Liquid mercury boils at 357°C (630 K) at 1 atm (101,325 N/m2). Calculate i t s approximate vapour pressure at 100°C (373 K), assuming that i t follows Trouton's rule.

To solve this problem, we will use the Clausius-Clapeyron equation and Trouton's rule.

Step 1: Trouton's rule states that the entropy of vaporization is approximately constant for many liquids and is approximately 85 J/mol*K. We can use this rule to find the enthalpy of vaporization (∆Hvap) using the formula:

∆Hvap = Tbp * ∆Svap

where Tbp is the boiling point in Kelvin and ∆Svap is the entropy of vaporization.

Step 2: Substitute the given values into the formula:

∆Hvap = 630 K * 85 J/mol*K = 53550 J/mol

Step 3: Now we will use the Clausius-Clapeyron equation, which is:

ln(P2/P1) = -∆Hvap/R * (1/T2 - 1/T1)

where P1 and P2 are the pressures at temperatures T1 and T2 respectively, and R is the ideal gas constant (8.314 J/mol*K).

Step 4: We know P1 (the pressure at the boiling point, which is 1 atm), T1 (the boiling point, which is 630 K), T2 (the temperature at which we want to find the pressure, which is 373 K), and ∆Hvap (which we calculated in step 2). We want to find P2, so we rearrange the equation to solve for P2:

P2 = P1 * e^(-∆Hvap/R * (1/T2 - 1/T1))

Step 5: Substitute the known values into the equation:

P2 = 1 atm * e^(-53550 J/mol / 8.314 J/mol*K * (1/373 K - 1/630 K))

Step 6: Solve the equation to find P2. This will give you the vapor pressure of mercury at 100°C.