Match List-I with List-II: List-I List-II (A) Isobaric (I) Delta Q=Delta W (B) Isochoric (II) Delta Q=Delta U (C) Adiabatic (III) Delta Q=zero (D) Isothermal (IV) Delta Q=Delta U+PDelta V Choose the correct answer from the options given below:

First Law of Thermodynamics: - **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 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). Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). 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. Class 11 Physics: Thermodynamics

Match List-I with List-II: List-I List-II (A) Isobaric (I) Delta Q=Delta W (B) Isochoric (II) Delta Q=Delta U (C) Adiabatic (III) Delta Q=zero (D) Isothermal (IV) Delta Q=Delta U+PDelta V Choose the c

First Law of Thermodynamics: - **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 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). Combining everything gives: (A)-(IV), (B)-(II), (C)-(III),

Match List-I with List-II: List-I List-II (A) Isobaric (I) Delta Q=Delta W (B) Isochoric (II) Delta Q=Delta U (C) Adiabatic (III) Delta Q=zero (D) Isothermal (IV) Delta Q=Delta U+PDelta V Choose the correct answer from the options given below:

First Law of Thermodynamics: - **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 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). Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). 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. Class 11 Physics: Thermodynamics

Match List-I with List-II: List-I List-II (A) Isobaric (I) Delta Q=Delta W (B) Isochoric (II) Delta Q=Delta U (C) Adiabatic (III) Delta Q=zero (D) Isothermal (IV) Delta Q=Delta U+PDelta V Choose the c

First Law of Thermodynamics: - **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 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). Combining everything gives: (A)-(IV), (B)-(II), (C)-(III),

Match List-I with List-II: List-I List-II (A) Isobaric (I) Delta Q=Delta W (B) Isochoric (II) Delta Q=Delta U (C)…

Year	2026	2025	2024	Total
Questions	11	19	6	36

Year

2026

2025

2024

Total

Questions

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

List-I

List-II

(A) Isobaric

(I)

Δ Q=Δ W

(B) Isochoric

(II)

Δ Q=Δ U

(III)

Δ Q=zero

(D) Isothermal

(IV)

Δ Q=Δ U+PΔ V

Solution & Explanation

Related Formula

First Law of Thermodynamics:

Δ Q = Δ U + Δ W

Core Logic

Isobaric: Pressure is constant, work done $Δ W = PΔ V$ $\Delta W = P\Delta V$. Thus, $Δ Q = Δ U + PΔ V$ $\Delta Q = \Delta U + P\Delta V$ (Matches IV).
Isochoric: Volume is constant, $Δ V = 0 Δ W = 0$ $\Delta V = 0 \implies \Delta W = 0$. Thus, $Δ Q = Δ U$ $\Delta Q = \Delta U$ (Matches II).
Adiabatic: No heat transfer, $Δ Q = 0$ $\Delta Q = 0$ (Matches III).
Isothermal: Temperature is constant, internal energy change $Δ U = 0$ $\Delta U = 0$ for an ideal gas. Thus, $Δ Q = Δ W$ $\Delta Q = \Delta W$ (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 ( $Q=0$ $Q=0$), isochoric implies rigid boundaries ( $W=0$ $W=0$). Identifying these two immediately isolates the correct match permutation in seconds.

Chapter Mix

Class 11 Physics: Thermodynamics

More Thermodynamics Previous-Year Questions — Page 8

Q jee_main_2024_31_jan_morning Isobaric Process

The given figure represents two isobaric processes for the same mass of an ideal gas, then

Isobaric Process diagram for Q38 - JEE Main 2024 Morning — A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.

A. $P₂\geq P₁$ $\mathrm{P}_2\geq \mathrm{P}_1$
B. $P₂ > P₁$ $\mathrm{P}_2 > \mathrm{P}_1$
C. $P₁ = P₂$ $\mathrm{P_1 = P_2}$
D. $P₁ > P₂$ $\mathrm{P}_1 > \mathrm{P}_2$

Solution

Related Formula

$PV = nRT$ $PV = nRT$

Core Logic

From the Ideal Gas Law:

V = ((nR)/(P)) T

In a $V-T$ $V-T$ graph, the equation of the line represents $y = mx$ $y = mx$, where the slope $m$ $m$ is:

Slope = (nR)/(P)

Slope \propto (1)/(P)

Thus, a higher slope corresponds to a lower pressure.

Step 2: Compare Slopes

From the given figure, the slope of line 2 is greater than the slope of line 1:

(Slope)₂ > (Slope)₁

Therefore, inversely: $P₂ < P₁$ $P_2 < P_1$ or $P₁ > P₂$ $P_1 > P_2$.

Pattern Recognition

In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Match Compilation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 2: Compare Slopes

Pattern Recognition

Chapter Mix

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Match List-I with List-II: List-IList-II(A) Isobaric(I) Δ Q=Δ W$\Delta Q=\Delta W$(B) Isochoric(II) Δ Q=Δ U$\Delta Q=\Delta U$(C) Adiabatic(III) Δ Q=zero$\Delta Q=\text{zero}$(D) Isothermal(IV) Δ Q=Δ U+PΔ V$\Delta Q=\Delta U+P\Delta V$ Choose the correct answer from the options given below:

Solution & Explanation

Related Formula

Core Logic

Step 1: Match Compilation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 2: Compare Slopes

Pattern Recognition

Chapter Mix

More Thermodynamics Questions — jee_main_2025_04_april_evening

Select Your Target Exam

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