Naphthalene, c10h8, melts at 80.2â°c. if the vapour pressure of the liquid is 1.3 kpa at 85.8â°c and 5.3 kpa at 119.3â°c, use the clausiusâ clapeyron equation to calculate

(a) the enthalpy of vaporization,

(b) the normal boiling point,

(c) the enthalpy of vaporization at the boiling point

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

P₁ = 1.3 kPaP₂ = 5.3 kPaT₁ = 85.8°C = 358.96 KT₂ = 119.3°C = 392.46 K

Solving for ΔHv:

ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)ΔHv = 49111.12 J/molK

(b) Normal boiling point means that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, solving for T₂:

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ] 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]1/T₂ = 2.049 * 10⁻³ K⁻¹T₂ = 488.1 K = 214.94 °C

(c) The enthalpy of vaporization was calculated in part (a), and it does not vary depending on temperature, meaning that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK.

It’s b. mass i searched it up

The answer is nitrogen-containing bases