochoric implies rigid boundaries (

W=0

). Identifying these two immediately isolates the correct match permutation in seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Q17 2025 Thermodynamic Processes

Match List-I with List-II.

List-I	List-II
(A) Isothermal	(I) $Delta W$ $\Delta W$ (work done) $=0$ $=0$
(B) Adiabatic	(II) $Delta Q$ $\Delta Q$ (supplied heat) $=0$ $=0$
(C) Isobaric	(III) $Delta U$ $\Delta U$ (change in internal energy) $ne0$ $\ne0$
(D) Isochoric	(IV) $Delta U=0$ $\Delta U=0$

Choose the correct answer from the options given below: [cite: 151, 152]

A. (A)-(III), (B)-(II), (C)-(I), (D)-(IV) [cite: 153]
B. (A)-(IV), (B)-(I), (C)-(III), (D)-(II) [cite: 154]
C. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 155]
D. (A)-(II), (B)-(IV), (C)-(I), (D)-(III) [cite: 156]

Solution

### Core Logic Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (

Delta T = 0

) implies that the internal energy change of an ideal gas is zero, so

Delta U = 0 implies text(IV)

[cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning

Delta Q = 0 implies text(II)

[cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously,

Delta U neq 0 implies text(III)

[cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (

Delta V = 0

) ensures work done

Delta W = PDelta V = 0 implies text(I)

[cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. ### Pattern Recognition Matching 'Isochoric' with zero work done (

Delta W=0

) or 'Adiabatic' with zero heat exchange (

Delta Q=0

) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Q6 2025 Adiabatic Processes

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statement, choose the correct answer from the options given below :

A. Both $(A)$ $(A)$ and $(R)$ $(R)$ are true and $(R)$ $(R)$ is the correct explanation of $(A)$ $(A)$
B. $(A)$ $(A)$ is true but $(R)$ $(R)$ is false
C. $(A)$ $(A)$ is false but $(R)$ $(R)$ is true
D. Both $(A)$ $(A)$ and $(R)$ $(R)$ are true but $(R)$ $(R)$ is NOT the correct explanation of $(A)$ $(A)$

Solution

### Related Formula

T_1 V_1^gamma-1 = T_2 V_2^gamma-1

### Core Logic Assertion (A) review: An insulated container means the process is adiabatic (

Q=0

). When the gas is shrunk (compressed) to half its volume (

V_2 = V_1 / 2

T_2 = T_1 left(fracV_1V_2 ight)^gamma-1 = T_1 (2)^gamma-1

Since

gamma > 1

T_2 > T_1

. Therefore, temperature increases during adiabatic compression, which makes Assertion (A) false. Reason (R) review: Free expansion occurs when a gas expands into a vacuum inside an insulated container. No work is done (

W=0

) and no heat is exchanged (

Q=0

), hence it is adiabatic. It cannot spontaneously reverse, so it is irreversible. Thus, Reason (R) is true. ### Pattern Recognition Adiabatic compression always raises temperature due to work being done on the gas, while free expansion keeps the temperature of an ideal gas constant (

dI=0

W=0, Q=0

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Q13 2025 Cyclic Processes

The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure

Cyclic semi-circular P-V process indicator diagram Q13 — The image shows a P-V cycle consisting of a horizontal line from C to A and a semicircular loop from A back to C via B.

) is (in SI unit)

A. $10pi$ $10\pi$
B. $5pi$ $5\pi$
C. zero
D. $40pi$ $40\pi$

Solution

### Related Formula From the first law of thermodynamics for a complete cycle:

Delta U = 0 implies Q = W = textArea of the loop

### Core Logic The graph shows a semicircle in a

Ptext-V

indicator diagram.

P-V cycle geometry calculation graph Q13 — The image shows a P-V cycle consisting of a horizontal line from C to A and a semicircular loop from A back to C via B.

- Pressure dimension diameter:

Delta P = 400 - 200 = 200\ mathrmkPa = 200 times 10^3\ mathrmPa

- Volume dimension diameter:

Delta V = 400 - 200 = 200\ mathrmcc = 200 times 10^-6\ mathrmm^3

Radius along Pressure axis:

R_P = 100 times 10^3\ mathrmPa

Radius along Volume axis:

R_V = 100 times 10^-6\ mathrmm^3

Area of the closed semicircular path:

W = frac12 pi R_P R_V

W = frac12 times pi times (100 times 10^3) times (100 times 10^-6)

W = frac10pi2 = 5pi\ mathrmJ

Since

Q = W

, the heat exchanged has a magnitude of

5pi\ mathrmJ

. ### Pattern Recognition For a cycle on an indicator chart with mismatched scales, use the elliptic area template

pi a b

(or

frac12pi a b

for a half-ellipse/semicircle). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development

PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Latest from the Blog

View all →

Loading articles...

Review & Tools

Mistakes To Revise

No conceptual mistakes detected.

🔥 Practice Session Live

Showing All Questions

Exam Questions

Test your concepts live. Choose options or enter numerical values, then verify your answer to reveal the double-box solution matrix.

Solution & Explanation

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 4

Solution

Solution

Solution

Solution

Solution

More Thermodynamics Questions — jee_main_2025_03_april_morning

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam