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Mechanical Properties of Fluids physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Ten

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 = 540 k cal kg^-1, the mechanical equivalent of heat J = 4.2 J cal^-1, density of water ρ_w = 103 kg/m³, Avagadro’s No N_A = 6.0 × 10²⁶ k mole^-1 and the molecular weight of water MA = 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= ${[\frac{M_{A}}{N_{A}} \times \frac{1}{ρ_{w}}]}^{1 / 3}$ and find its value.

(c) 1 g of water in the vapor state at 1 atm occupies 1601cm³. Estimate the intermolecular 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.

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) L_v=540 kcal kg^-1

= 540 $\times 10^{3} c a l$ kg^-1 = 540 $\times 10^{3} \times$ 4.2jkg^-1

Energy required to evaporate 1kg of water = L_v kcal

And M_A kg of water requires M_AL_V kcal

Since there are N_A molecules in M_A kg of water the energy required for 1 molecule to evaporate

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

= $\frac{18 \times 540 \times 4.2 \times 10^{3}}{6 \times 10^{26}} J$

=90 $\times 18 \times 4.2 \times 10^{- 23} J$

= 6.8 $\times 10^{- 20} J$

(b) Let the water molecules to be points and are separated at a distance d from each other

volume of N_A molecule of water = $\frac{M_{A}}{ρ_{w}}$

thus the volume of one molecule is = $\frac{M_{A}}{N_{A} ρ_{w}}$

the volume around one molecule is d³= $\frac{M_{A}}{N_{A} ρ_{w}}$

d= $({\frac{M_{A}}{N_{A} ρ_{w}})}^{3} = ({\frac{18}{6 \times 10^{26} \times 10^{3}})}^{1 / 3}$

d= 3.1 $\times 10^{- 10} m$

(c) 1 kg of vapour occupies volume =1601 $\times 10^{- 3}$ m³

18 kg of vapour occupies 18 $\times 1601 \times 10^{- 3}$ m³

6 $\times 10^{26}$ molecules occupies 18 $\times 1601 \times 10^{- 3}$ m³

1 mo

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This is a long answer type question as classified in NCERT Exemplar

(a) L_v=540 kcal kg^-1

= 540 $\times 10^{3} c a l$ kg^-1 = 540 $\times 10^{3} \times$ 4.2jkg^-1

Energy required to evaporate 1kg of water = L_v kcal

And M_A kg of water requires M_AL_V kcal

Since there are N_A molecules in M_A kg of water the energy required for 1 molecule to evaporate

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

= $\frac{18 \times 540 \times 4.2 \times 10^{3}}{6 \times 10^{26}} J$

=90 $\times 18 \times 4.2 \times 10^{- 23} J$

= 6.8 $\times 10^{- 20} J$

(b) Let the water molecules to be points and are separated at a distance d from each other

volume of N_A molecule of water = $\frac{M_{A}}{ρ_{w}}$

thus the volume of one molecule is = $\frac{M_{A}}{N_{A} ρ_{w}}$

the volume around one molecule is d³= $\frac{M_{A}}{N_{A} ρ_{w}}$

d= $({\frac{M_{A}}{N_{A} ρ_{w}})}^{3} = ({\frac{18}{6 \times 10^{26} \times 10^{3}})}^{1 / 3}$

d= 3.1 $\times 10^{- 10} m$

(c) 1 kg of vapour occupies volume =1601 $\times 10^{- 3}$ m³

18 kg of vapour occupies 18 $\times 1601 \times 10^{- 3}$ m³

6 $\times 10^{26}$ molecules occupies 18 $\times 1601 \times 10^{- 3}$ m³

1 molecule occupies = 18 $\times 1601 \times 10^{- 3}$ /6 $\times 10^{26}$

If d is the inter molecular distance then

$d_{1}^{3} = 3 \times 1601 \times 10^{- 29}$ m³

d₁= (30 $\times 1601)$ ^{1/3
$\times 10^{- 10} m$}

= 36.3 $\times 10^{- 10}$ m

(d) Work done to change the distance from d to d₁ is = F(d₁-d)

this work done is equal to energy required to evaporate 1 molecule

F(d₁-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 $\times 10^{- 11} N$

(e) Surface tension =f/d= $\frac{2.05 \times 10^{- 11}}{3.1 \times 10^{- 10}} = 6.6 \times 10^{- 2} N / m$

less

This is a long answer type question as classified in NCERT Exemplar(a) Lv=540 kcal kg-1= 540 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mi>c</mi> <mi>a</mi> <mi>l</mi> </math> kg-1 = 540 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>×</mo> </math> 4.2jkg-1Energy required to evaporate 1kg of water = Lv kcalAnd MA kg of water requires MALV kcalSince there are NA molecules in MA kg of water the energy required for 1 molecule to evaporateIsU= <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mi>J</mi> </math> = <math> <mfrac> <mrow> <mrow> <mn>18</mn> <mo>×</mo> <mn>540</mn> <mo>×</mo> <mn>4.2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mi>J</mi> </math> =90 <math> <mo>×</mo> <mn>18</mn> <mo>×</mo> <mn>4.2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mi>J</mi> </math> = 6.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20</mn> </mrow> </mrow> </msup> <mi>J</mi> </math> (b) Let the water molecules to be points and are separated at a distance d from each othervolume of NA molecule of water = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> thus the volume of one molecule is = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> the volume around one molecule is d3= <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> d= <math> <mo>(</mo> <msup> <mrow> <mrow> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>)</mo> <mi> </mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>=</mo> <mo>(</mo> <msup> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> d= 3.1 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> (c) 1 kg of vapour occupies volume =1601 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> m318 kg of vapour occupies 18 <math> <mo>×</mo> <mn>1601</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> m36 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </math> molecules occupies 18 <math> <mo>×</mo> <mn>1601</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> m31 molecule occupies = 18 <math> <mo>×</mo> <mn>1601</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> /6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </math> If d is the inter molecular distance then <math> <msubsup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msubsup> <mo>=</mo> <mn>3</mn> <mo>×</mo> <mn>1601</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>29</mn> </mrow> </mrow> </msup> </math> m3d1= (30 <math> <mo>×</mo> <mn>1601</mn> <mo>)</mo> </math> 1/3 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> = 36.3 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </math> m(d) Work done to change the distance from d to d1 is = F(d1-d)this work done is equal to energy required to evaporate 1 moleculeF(d1-d)=6.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20</mn> </mrow> </mrow> </msup> </math> F= <math> <mfrac> <mrow> <mrow> <mn>6.8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">d</mi> <mn>1</mn> <mo>-</mo> <mi mathvariant="normal">d</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>6.8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>36.3</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> <mo>-</mo> <mn>3.1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 2.05 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>11</mn> </mrow> </mrow> </msup> <mi>N</mi> </math> (e) Surface tension =f/d= <math> <mfrac> <mrow> <mrow> <mn>2.05</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>11</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3.1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>6.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mi>N</mi> <mo>/</mo> <mi>m</mi> </math>

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