JEE Main · Physics → Steady

Thermodynamics appeared 36 times across 3 years — 4.2% of Physics. This question is from First Law of Thermodynamics.

Year 2026 2025 2024 Total
Questions 11 19 6 36

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)

Solution & Explanation

Related Formula
Δ Q = Δ U + W Δ U = n Cv Δ T W = P Δ V = P (A · Δ x)
Core Logic

For a monoatomic gas like Helium, Cv = (3)/(2)R. Given internal energy is 3nRT, but standard change is Δ U = n Cv Δ 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: Δ U = 3nRΔ T is directly given.

Step 1: Calculate Change in Internal Energy

Using the given relation:

Δ U = 3nR Δ T

Substitute n=1, R = (25)/(3) J/mol · K, and Δ T = 4°C:

Δ U = 3 × 1 × (25)/(3) × 4 = 100 J
Step 2: Apply First Law of Thermodynamics

We know Δ Q = 126 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 · A · Δ x 26 = 10⁵ × 17 × 10⁻⁴ × Δ x 26 = 170 · Δ x Δ x = (26)/(170) m = (26)/(170) × 100 cm = 15.3 cm

Closest given option is 15.5 cm.

Pattern Recognition

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.

Chapter Mix

Class 11 Physics: Thermodynamics

Reference Study Guides

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≥ P₁
  • B. P₂ > P₁
  • C. P₁ = P₂
  • D. P₁ > P₂

Solution

Related Formula

PV = nRT

Core Logic

From the Ideal Gas Law:

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

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

Slope = (nR)/(P) Slope ∝ (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 or P₁ > P₂.

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

More Thermodynamics Questions — jee_main_2026_23_january_evening

Practice all Thermodynamics previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)