Question

The normal boiling point for acetone is 56.5°C. At an elevation of 5700 ft, the atmospheric pressure is 623 torr. What would be the boiling point of acetone (ΔH_vap = 32.0 kJ/mol) at this elevation?
_____ °C

What would be the vapor pressure of acetone at 25.0°C at this elevation?
_____ torr

Answer 1

We know, normal atmospheric pressure = 1 atm = 760 torr. So P₁ = 760 torr

Given, P₂ = 623 torr , T₁ = 56.5⁰C = (56.5+273)K = 329.5 K and H = 32.0 KJ/mol = 32000 J/mol

Now, from Clasius-Clapeyron equation we have,

ln (P₁ /P₂) = (H/R) ( 1/T₂ - 1/T₁)

=> ln(760 torr/ 623 torr) = (32000 J/mol / 8.314 J/k.mol) ( 1/T₂ - 1/329.5 K)

=> ln(1.2199036392) = 3848.929516 (329.5 -T₂ / T₂ x 329.5)

=> 0.1987719146( T₂ x 329.5) = 1268222.276-3848.929516T₂

=> 0.1987719146 T₂ x 65.49534586 = 1268222.276 - 3848.929516T₂

=> 13.01863529T₂ + 3848.929516 T₂ = 1268222.276

=> 3861.948151 T₂ = 1268222.276

=> T₂ =(1268222.276/3861.948151) K

=> T₂ = 328.4 K = (328.4-273)⁰C = 55.4 ⁰C

Hence, boiling point of acetone = 55.4 ⁰C.

Again, we know the boiling point of acetone at this elevation so we can figure the vapor pressure as:

We need to find P₁ here.

Here T₂ = 25.0⁰C = (25+273)K = 298 K

ln (P₁ /P₂) = (H/R) ( 1/T₂ - 1/T₁)

=> ln (P₁ /623) = (32000 J/mol / 8.314 J/k.mol) ( 1/298 K - 1/329.5 K)

=> ln (P₁ /623) = 3848.929516 ( 329.5-298/ 298 x 329.5)

=> ln (P₁ /623) = 3848.929516 x ( 31.5 / 98191)

=>  ln (P₁ /623) =121241.2798/98191

=>  ln (P₁ /623) =1.234749414

=> P₁ / 623 = e^-1.234749414

=> P₁ / 623 = 0.2909076506

=> P₁ = 623 x 0.2909076506 torr = 181.2354663 torr = 181 torr

Hence vapor pressure of acetone at 25.0⁰C = 181 torr