Question

Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lungs, causing signifiant health risks. The enthalpy of vaporization of mercury is 59.1 kJ/mol. The normal boiling point of mercury is 357°C. What is the vapor pressure of mercury at 43°C? Assume the enthalpy of vaporization does not depend on temperature.

Answer 1

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

P₂ = 0.0166 mm Hg

Hope this helps