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NCERT Exemplar · Q22

Q.Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water Lv=540L_v = 540 k cal kg−1^{-1}, the mechanical equivalent of heat J=4.2J = 4.2 J cal−1^{-1}, density of water ρw=103\rho_w = 10^{3} kg l−1l^{-1}, Avagadro's No NA=6.0×1026N_A = 6.0 \times 10^{26} k mole−1^{-1} and the molecular weight of water MA=18M_A = 18 kg for 1 k mole.

(a) estimate the energy required for one molecule of water to evaporate.
(b) show that the inter–molecular distance for water is d=[MANA×1ρw]1/3d = \left[ \dfrac{M_A}{N_A} \times \dfrac{1}{\rho_w} \right]^{1/3} and find its value.
(c) 1 g of water in the vapor state at 1 atm occupies 1601 cm3^{3}. Estimate the intermolecular distance at boiling point, in the vapour state.
(d) During vaporisation a molecule overcomes a force FF, assumed constant, to go from an inter-molecular distance dd to d′d'. Estimate the value of FF.
(e) Calculate F/dF/d, which is a measure of the surface tension.
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Dividing the latent heat of vaporisation by the number of molecules gives the energy each molecule needs to escape the liquid. The liquid and vapour densities fix the two molecular spacings, and equating the escape energy to work done against a constant force over the change in spacing gives the force, whose ratio to the liquid spacing, F/dF/d, comes out close to the real surface tension of water: about 6.6×10−26.6\times10^{-2} N m−1^{-1}.

(a) Energy to evaporate one molecule

Convert the latent heat to SI units:

Lv=540 kcal kg−1=540×103×4.2 J kg−1=2.268×106 J kg−1L_v = 540\text{ kcal kg}^{-1} = 540\times10^3\times4.2\text{ J kg}^{-1} = 2.268\times10^6\text{ J kg}^{-1}

Mass of one water molecule:

m1=MANA=186.0×1026=3.0×10−26 kgm_1 = \frac{M_A}{N_A} = \frac{18}{6.0\times10^{26}} = 3.0\times10^{-26}\text{ kg}

Energy needed for one molecule to evaporate:

E=Lv×m1=2.268×106×3.0×10−26≈6.8×10−20 JE = L_v\times m_1 = 2.268\times10^6\times3.0\times10^{-26} \approx 6.8\times10^{-20}\text{ J}

(b) Intermolecular distance in liquid water

Model each molecule as occupying a small cube of side dd (the intermolecular spacing), so the volume per molecule equals d3d^3:

d3=MANAρw  ⟹  d=[MANA×1ρw]1/3d^3 = \frac{M_A}{N_A\rho_w} \;\Longrightarrow\; d=\left[\frac{M_A}{N_A}\times\frac{1}{\rho_w}\right]^{1/3}

With ρw=103\rho_w=10^3 kg m−3^{-3}:

d=[18(6.0×1026)(103)]1/3=(3.0×10−29)1/3≈3.1×10−10 md = \left[\frac{18}{(6.0\times10^{26})(10^3)}\right]^{1/3} = \left(3.0\times10^{-29}\right)^{1/3} \approx 3.1\times10^{-10}\text{ m}

(c) Intermolecular distance in the vapour (boiling point)

1 g of steam contains

10−3 kg18 kg per kmol×6.0×1026≈3.3×1022 molecules\frac{10^{-3}\text{ kg}}{18\text{ kg per kmol}}\times6.0\times10^{26} \approx 3.3\times10^{22}\text{ molecules}

and occupies 1601 cm3=1.601×10−3 m31601\text{ cm}^3=1.601\times10^{-3}\text{ m}^3. Volume per molecule:

1.601×10−33.3×1022≈4.8×10−26 m3\frac{1.601\times10^{-3}}{3.3\times10^{22}} \approx 4.8\times10^{-26}\text{ m}^3

So the vapour spacing is

d′=(4.8×10−26)1/3≈3.6×10−9 md' = \left(4.8\times10^{-26}\right)^{1/3} \approx 3.6\times10^{-9}\text{ m}

Watch out

Keep track of powers of ten carefully here: the vapour's volume-per-molecule (≈4.8×10−26\approx4.8\times10^{-26} m3^3) is about a thousand times larger than the liquid's (≈3.0×10−29\approx3.0\times10^{-29} m3^3), so d′d' should come out roughly ten times dd (since 10001/3=101000^{1/3}=10) — a useful order-of-magnitude check.

(d) The constant force FF …

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