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NEET PHYSICSKinetic TheoryMedium

Question

The volume occupied by the molecules contained in 4.5 kg4.5 \text{ kg} water at STP, if the molecular forces vanish away, is:

A

5.6 m35.6 \text{ m}^3

B

5.6×106 m35.6 \times 10^6 \text{ m}^3

C

5.6×103 m35.6 \times 10^3 \text{ m}^3

D

5.6×103 m35.6 \times 10^{-3} \text{ m}^3

Step-by-Step Solution

When intermolecular forces vanish, water vapour behaves as an ideal gas. We can calculate the volume using the Ideal Gas Law or the molar volume at STP.

  1. Calculate Moles (nn): Molar mass of water (H2OH_2O) = 18 g/mol18 \text{ g/mol}. Given mass = 4.5 kg=4500 g4.5 \text{ kg} = 4500 \text{ g}. n=MassMolar Mass=450018=250 moln = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{4500}{18} = 250 \text{ mol}.

  2. Calculate Volume (VV): At STP (Standard Temperature and Pressure), one mole of an ideal gas occupies 22.4 Litres22.4 \text{ Litres} (standard approximation for NEET calculations; older definition). V=n×22.4 L/molV = n \times 22.4 \text{ L/mol} V=250×22.4=5600 LitresV = 250 \times 22.4 = 5600 \text{ Litres}.

  3. Convert to SI Units (m3m^3): Since 1 m3=1000 Litres1 \text{ m}^3 = 1000 \text{ Litres}: V=56001000 m3=5.6 m3V = \frac{5600}{1000} \text{ m}^3 = 5.6 \text{ m}^3.

(Note: Using the modern IUPAC STP molar volume of 22.7 L22.7 \text{ L} would yield 5.675 m35.675 \text{ m}^3, which is closest to option A).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Kinetic Theory. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSKinetic Theoryvolumeoccupiedmoleculescontainedmolecular

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