Physics · Kinetic Theory · NEET
No. When we say a gas occupies 22.4 litres at STP, that 22.4 litres is the volume of the container the gas spreads into, not the space taken by the solid molecules. The molecules are tiny and far apart. If you added up the actual volume of all the molecules alone, it would be only about one-thousandth of the container. So the number you see for a gas volume is mostly empty space with a few molecules moving in it.
Because measurements and NCERT's own estimate show the molecules fill only about 0.1% of the container. In liquid water the molecules are packed and fill almost all the space. When water turns to vapour at 100°C and 1 atm, the same molecules spread out over about 1000/0.6 ≈ 1670 times more volume. So the fraction actually filled by molecules drops to about 6×10⁻⁴. Since this fraction is so small, the ideal gas model treats the molecular volume as zero (a point particle).
NCERT Example 12.1 uses density. Density of liquid water = 1000 kg/m³, density of water vapour at 100°C and 1 atm = 0.6 kg/m³. For the same mass, volume is inversely proportional to density, so the vapour volume is 1000/0.6 larger than the liquid volume. In liquid, the molecules almost fill the space (fraction ≈ 1). In vapour the total volume grew by that factor, so the molecular fraction shrinks by the same factor to about 6×10⁻⁴ (roughly 0.06%). This is the tiny real volume the molecules occupy.
The V in PV = nRT is the volume available to the gas, meaning the container. It is the room the molecules move around in, not the room they physically block. Ideal gas assumes molecular size is zero, so the whole container volume V is free for motion. A real gas correction (van der Waals b term) subtracts a small volume because molecules do have a tiny size, but for NEET at normal conditions this correction is small and V is taken as the container volume.
Because the question says the intermolecular forces vanish, so the water behaves as an ideal gas. For an ideal gas at STP, one mole occupies 22.4 litres = 22.4×10⁻³ m³, regardless of what the substance is. You find the moles from mass and molar mass, then multiply by 22.4 litres. This gives the volume the gas spreads into, which is the answer 5.6 m³, not the tiny real volume of the molecules themselves.
The volume occupied by the molecules contained in 4.5 kg of water at STP, if the intermolecular forces vanish away, is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the combined solid volume of all the molecules themselves, not the container. In a gas at normal conditions this is only about 0.1% (near 10⁻³) of the total volume, so it is treated as negligible in the ideal gas model.
Because the molecules are very small (diameter around 10⁻¹⁰ m) and sit far apart. NCERT's estimate shows the molecules fill only about 6×10⁻⁴ of the container in water vapour, so most of the space between them is empty.
Yes. In PV = nRT the molecules are treated as points with zero size, so V is the full container volume. Real gas models like van der Waals add a small volume correction (the b term), but it is small at normal NEET conditions.
First find moles: n = mass / molar mass. Then multiply by the molar volume at STP, 22.4 L per mole. Example: 4.5 kg water = 4500/18 = 250 mol, so V = 250 × 22.4 L = 5600 L = 5.6 m³.
It is the container volume that one mole of an ideal gas spreads into at STP. The molecules inside fill only a tiny fraction of that 22.4 litres; the rest is empty space.
