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 , then the piston will move ____ cm. (atmospheric pressure = 105 , Pa)

For a monoatomic gas like Helium, Cv = (3/2)R. Given internal energy is 3nRT, but standard change is Delta U = n Cv Delta T. Wait, the problem states internal energy U = 3nRT (which implies Cv = 3R for this specific state/setup, or it's a typo in the paper and they mean U = (3/2)nRT). Let's follow the solution's lead: Delta U = 3nRDelta T is directly given. Using the given relation: Substitute n=1, R = (25/3) , J/mol cdot K, and Delta T = 4: We know Delta Q = 126 , J. Work done is isobaric since the piston is freely moving: Closest given option is 15.5 , cm. When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified U definitions given in the prompt, replacing standard Cv logic. Class 11 Physics: 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

For a monoatomic gas like Helium, Cv = (3/2)R. Given internal energy is 3nRT, but standard change is Delta U = n Cv Delta T. Wait, the problem states internal energy U = 3nRT (which implies Cv = 3R for this specific state/setup, or it's a typo in the paper and they mean U = (3/2)nRT). Let's follow the solution's lead: Delta U = 3nRDelta T is directly given. Using the given relation: Substitute n=1, R = (25/3) , J/mol cdot K, and Delta T = 4: We know Delta Q = 126 , J. Work done is isobaric since

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 , then the piston will move ____ cm. (atmospheric pressure = 105 , Pa)

For a monoatomic gas like Helium, Cv = (3/2)R. Given internal energy is 3nRT, but standard change is Delta U = n Cv Delta T. Wait, the problem states internal energy U = 3nRT (which implies Cv = 3R for this specific state/setup, or it's a typo in the paper and they mean U = (3/2)nRT). Let's follow the solution's lead: Delta U = 3nRDelta T is directly given. Using the given relation: Substitute n=1, R = (25/3) , J/mol cdot K, and Delta T = 4: We know Delta Q = 126 , J. Work done is isobaric since the piston is freely moving: Closest given option is 15.5 , cm. When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified U definitions given in the prompt, replacing standard Cv logic. Class 11 Physics: 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

For a monoatomic gas like Helium, Cv = (3/2)R. Given internal energy is 3nRT, but standard change is Delta U = n Cv Delta T. Wait, the problem states internal energy U = 3nRT (which implies Cv = 3R for this specific state/setup, or it's a typo in the paper and they mean U = (3/2)nRT). Let's follow the solution's lead: Delta U = 3nRDelta T is directly given. Using the given relation: Substitute n=1, R = (25/3) , J/mol cdot K, and Delta T = 4: We know Delta Q = 126 , J. Work done is isobaric since

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section…

Year	2026	2025	2024	Total
Questions	11	19	6	36

Q6 jee_main_2025_24_jan_evening 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

T₁ V₁γ-1 = T₂ V₂γ-1

Core Logic

Assertion (A) review: An insulated container means the process is adiabatic ( $Q=0$ $Q=0$). When the gas is shrunk (compressed) to half its volume ( $V₂ = V₁ / 2$ $V_2 = V_1 / 2$),

T₂ = T₁ ((V₁)/(V₂))γ-1 = T₁ (2)γ-1

Since $γ > 1$ $\gamma > 1$, $T₂ > T₁$ $T_2 > T_1$. 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 ( $W=0$ $W=0$) and no heat is exchanged ( $Q=0$ $Q=0$), 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 ( $dI=0$ $dI=0$ as $W=0, Q=0$ $W=0, Q=0$).

Chapter Mix

Class 11 Physics: Thermodynamics

Q13 jee_main_2025_24_jan_evening 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. $10π$ $10\pi$
B. $5π$ $5\pi$
C. zero
D. $40π$ $40\pi$

Solution

Related Formula

From the first law of thermodynamics for a complete cycle:

Δ U = 0 Q = W = Area of the loop

Core Logic

The graph shows a semicircle in a $P-V$ $P\text{-}V$ 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: $Δ P = 400 - 200 = 200 kPa = 200 \times 10³ Pa$
Volume dimension diameter: $Δ V = 400 - 200 = 200 cc = 200 \times 10⁻⁶ m³$

Radius along Pressure axis: $RP = 100 \times 10³ Pa$ $R_P = 100 \times 10^3\ \mathrm{Pa}$ Radius along Volume axis: $RV = 100 \times 10⁻⁶ m³$

Area of the closed semicircular path:

W = (1)/(2) π RP RV

W = (1)/(2) \times π \times (100 \times 10³) \times (100 \times 10⁻⁶)

W = (10π)/(2) = 5π J

Since $Q = W$ $Q = W$, the heat exchanged has a magnitude of $5π J$ $5\pi\ \mathrm{J}$.

Pattern Recognition

For a cycle on an indicator chart with mismatched scales, use the elliptic area template $π a b$ $\pi a b$ (or $(1)/(2)π a b$ $\frac{1}{2}\pi a b$ for a half-ellipse/semicircle).

Chapter Mix

Class 11 Physics: Thermodynamics

Q20 jee_main_2025_24_jan_morning Thermodynamic Processes

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 :-

A. A, B, C, D Only
B. A, D, E Only
C. E Only
D. A, C Only

Solution

Related Formula

From the Ideal Gas Law:

$PV = nRT$ $PV = nRT$

According to the First Law of Thermodynamics:

Δ Q = Δ U + W

Core Logic

The question states that pressure increases linearly with temperature, which means their ratio is constant :

P = kT (P)/(T) = constant

Since $(P)/(T) = (nR)/(V)$ $\frac{P}{T} = \frac{nR}{V}$, the volume $V$ $V$ must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true).

Step 1: Evaluate All Statements

Statement A: True. In an isochoric process, $dV = 0 W = \int P dV = 0$ $dV = 0 \implies W = \int P dV = 0$.
Statement B: False. Since work is zero, the First Law simplifies to $Δ Q = Δ U$ $\Delta Q = \Delta U$, 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.

Step 2: Final Selection

Gathering the true statements (A, D, and E) points directly to Option (2).

Pattern Recognition

A linear $P-T$ $P-T$ 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.

Chapter Mix

Class 11 Physics: Thermodynamics

Q25 jee_main_2025_24_jan_morning Specific Heat Capacities of Gases

The temperature of 1 mole of an ideal monoatomic gas is increased by

50°C

at constant pressure. The total heat added and change in internal energy are

E₁

and

E₂

, respectively. If

E₁E₂=(x)/(9)

then the value of x is

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

For an ideal gas thermodynamic process:

Total heat added at constant pressure (isobaric process) is :

E₁ = n CP Δ T

Total change in internal energy is given by :

E₂ = n CV Δ T

The ratio of specific heat capacities is defined as:

γ = CPCV

Core Logic

Taking the ratio of the two energy expressions[cite: 179, 823]:

E₁E₂ = n CP Δ Tn CV Δ T = CPCV = γ

Step 1: Evaluating for a Monoatomic Gas

For an ideal monoatomic gas, the degrees of freedom are $f = 3$ $f = 3$. This gives an adiabatic index of :

γ = 1 + (2)/(f) = 1 + (2)/(3) = (5)/(3)

Equating this value to the given ratio expression [cite: 179, 826]:

(5)/(3) = (x)/(9)

x = (5 \times 9)/(3) = 15

Pattern Recognition

The ratio of heat added to the change in internal energy during an isobaric process is always equal to the adiabatic exponent $γ$ $\gamma$ of the gas.

Chapter Mix

Class 11 Physics: Thermodynamics

Q jee_main_2025_29_jan_morning Adiabatic Process

The workdone in an adiabatic change in an ideal gas depends upon only :

A. change in its pressure
B. change in its specific heat
C. change in its volume
D. change in its temperature

Solution

Related Formula

Δ W = -Δ U = -n Cv Δ T

Core Logic

In an adiabatic system, no heat exchange occurs ( $Q=0$ $Q=0$). By the first law of thermodynamics, $Δ W = -Δ U$ $\Delta W = -\Delta U$. Since internal energy $U$ $U$ depends explicitly on temperature metrics, the total work output shifts uniquely based on temperature variation $Δ T$ $\Delta T$.

Chapter Mix

Class 11 Physics: Thermodynamics

Solution & Explanation

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

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate All Statements

Step 2: Final Selection

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating for a Monoatomic Gas

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

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