Because of its high dielectric strength, SF6 (sulfur hexafluoride) gas is widely used as an insulator and a dielectric in HV applications such as HV transformers, switches, circuit breakers, transmission lines, and even HV capacitors. The SF6 gas at 1 atm and at room temperature has a dielectric constant of 1.0015. The number of SF6 molecules per unit volume Ncan be found by the gas law, P= (N/NA)RT. Required:
Calculate the electronic polarizability αeof the SF6 molecule.

5.31 × 10⁻⁴⁰ Fm²

Explanation:

Using the Clausius-Mossotti equation

(ε - 1)/(ε + 2) = Nα/3ε₀ where ε = dielectric constant = 1.0015, ε₀ = 8.854 × 10⁻¹² F/m, N = number of molecules of SF6 per unit volume and α = electronic polarizability

So, α = (ε - 1)3ε₀/[(ε + 2)N]

Also

P = (N/NA)RT where P = pressure = 1 atm = 1.013 × 10⁵ Pa, N = number of molecules of SF6 per unit volume, NA = Avogadro's constant = 6.022 × 10²³ /mol, R = molar gas constant = 8.314 J/mol-K and T = room temperature = 25° = 273 + 25 = 293 K

So, N = PNA/RT

= 1.013 × 10⁵ Pa × 6.022 × 10²³ /mol ÷ (8.314 J/mol-K × 293 K)

= 6.100286 × 10²⁸ Pa/mol ÷ (2436.002 J/mol)

= 0.0025 × 10²⁸ molecules/m³

= 2.5 × 10²⁵ molecules/m³

So,

α = (ε - 1)3ε₀/[(ε + 2)N]

α = (1.0015 - 1)× 3 × 8.854 × 10⁻¹² F/m/[(1.0015 + 2)2.5 × 10²⁵ molecules/m³]

α = (0.0015)26.562 × 10⁻¹² F/m/[(3.0015)2.5 × 10²⁵ molecules/m³]

α = 0.039843 × 10⁻¹² F/m/[7.50375 × 10²⁵ molecules/m³]

α = 0.00531 × 10⁻³⁷ Fm²

α = 5.31 × 10⁻⁴⁰ Fm²

i mean i gues

explanation:

answer:

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