A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the…

Year	2026	2025	2024	Total
Questions	11	19	6	36

Q39 jee_main_2026_23_january_evening Isothermal and Adiabatic Processes

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature

27°

C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from

27°

C doing the same amount of work W, then its final temperature will be (close to) ____

°

C.

A. $-189$ $-189$
B. $-56$ $-56$
C. $-30$ $-30$
D. $-117$ $-117$

Solution

Related Formula

Wisothermal = nRT ln ((V₂)/(V₁))

Wadiabatic = (nR(T₁ - T₂))/(γ - 1)

Core Logic

For a diatomic gas, $γ = 1.4$ $\gamma = 1.4$. Given initial temperature $T = 27°C = 300 K$ . Since $Wisothermal = Wadiabatic$ , we equate the two expressions.

Step 1: Compute Work Done in Isothermal Expansion

Wisothermal = (1) \cdot R \cdot 300 \cdot ln(2)

Wisothermal = 300R(0.693) --- (1)

Step 2: Equate and Solve for Final Temperature

Wadiabatic = (nR(T₁ - T₂))/(γ - 1)

Equating (1) to adiabatic work:

((1)R(300 - Tf))/(1.4 - 1) = 300R(0.693)

(300 - Tf)/(0.4) = 300(0.693)

300 - Tf = 0.4 \times 300 \times 0.693

300 - Tf = 120 \times 0.693 = 83.16

Tf = 300 - 83.16 = 216.84 K

Step 3: Convert to Celsius

Tf (in °C) = 216.84 - 273.15 = -56.31°C

Closest option is $-56$ $-56$.

Pattern Recognition

Equating the work of two different processes links the log expansion to the temperature drop. Recognize diatomic gas instantly means $γ = 1.4$ $\gamma = 1.4$, providing the $0.4$ $0.4$ divisor.

Chapter Mix

Class 11 Physics: Thermodynamics

Q29 jee_main_2026_24_january_morning First Law of Thermodynamics

Density of water at

4°

C and

20°

C are

1000 kg/m³

and

998 kg/m³

respectively. The increase in internal energy of 4 kg water when it is heated from

4°

C to

20°

C is ____J. (Specific heat capacity of water =

4.2 J/g°C

and 1 atmospheric pressure =

10⁵

Pa)

A. $315826.2$ $315826.2$
B. $234699.2$ $234699.2$
C. $258700.8$ $258700.8$
D. $268799.2$ $268799.2$

Solution

Related Formula

Δ Q = mcΔ T

$W = P Δ V$ $W = P \Delta V$

Δ U = Δ Q - W

Core Logic

First, calculate the heat supplied $Δ Q$ $\Delta Q$:

Q = mSΔ T = 4 \times 4200 \times 16 = 268800 J

Next, calculate the work done $W$ $W$ against atmospheric pressure due to volume change: $W = P Δ V$ $W = P \Delta V$

Δ V = ( (m)/(ρf) - (m)/(ρᵢ) ) = 4 [ (1)/(998) - (1)/(1000) ]

Given $P = 10⁵ Pa$ $P = 10^5 \text{ Pa}$:

W = 10⁵ \times 4 \times [ (1000 - 998)/(998 \times 1000) ] = (8 \times 10⁵)/(998 \times 10³) \approx 0.8 J

Step 1: Internal Energy Calculation

Using the First Law of Thermodynamics:

Δ U = Q - W

Δ U = 268800 - 0.8 = 268799.2 J

Pattern Recognition

When heating solids or liquids, the expansion is so small that $Δ V$ $\Delta V$ is negligible, meaning $W \approx 0$ $W \approx 0$, making $Δ U \approx Δ Q$ $\Delta U \approx \Delta Q$. Subtracting the exact $0.8J$ $0.8\text{J}$ provides the highly precise answer required here.

Chapter Mix

Class 11 Physics: Thermodynamics Class 11 Physics: Thermal Properties of Matter

Q43 jee_main_2026_24_january_evening Isochoric Process and Heat Transfer

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from

P₁

to

P₂

is

α

Joule (

P₁ = 21.7Pa

and

P₂ = 30Pa

,

Cv = 21J / K.mol

,

R = 8.3J / mol.K

). The value of

α

is ____.

Isochoric Process and Heat Transfer diagram for Q43 - JEE Main 2026 Evening — A Pressure-Volume graph displaying an isochoric rise from P1 to P2 at constant V=1.

A. $24$ $24$
B. $15$ $15$
C. $21$ $21$
D. $28$ $28$

Solution

Related Formula

Δ Q = n Cv Δ T

From Ideal Gas Law for Isochoric Process:

n R Δ T = Δ P \cdot V

Core Logic

From the P-V graph, the process from $P₁$ $P_1$ to $P₂$ $P_2$ takes place at a constant volume $V = 1 m³$ $V = 1 \text{ m}^3$.

Step 1: Calculate Heat Involved

Δ Q = n Cv Δ T

Multiply and divide by $R$ $R$:

Δ Q = (Cv)/(R) \cdot (nRΔ T) = (Cv)/(R) \cdot (P₂ - P₁)V

Step 2: Substitution

Given $Cv = 21 J/K\cdot mol$ $C_v = 21 \text{ J/K\cdot mol}$, $R = 8.3 J/mol\cdot K$ $R = 8.3 \text{ J/mol\cdot K}$, $P₂ = 30$ $P_2 = 30$, $P₁ = 21.7$ $P_1 = 21.7$, and $V = 1$ $V = 1$:

Δ Q = (21)/(8.3) \times (30 - 21.7) \times 1

Δ Q = (21)/(8.3) \times 8.3 = 21 J

Pattern Recognition

For constant volume processes, $Δ Q$ $\Delta Q$ strictly scales with $Δ P$ $\Delta P$. Re-writing $n Δ T$ $n \Delta T$ as $V Δ P / R$ $V \Delta P / R$ circumvents calculating absolute temperatures entirely.

Chapter Mix

Class 11 Physics: Thermodynamics

Q46 jee_main_2026_24_january_evening Specific Heat of Gases

When 300 J of heat given to an ideal gas with

Cₚ=(7)/(2)R

its temperature raises from 20 °C to 50 °C keeping its volume constant. The mass of the gas is (approximately) ____ g. (R = 8.314 J/mol.K).

Numerical Answer. Answer: 481 to 481

Solution

Related Formula

$Cv = CP - R$ $C_v = C_P - R$

Δ Q = n CV Δ T

Core Logic

Cv = (7)/(2)R - R = (5)/(2)R

For an isochoric process (constant volume), the heat given is strictly equal to internal energy change:

300 = n \times (5)/(2) \times 8.314 \times (50 - 20)

Step 1: Find moles (n)

300 = n \times (5)/(2) \times 8.314 \times 30

300 = n \times 75 \times 8.314

n = 0.48 moles

(m)/(M) = 0.48

Step 2: Evaluating the output

Our Ans. (Bonus) We cannot definitively find the mass ( $m$ $m$) in grams because the molar mass ( $M$ $M$) of the specific ideal gas is not given in the problem statement. (Note: NTA official answer is 481. This implies assuming $M \approx 1000$ $M \approx 1000$ or an error in problem specification).

Pattern Recognition

Always extract $Cv$ $C_v$ from $Cₚ$ $C_p$ using Mayer's relation for constant volume heat transfers. If an exam sets a numerical without giving molar mass, it relies on partial contextual hints or is marked as bonus.

Chapter Mix

Class 11 Physics: Thermodynamics

Q37 jee_main_2026_28_january_morning Work Done in Thermodynamic Processes

In the following

p - V

diagram the equation of state along the curved path is given by

(V - 2)² = 4ap

where

a

is a constant. The total work done in the closed path is

p-V diagram showing a cyclic process with a parabolic path — A p-V graph indicating a cyclic process clockwise with points A at V=1, C at V=3 and a parabolic curve below.

A. $-(1)/(a)$ $-\frac{1}{a}$
B. $+(1)/(3a)$ $+\frac{1}{3a}$
C. $(1)/(2a)$ $\frac{1}{2a}$
D. $-(1)/(3a)$ $-\frac{1}{3a}$

Solution

Related Formula

W = \int p dV

Core Logic

The total work done in a cyclic process is the enclosed area. The cycle is clockwise, but wait—the arrows in the diagram point anti-clockwise (A to B to C and back to A via horizontal top line). Anti-clockwise cycle means negative work. The enclosed area is bounded by a parabola.

Step 1: Identifying the Parabola Parameters

Equation of curve: $(V - 2)² = 4ap$ $(V - 2)^2 = 4ap$ This is an upward-opening parabola with vertex at $V = 2$ $V = 2$, $p = 0$ $p = 0$ (assuming $a>0$ $a>0$). Let's check points A ( $V=1$ $V=1$) and C ( $V=3$ $V=3$). At $V=1$ $V=1$: $(1 - 2)² = 4ap \Rightarrow 1 = 4ap₀ \Rightarrow p₀ = (1)/(4a)$ At $V=3$ $V=3$: $(3 - 2)² = 4ap \Rightarrow 1 = 4ap₀ \Rightarrow p₀ = (1)/(4a)$ So the top horizontal line is at $p₀ = (1)/(4a)$ $p_0 = \frac{1}{4a}$.

Step 2: Area Calculation

The enclosed shape is bounded by the line $p = p₀$ $p = p_0$ and the parabola $p = ((V - 2)²)/(4a)$ $p = \frac{(V - 2)^2}{4a}$. The area of a parabolic segment (bounded by a chord) is $(2)/(3) \times Area of enclosing rectangle$ . Rectangle width = $Δ V = 3 - 1 = 2$ $\Delta V = 3 - 1 = 2$. Rectangle height = $p₀ - 0 = p₀$ $p_0 - 0 = p_0$. Area = $(2)/(3) \times (p₀) \times (2) = (4p₀)/(3)$ .

Step 3: Substitute and Determine Sign

Substitute $p₀ = (1)/(4a)$ $p_0 = \frac{1}{4a}$:

|W| = (4)/(3) ((1)/(4a)) = (1)/(3a)

The cycle is counter-clockwise, so the net work done by the gas is negative.

Wgas = -(1)/(3a)

Pattern Recognition

For parabolic curves $y = kx²$ $y = kx^2$, the area inside the 'bowl' up to a chord is $2/3$ $2/3$ of the bounding rectangle. Direction of arrows dictates the sign: clockwise = positive, anti-clockwise = negative.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Work Done in Process AB

Step 2: Calculate Work Done in Process BC

Step 3: Total Work and Unit Conversion

Pattern Recognition

Chapter Mix

More Thermodynamics Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Compute Work Done in Isothermal Expansion

Step 2: Equate and Solve for Final Temperature

Step 3: Convert to Celsius

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Internal Energy Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Heat Involved

Step 2: Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find moles (n)

Step 2: Evaluating the output

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identifying the Parabola Parameters

Step 2: Area Calculation

Step 3: Substitute and Determine Sign

Pattern Recognition

Chapter Mix

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