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Physics · Kinetic Theory of Gases

NEET (UG) 2025 — Question 27

A container has two chambers of volumes V1=2V_{1}=2 litres and V2=3V_{2}=3 litres separated by a partition made of a thermal insulator. The chambers contains n1=5n_{1}=5 and n2=4n_{2}=4 moles of ideal gas at pressures p1=1 atmp_{1}=1 \mathrm{~atm} and p2=2 atmp_{2}=2 \mathrm{~atm}, respectively. When the partition is removed, the mixture attains an equilibrium pressure of :

  1. Option A:

    1.3 atm

  2. Option B:

    1.6 atm

    Correct
  3. Option C:

    1.4 atm

  4. Option D:

    1.8 atm

Answer: B

Step-by-step solution

By energy conservation

E1+E2=Emix \mathrm{E}_{1}+\mathrm{E}_{2}=\mathrm{E}_{\text {mix }}

n1f1RT12+n2f2RT22=(n1×n2)fRTmix 2\frac{\mathrm{n}_{1} \mathrm{f}_{1} \mathrm{RT}_{1}}{2}+\frac{\mathrm{n}_{2} \mathrm{f}_{2} \mathrm{RT}_{2}}{2}=\frac{\left(\mathrm{n}_{1} \times \mathrm{n}_{2}\right) \mathrm{fRT}_{\text {mix }}}{2}

⇒32P1 V1+32P2 V2=32Pmix (V1+V2)\Rightarrow \frac{3}{2} \mathrm{P}_{1} \mathrm{~V}_{1}+\frac{3}{2} \mathrm{P}_{2} \mathrm{~V}_{2}=\frac{3}{2} \mathrm{P}_{\text {mix }}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)

32×1×2+32×2×3=32×Pmix (5)\frac{3}{2} \times 1 \times 2+\frac{3}{2} \times 2 \times 3=\frac{3}{2} \times \mathrm{P}_{\text {mix }}(5)

∴Pmix =85\therefore \mathrm{P}_{\text {mix }}=\frac{8}{5}

Pmix =1.6 atm\mathrm{P}_{\text {mix }}=1.6 \mathrm{~atm}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2025
Paper
2025 paper
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems

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