Ncert Exercises & Solutions

Q. 22 Surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with an increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water $L_{v} = 540 k c a l k g^{- 1}$ , the mechanical equivalent of heat, $J = 4.2 J c a l^{- 1}$ , density of water, $ρ_{w} = 10^{3} k g l^{- 1}$ , Avogadro's number, $N_{A} = 6.0 \times 10^{26} k m o l e^{- 1}$ and the molecular weight of water, $M_{A}$ = 10 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= ${[\frac{M_{A}}{N_{A}} \times \frac{1}{ρ_{w}}]}^{1 / 3}$ and find its value.

(c) 1g of water in the vapour state at 1 atm occupies 1601 ${cm}^{3}$ . Estimate the inter-molecular distance at boiling point in the vapour state.

(d) During vaporisation, a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d’. Estimate the value of F.

(e) Calculate F/d which is a measure of the surface tension.

Hint: Use the unitary method to find the energy required to evaporate a molecule of water.

Step 1: Find the energy required to evaporate a molecule of water.

(a) Given,

L_{v} = 540 k c a l k g^{- 1} = 540 \times 10^{3} c a l k g^{- 1} = 540 \times 10^{3} \times 4.2 J k g^{- 1}

∵

The energy required to evaporate 1 kg of water =

L_{v}

kcal

∴

The energy required to evaporate

M_{A}

kg of water =

M_{A} L_{V}

kcal [

∵

Q=mL]

Since there are

N_{A}

molecules in

M_{A}

kg of water, the energy required for a molecule to evaporate:

U = \frac{M_{A} L_{V}}{N_{A}} J

[where

N_{A} = 6 \times 10^{26} =

Avogadro number]

= \frac{18 \times 540 \times 4.2 \times 10^{3}}{6 \times 10^{26}} J = 90 \times 18 \times 4.2 \times 10^{- 23} J = 68 \times 10^{- 20} J

Step 2: Find the inter-molecular distance for water.

(b) Let the water molecules be points and are separated at distance d from each other. The volume of

N_{A}

molecules of water =

\frac{M_{A}}{ρ_{w}}

[∵ V = \frac{M}{ρ}]

Thus, the volume around one molecule =

\frac{M_{A}}{N_{A} ρ_{w}}

The volume around one molecule,

d^{3} = (M_{A} / N_{A} ρ_{w}) ∴ d = {(\frac{M_{A}}{N_{A} ρ_{w}})}^{1 / 3} = {(\frac{18}{6 \times 10^{26} \times 10^{3}})}^{1 / 3} = {(30 \times 10^{- 30})}^{\frac{1}{3}} m \approx 3.1 \times 10^{- 10} m

Step 3: Find the inter-molecular distance at boiling point in the vapour state.

(c)

∵

1 kg of vapour occupies volume = 1601

\times 10^{- 3} m^{3}

∴

18 kg of vapour occupies volume =

18 \times 1601 \times 10^{- 3} m^{3}

6 \times 10^{26}

molecules occupy volume = 18

\times 1601 \times 10^{- 3} m^{3}

∴

1 molecule occupies volume =

\frac{18 \times 1601 \times 10^{- 3}}{6 \times 10^{26}} m^{3}

d_{1}

is the inter-molecular distance, then,

d_{1}^{3} = (3 \times 1601 \times 10^{- 29}) m^{3}

∴ d_{1} = {(30 \times 1601)}^{1 / 3} \times 10^{- 10} m

= 36.3 \times 10^{- 10} m

Step 4: Find the force F that is to be overcome by the molecules to eveporate.

(d) Work done to change the distance from d to

d_{1} i s = F (d_{1} - d)

This work done is equal to the energy required to evaporate 1 molecule.

∴ F (d_{1} - d) = 6.8 \times 10^{- 20}

F = \frac{6.8 \times 10^{- 20}}{d_{1} - d}

= \frac{6.8 \times 10^{- 20}}{(36.3 \times 10^{- 10} - 3.1 \times 10^{- 10})}

= 2.05 \times 10^{- 11} N

Step 5: Find the surface tension.

(e) Surface tension

= \frac{F}{d} = \frac{2.05 \times 10^{- 11}}{3.1 \times 10^{- 10}} = 6.6 \times 10^{- 2} N / m

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