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248 PHYSICS Answer For a given mass of water molecules, the density is less if volume is large. So the number of molecules and (ii) mass density -4 of neon and oxygen in the vessel. Atomic volume of the vapour is 1000/0.6 = 1/(6 × 10 ) mass of Ne = 20.2 u, molecular mass of O times larger. If densities of bulk water and water 2 molecules are same, then the fraction of = 32.0 u. molecular volume to the total volume in liquid Answer Partial pressure of a gas in a mixture is state is 1. As volume in vapour state has the pressure it would have for the same volume increased, the fractional volume is less by the and temperature if it alone occupied the vessel. same amount, i.e. 6×10 .-4 ⊳ (The total pressure of a mixture of non-reactive ⊳ gases is the sum of partial pressures due to its Example 12.2 Estimate the volume of a water molecule using the data in Example constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to 12.1. the two gases, we have P V = µ RT and P V = 1 1 2 Answer In the liquid (or solid) phase, the µ2 RT, i.e. (1 /2 ) =µ1/ µ 2. Here 1 and 2 refer molecules of water are quite closely packed. The to neon and oxygen respectively. Since (P 1P 2 = (3/2) (given), µ /µ ) = 3/2. density of water molecule may therefore, be 1 2 regarded as roughly equal to the density of bulk (i) By definition µ1 = (N1/N A and µ 2= (N2/N A water = 1000 kg m . To estimate the volume of where N 1nd N a2e the number of molecules of 1 and 2, and N is the Avogadro’s number. a water molecule, we need to know the mass of A a single water molecule. We know that 1 mole Therefore, (N1/N 2 = µ 1/ µ2) = 3/2. of water has a mass approximately equal to (ii) We can also write µ1 = (m 1M )1and µ2 = (m /M ) where m and m are the masses of (2 + 16)g = 18 g = 0.018 kg. 2 2 1 2 Since 1 mole contains about 6 × 10 23 1 and 2; and M 1nd M ar2 their molecular molecules (Avogadro’s number), the mass of masses. (Both m a1d M ; a1 well as m an2 23 M should be expressed in the same units). a molecule of water is (0.018)/(6 × 10 ) kg = 2 3 × 10–26kg. Therefore, a rough estimate of the Ifρ1 and ρ2 are the mass densities of 1 and 2 respectively, we have volume of a water molecule is as follows : Volume of a water molecule ρ m /V m µ  M  = (3 × 1026kg)/ (1000 kg m )3 1 = 1 = 1 = 1 × 1 –29 3 ρ2 m 2V m 2 µ2 M 2 = 3 × 10 m = (4/3)π (Radius)3 3 20.2 Hence, Radius ≈ 2 ×10 -10 m = 2 Å ⊳ = × = 0.947 2 32.0 ⊳ ⊳ Example 12.3 What is the average distance between atoms (interatomic 12.4 KINETIC THEORY OF AN IDEAL GAS distance) in water? Use the data given in Examples 12.1 and 12.2. Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a collection of a large number of molecules Answer : A given mass of water in vapour state has 1.67×10 times the volume of the same mass (typically of the order of Avogadro’s number) that of water in liquid state (Ex. 12.1). This is also are in incessant random motion. At ordinary pressure and temperature, the average distance the increase in the amount of volume available for each molecule of water. When volume between molecules is a factor of 10 or more than increases by 10 times the radius increases by the typical size of a molecule (2 Å). Thus, 1/3 interaction between molecules is negligible and V or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å. ⊳ we can assume that they move freely in straight lines according to Newton’s first law. However, ⊳ occasionally, they come close to each other, Example 12.4 A vessel contains two non- reactive gases : neon (monatomic) and experience intermolecular forces and their oxygen (diatomic). The ratio of their partial velocities change. These interactions are called pressures is 3:2. Estimate the ratio of (i) collisions. The molecules collide incessantly against each other or with the walls and change Reprint 2026-27