Start free

Physics · Thermodynamics

JEE Main 2026 — 23 January, Evening Shift — Question 26

The internal energy of a monoatomic gas is 3 nRT . One mole of helium is kept in a cylinder having internal cross section area of 17 cm217 \mathrm{~cm}^{2} and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4∘C4{ }^{\circ} \mathrm{C}, then the piston will move ____\_\_\_\_ cm. (atmospheric pressure =105 Pa=10^{5} \mathrm{~Pa} )

  1. Option A:

    14.5

  2. Option B:

    1.55

  3. Option C:

    15.5

    Correct
  4. Option D:

    1.45

Answer: C

Step-by-step solution

ΔU=3nRΔT\Delta \mathrm{U}=3 \mathrm{nR} \Delta \mathrm{T} ΔU=3×1×253×4=100\Delta \mathrm{U}=3 \times 1 \times \frac{25}{3} \times 4=100 Joule ΔQ=126\Delta \mathrm{Q}=126 W=26=PΔV\mathrm{W}=26=\mathrm{P} \Delta \mathrm{V} 26=105×17×10−4Δx26=10^{5} \times 17 \times 10^{-4} \Delta \mathrm{x} Δx=26170=15.3\Delta \mathrm{x}=\frac{26}{170}=15.3

Answer key and solution verified before publishing.

Practise Thermodynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy

More Thermodynamics questions from this paper

The internal energy of a monoatomic gas is 3 nRT . One mole of helium… | JEE Main 2026 PYQ with Solution · DhiX AI