Q14 of 14 Page 333

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

































Substance



Atomic Mass (u)



Density (103 Kg m-3)



Carbon (diamond)



12.01



2.22



Gold



197.00



19.32



Nitrogen (liquid)



14.01



1.00



Lithium



6.94



0.53



Fluorine (liquid)



19.00



1.14



[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Let us suppose that the atoms are ‘tightly packed’ in a solid or liquid phase and


The Density of the substance = D


Atomic mass of the substance = M


Avogadro’s Number, N = 6.022 × 1023


Radius of an atom of the substance = r


Volume of an atom =


Therefore, the volume of the N molecules = ……(1)


Also, the volume of one mole of the substance = …….(2)


Equating the equation (1) and (2) we get :


=


⇒


Thus,


(i) Carbon


Given,


Atomic mass of carbon, M = 12.01× 10-3 kg


The Density of the carbon, D = 2.22× 103 kg / m3


⇒


⇒ r = 1.29× 10-10 m = 1.29 Å


The radius of the carbon atom is 1.29Å.


(ii) Gold


Given,


Atomic mass of Gold atom, M = 197.0× 10-3 kg


The Density of the Gold atom, D = 19.32× 103 kg / m3


⇒


⇒ r = 1.59× 10-10 m = 1.59 Å


The radius of the Gold atom is 1.59Å.


(iii) Liquid Nitrogen


Given,


Atomic mass of Liquid nitrogen, M = 14.01× 10-3 kg


The Density of the Liquid nitrogen,D = 1.0× 103 kg / m3


⇒


⇒ r = 1.77× 10-10 m = 1.77 Å


The radius of the Liquid nitrogen atom is 1.77 Å.


(iv) Lithium


Given,


Atomic mass of Lithium, M = 6.94× 10-3 kg


The Density of the Lithium, D = 0.53× 103 kg / m3


⇒


⇒ r = 1.73× 10-10 m = 1.73 Å


The radius of the Lithium atom is 1.73 Å.


(v) Liquid Fluorine


Given,


Atomic mass of Liquid flourine, M = 19.0× 10-3 kg


The Density of the Liquid flourine,D = 1.14× 103 kg / m3


⇒


⇒ r = 1.88× 10-10 m = 1.88 Å


The radius of the Liquid flourine atom is 1.88 Å.


The Rough estimates of the size of the atoms thus are:




























Substance



Radius (Å)



Carbon (diamond)



1.29



Gold



1.59



Nitrogen (liquid)



1.77



Lithium



1.73



Fluorine (liquid)



1.88



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A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

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From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas.

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A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres

n2 = n1 exp [ -mg (h2 – h1)/ kBT]


Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:


n2 = n1 exp [ -mg NA(ρ - P′) (h2 – h1)/ (ρ RT)]


Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.


[NA is Avogadro’s number, and R the universal gas constant.]