Answer:
P = 0.0166 mm Hg
Explanation:
To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:
Clausius Clapeyron equation:
Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁) (1)
Where:
R: universal constant of gases (8.314 J / K.mol)
P₂: Vapour pressure at 43°C (or 316 K)
P₁: Pressure of mercury at the boiling point (1 atm)
T₂: temperature at 43 °C
T₁: Boiling point of mercury (357 °C or 630 K)
As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:
Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)
Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)
Ln P₂ = -7108.49 * (1.51x10⁻³)
Ln P₂ = -10.7338
P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾
P₂ = 2.18x10⁻⁵ atm
We can express this value in mm Hg and it will be:
P₂ = 2.18x10⁻⁵ * 760
<h2>
P₂ = 0.0166 mm Hg</h2>
Hope this helps
