Question

1) We'll start with a pair of problems dealing with the Clapeyron equation that are adapted from Atkins & de Paula: a. Prior to the discovery that Freon-12 (CF2Cl2) was harmful to the Earth's ozone layer, it was frequently used as the dispersing agent in spray cans for hair spray, etc. Its enthalpy of vaporization at its normal boiling point of -29.2 °C is 20.25 kJ/mol. Estimate the pressure that a can of hair spray using Freon-12 has to withstand at 105 °F, the temperature of a can left sitting out on the hottest day in Austin this summer. Assume that A Hvap is equal to its value at -29.2 °C over the temperature range involved. b. The enthalpy of fusion of Hg is 2.292 kJ/mol at its normal freezing temperature of 234.3 K. The molar volume of Hg changes 0.517 cm3/mol on melting. At what temperature will the bottom of a column of a 15.0 m column of liquid mercury (density of 13.6 g/cm) be expected to freeze?

section a and b please full procedure, thank you

Answer 1

(a). Vapour pressure of a liquid at its normal boiling point is equal to 1 atm (that is atmospheric pressure).

So, we can take P₁ = 1.00 atm, T₁ = - 29.2 ⁰C = (273.15 − 29.2) K = 243.95 K,

T₂ = 105 ⁰F = (105°F − 32) × 5/9 + 273.15= 313.706 K, P₂ = ?

Using Clausius Clapeyron equation:

ln (P₂ / P₁) = ($\Delta$_vapH° / R) X (1/T₁ - 1/T₂)

$\Rightarrow$ ln (P₂ / P₁) = (20250 J mol^-1 / 8.3145 J K^-1 mol^-1) X (1/243.95 K - 1/313.706 K)

$\Rightarrow$ ln (P₂ / P₁) = (2435.504 K) X (0.0041 - 0.0032) K^-1 = 2435.504 X 0.0009

$\Rightarrow$ ln (P₂ / P₁) = 2.1919 $\Rightarrow$   (P₂ / P₁) = e^2.1919 = 8.9522

$\Rightarrow$ P₂ = 8.9522 X 1 atm = 8.9522 atm

(b) Change in molar volume = $\Delta$V = 0.517 cm³/mol = 5.17 X 10^-7 m³/mol, T = 234.3 K,  $\Delta$_Fus H = 2.292 kJ/mol

Pressure = h ρ g

where h= height od the column = 15 m, ρ = density = 13.6 g/cm³ = 13600 kg/ m³ , g = 9.8 ms^-2 = 9.8 N kg^-1

So, P₂ = 15 m X 13600 kg/ m³ X 9.8 N kg^-1 = 1999200 Pa = 1999.2 kPa

P₁ = 1 atm = 101.325 kPa

$\Delta$P = 1999.2 kPa - 101.325 kPa = 1897.875 kPa

Using Clapeyron Equation:

$\Delta$T = (T $\Delta$V $\Delta$P) / $\Delta$_Fus H

$\Delta$T = (234.3 K X 5.17 X 10^-7 m³/mol X 1897.875 kPa) / 2.292 kJ mol^-1

$\Delta$T = (0.2299/ 2.292 ) K = 0.1003 K

T₂ = T₁ + $\Delta$T = 234.3 K + 0.1003 K = 234.4003 K

Hope this helped! Kindly upvote :)