An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below :-

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. Gathering the true statements (A, D, and E) points directly to Option (2). A linear P-T line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. Class 11 Physics: Thermodynamics

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat add

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning he

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below :-

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. Gathering the true statements (A, D, and E) points directly to Option (2). A linear P-T line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. Class 11 Physics: Thermodynamics

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat add

From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: The question states that pressure increases linearly with temperature, which means their ratio is constant : Since (P/T) = (nR/V), the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). * Statement A: True. In an isochoric process, dV = 0 W = P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning he

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with…

Year	2026	2025	2024	Total
Questions	11	19	6	36

Q49 jee_main_2026_21_jan_morning Internal Energy

10 mole of oxygen is heated at constant volume from

30°C

to

40°C

. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure,

Cₚ = 7 cal./mol °C

and

R = 2 cal./mol °C

.)

Numerical Answer. Answer: 500 to 500

Solution

Related Formula

Δ U = n Cv Δ T

$Cv = Cₚ - R$ $C_v = C_p - R$

Core Logic

Given values: $n = 10 moles$ $n = 10\text{ moles}$ $Δ T = 40°C - 30°C = 10°C$ $Cₚ = 7 cal/mol\cdot°C$ $R = 2 cal/mol\cdot°C$

First, find $Cv$ $C_v$ using Mayer's relation:

Cv = Cₚ - R = 7 - 2 = 5 cal/mol\cdot°C

Step 1: Calculate Internal Energy Change

Δ U = n Cv Δ T

Δ U = 10 \times (7 - 2) \times (40 - 30)

Δ U = 10 \times 5 \times 10 = 500 cal

Pattern Recognition

Change in internal energy of an ideal gas is ALWAYS $Δ U = n Cv Δ T$ $\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use $Cv = Cₚ - R$ $C_v = C_p - R$ when $Cₚ$ $C_p$ is given.

Chapter Mix

Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

Q50 jee_main_2026_21_jan_evening Isobaric Process

A diatomic gas (

γ = 1.4

) does

100 J

of work when it is expanded isobarically. Then the heat given to the gas ________ J.

Numerical Answer. Answer: 350 to 350

Solution

Related Formula

W = PΔ V = nRΔ T

Q = nCₚΔ T

Cₚ = ((f)/(2) + 1)R

Core Logic

For an isobaric (constant pressure) process, work done is given by:

W = nRΔ T = 100 J

For a diatomic gas, the degrees of freedom $f = 5$ $f = 5$. Thus, the molar heat capacity at constant pressure is:

Cₚ = ((5)/(2) + 1)R = (7)/(2)R

Step 1: Calculating Heat Transfer

The heat supplied to the gas is:

Q = nCₚΔ T = n ((7)/(2)R)Δ T = (7)/(2) (nRΔ T)

Step 2: Final Conclusion

Substitute the value of work done:

Q = (7)/(2) \times (100) = 350 J

Pattern Recognition

In an isobaric process, the ratio of Work : Internal Energy Change : Heat Added ( $W : Δ U : Q$ $W : \Delta U : Q$) is always $2 : f : (f+2)$ $2 : f : (f+2)$. For diatomic gases, $f=5$ $f=5$, so the ratio is $2:5:7$ $2:5:7$. Thus $Q = (7)/(2)W$ $Q = \frac{7}{2}W$.

Chapter Mix

Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

Q41 jee_main_2026_22_january_morning Adiabatic Process and Atomicity

The volume of an ideal gas increases 8 times and temperature becomes

(1/4)th

of initial temperature during a reversible change. If there is no exchange of heat in this process (

Δ Q = 0

) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

A. $CO₂$ $\mathrm{CO}_{2}$
B. $O₂$ $\mathrm{O}_{2}$
C. $NH₃$ $\mathrm{NH}_{3}$
D. He

Solution

Related Formula

TVγ-1 = constant

Core Logic

Solution adiabatic process diagram for Q41 - JEE Main 2026 Morning

Using adiabatic relation $TVγ-1 = constant$ $TV^{\gamma-1} = \text{constant}$:

T₁ V₁γ-1 = T₂ V₂γ-1

T(V)γ-1 = ((T)/(4))(8V)γ-1

4 = 8γ-1 2² = 23(γ-1) γ = (5)/(3)

Since $γ = 5/3$ $\gamma = 5/3$, the gas is monoatomic (Helium / He).

Pattern Recognition

Sees: Adiabatic expansion with volume and temperature changes. Shortcut: Apply $TVγ-1$ $TV^{\gamma-1}$ relation to determine adiabatic exponent $γ$ $\gamma$. Check: Matches option (4). ✓

Chapter Mix

Class 11 Physics: Thermodynamics

Q50 jee_main_2026_22_january_evening Ideal Gas Law and Gas Compression

An insulated cylinder of volume

60 ~cm³

is filled with a gas at

27^

and 2 atmospheric pressure. Then the gas is compressed making the final volume as

20 ~cm³

while allowing the temperature to rise to

77^

. The final pressure is ____ atmospheric pressure.

Numerical Answer. Answer: 7 to 7

Solution

Related Formula

(P₁ V₁)/(T₁) = (P₂ V₂)/(T₂)

T(K) = T(^ ) + 273

Core Logic

Converting initial and final temperatures to Kelvin:

T₁ = 27^ + 273 = 300 ~K

T₂ = 77^ + 273 = 350 ~K

Applying ideal gas law relation:

(2 \times 60)/(300) = (P₂ \times 20)/(350)

(120)/(300) = (20 P₂)/(350)

(2)/(5) = (20 P₂)/(350) 20 P₂ = 140 P₂ = 7 ~atm

Step 1: Final Conclusion

The final pressure is $7$ $7$ atmospheric pressure.

Pattern Recognition

Combined Gas Law: $P₂ = P₁ ((V₁)/(V₂)) ((T₂)/(T₁))$ . $P₂ = 2 \times ((60)/(20)) \times ((350)/(300)) = 2 \times 3 \times (7)/(6) = 7 atm$ .

Chapter Mix

Class 11 Physics: Thermodynamics

Q26 jee_main_2026_23_january_evening First Law of Thermodynamics

The internal energy of a monoatomic gas is

3nRT

. One mole of helium is kept in a cylinder having internal cross section area of

17 cm²

and fitted with a light movable frictionless piston. The gas is heated slowly by suppling

126 J

heat. If the temperature rises by

4°C

, then the piston will move ____ cm. (atmospheric pressure =

10⁵ Pa

)

A. $14.5$ $14.5$
B. $1.55$ $1.55$
C. $15.5$ $15.5$
D. $1.45$ $1.45$

Solution

Related Formula

Δ Q = Δ U + W

Δ U = n Cv Δ T

W = P Δ V = P (A \cdot Δ x)

Core Logic

For a monoatomic gas like Helium, $Cv = (3)/(2)R$ $C_v = \frac{3}{2}R$. Given internal energy is $3nRT$ $3nRT$, but standard change is $Δ U = n Cv Δ T$ $\Delta U = n C_v \Delta T$. Wait, the problem states internal energy $U = 3nRT$ $U = 3nRT$ (which implies $Cv = 3R$ $C_v = 3R$ for this specific state/setup, or it's a typo in the paper and they mean $U = (3)/(2)nRT$ $U = \frac{3}{2}nRT$). Let's follow the solution's lead: $Δ U = 3nRΔ T$ $\Delta U = 3nR\Delta T$ is directly given.

Step 1: Calculate Change in Internal Energy

Using the given relation:

Δ U = 3nR Δ T

Substitute $n=1$ $n=1$, $R = (25)/(3) J/mol \cdot K$ , and $Δ T = 4°C$ $\Delta T = 4^{\circ}\mathrm{C}$:

Δ U = 3 \times 1 \times (25)/(3) \times 4 = 100 J

Step 2: Apply First Law of Thermodynamics

We know $Δ Q = 126 J$ $\Delta Q = 126 \, \mathrm{J}$.

W = Δ Q - Δ U

W = 126 - 100 = 26 J

Step 3: Calculate Piston Displacement

Work done is isobaric since the piston is freely moving:

W = P Δ V = P \cdot A \cdot Δ x

26 = 10⁵ \times 17 \times 10⁻⁴ \times Δ x

26 = 170 \cdot Δ x

Δ x = (26)/(170) m = (26)/(170) \times 100 cm = 15.3 cm

Closest given option is $15.5 cm$ $15.5 \, \mathrm{cm}$.

Pattern Recognition

When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified $U$ $U$ definitions given in the prompt, replacing standard $Cv$ $C_v$ logic.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate All Statements

Step 2: Final Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Internal Energy Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Heat Transfer

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Change in Internal Energy

Step 2: Apply First Law of Thermodynamics

Step 3: Calculate Piston Displacement

Pattern Recognition

Chapter Mix

More Thermodynamics Questions — jee_main_2025_24_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back