Physics · Kinetic Theory of Gases

JEE Main 2024 — 29 January, Shift 2 — Question 34

The temperature of a gas having 2.0×10252.0 \times 10^{25} molecules per cubic meter at 1.38 atm (Given, k=\mathrm{k}= 1.38×10−23JK−1)\left.1.38 \times 10^{-23} \mathrm{JK}^{-1}\right) is :

  1. Option A:

    500 K

    Correct
  2. Option B:

    200 K

  3. Option C:

    100 K

  4. Option D:

    300 K

Answer: A

Step-by-step solution

PV=nRT\mathrm{PV}=\mathrm{nRT}

PV=NNART\mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT}

N=\mathrm{N}=

Total no. of molecules P=NVkT\mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT}

1.38×1.01×105=2×1025×1.38×10−23×T1.38 \times 1.01 \times 10^{5}=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T}

1.01×105=2×102×T1.01 \times 10^{5}=2 \times 10^{2} \times \mathrm{T}

T=1.01×1032≈500 K\mathrm{T}=\frac{1.01 \times 10^{3}}{2} \approx 500 \mathrm{~K}

Answer key and solution verified before publishing.

Practise Kinetic Theory of Gases

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

Exam
JEE Main 2024
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
The temperature of a gas having 2.0 × 10 25 molecules per cubic meter… | JEE Main 2024 PYQ with Solution · DhiX AI