(a) The volume of a certain gas is found to be 5.0×10⁻⁴ m³ mol⁻¹ at 273 K and 3.0×10⁶ Pa. This gas obeys the van der Waals' equation with a = 0.50 m⁶ Pa mol⁻².
(i) Calculate the other van der Waals' constant, b.
(ii) What is the compression factor fo

Question

(a) The volume of a certain gas is found to be 5.0×10⁻⁴ m³ mol⁻¹ at 273 K and 3.0×10⁶ Pa. This gas obeys the van der Waals' equation with a = 0.50 m⁶ Pa mol⁻².
(i) Calculate the other van der Waals' constant, b.
(ii) What is the compression factor for this gas at the same temperature and pressure?
(iii) Comment on the nature of the molecular interactions of this gas. 20

(b) A 15 cm drinking straw is inserted in a glass of water at an angle 45°. What pressure difference (in torr) must be generated across the length of the straw to drink water? The density of water is 1.0 g cm⁻³. 10

(c) A capillary tube of radius r is inserted into a liquid to blow a bubble of the same radius r. If the excess pressure required to blow the bubble is 2·16 torr, then what is the diameter of the capillary tube in cm? The surface tension of the liquid is 0·072 N m⁻¹. 10

(d) Pyroxenes, amphiboles and phyllosilicates are well-known groups of silicates that occur in crust of the earth. Write the empirical formulae and draw the basic structural units of the above-mentioned silicates. 10

UPSC Answer Check · Accepted Answer

(a) 
For one mole, the van der Waals equation is
(P + a/Vₘ²)(Vₘ − b) = RT.
Given: Vₘ = 5.0×10⁻⁴ m³ mol⁻¹, T = 273 K, P = 3.0×10⁶ Pa, a = 0.50 m⁶ Pa mol⁻², R = 8.314 J mol⁻¹ K⁻¹ = 8.314 Pa m³ mol⁻¹ K⁻¹.
RT = 8.314 × 273 = 2269.722 Pa m³ mol⁻¹.
a/Vₘ² = 0.50/(5.0×10⁻⁴)² = 0.50/(2.5×10⁻⁷) = 2.0×10⁶ Pa.
Therefore P + a/Vₘ² = 3.0×10⁶ + 2.0×10⁶ = 5.0×10⁶ Pa.

(i) 
Using the van der Waals equation,
Vₘ − b = RT/(P + a/Vₘ²)
= 2269.722/(5.0×10⁶)
= 4.53944×10⁻⁴ m³ mol⁻¹.
Thus
b = Vₘ − (Vₘ − b)
= 5.0×10⁻⁴ − 4.53944×10⁻⁴
= 4.6056×10⁻⁵ m³ mol⁻¹.
 b = 4.61×10⁻⁵ m³ mol⁻¹ = 46.1 cm³ mol⁻¹.

(ii) 
The compression factor is defined as
Z = PVₘ/(RT).
So
Z = (3.0×10⁶ × 5.0×10⁻⁴)/2269.722
= 1500/2269.722
= 0.66087.
 Z = 0.661, dimensionless.

(iii) 
Since Z < 1, the gas is more compressible than an ideal gas under the same conditions. The attractive-pressure term a/Vₘ² = 2.0×10⁶ Pa is comparable with the applied pressure P = 3.0×10⁶ Pa, so attractive intermolecular forces are significant and dominate over repulsive excluded-volume effects at this temperature and pressure. The positive value of b = 4.61×10⁻⁵ m³ mol⁻¹ shows that the molecules do have finite size and short-range repulsion, but the net molecular interaction here is predominantly attractive.

(b) 
Let L = 15 cm = 0.15 m and the straw make an angle θ = 45° with either the horizontal or vertical. Since sin 45° = cos 45°, the vertical rise of water is
h = L sin 45° = 0.15/√2 = 0.106066 m.
Density of water ρ = 1.0 g cm⁻³ = 1000 kg m⁻³.
Taking g = 9.8 m s⁻², the hydrostatic pressure difference is
ΔP = ρgh
= 1000 × 9.8 × 0.106066
= 1.03945×10³ Pa.
Now 1 torr = 133.322 Pa, so
ΔP = 1.03945×10³/133.322 = 7.796 torr.
 ΔP ≈ 7.80 torr. 
Condition: static water column, neglecting viscosity and dynamic suction losses.

(c) 
For a bubble formed inside a liquid, there is one liquid–gas interface, so the Young–Laplace excess pressure is
ΔP = 2γ/r.
Given ΔP = 2.16 torr = 2.16 × 133.322 = 287.976 Pa, and γ = 0.072 N m⁻¹.
Thus
r = 2γ/ΔP
= (2 × 0.072)/287.976
= 0.144/287.976
= 5.000×10⁻⁴ m.
Diameter d = 2r = 1.000×10⁻³ m = 0.100 cm.
 d = 0.100 cm. 
Condition: bubble radius equals capillary radius, and the bubble is inside the liquid.

(d) 
All these silicates are built from SiO₄ tetrahedra. The basic tetrahedral unit has Si at the centre and four O atoms at the corners. Condensation occurs by sharing corner O atoms.
Pyroxenes: single-chain silicates. Empirical silicate formula: (SiO₃)ₙ²ⁿ⁻, commonly written [SiO₃]²⁻; a two-tetrahedron repeat is [Si₂O₆]⁴⁻. General mineral formula approximately M²⁺SiO₃, e.g. MgSiO₃. Structural unit: each SiO₄ tetrahedron shares two corner O atoms with neighbouring tetrahedra, forming an infinite single chain. Schematic connectivity: –O–Si(O₂)–O–Si(O₂)–O–, with non-bridging O atoms above and below the chain.
Amphiboles: double-chain silicates. Empirical silicate formula: (Si₄O₁₁)ₙ⁶ⁿ⁻, commonly written [Si₄O₁₁]⁶⁻. General formula approximately A₀₋₁B₂C₅T₈O₂₂(OH)₂. Structural unit: two single SiO₄ chains are cross-linked by sharing additional corner O atoms, forming an infinite double chain. Schematic connectivity: two parallel –O–Si(O₂)–O–Si(O₂)–O– chains joined by bridging O atoms.
Phyllosilicates: sheet silicates. Empirical silicate formula: (Si₂O₅)ₙ²ⁿ⁻, commonly written [Si₂O₅]²⁻. Examples: micas, clays, talc. Structural unit: each SiO₄ tetrahedron shares three corner O atoms, forming an infinite two-dimensional hexagonal sheet. Schematic connectivity: a hexagonal net of SiO₄ tetrahedra, with one non-bridging O per tetrahedron.

(a) The volume of a certain gas is found to be 5.0×10⁻⁴ m³ mol⁻¹ at 273 K and 3.0×10⁶ Pa. This gas obeys the van der Waals'…

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