Question

The vapor pressure of a liquid doubles when the temperature is raised from 77

°

C to 85

°

C. At what temperature will the vapor pressure be seven times the value at 77

°

C?​

Answer 1

Knowing that the vapor pressure of a liquid doubles when temperature changes from T₁ to T₂ (known values), it's required to calculate the temperature T₂ if vapor pressure P₂ is 7 times the P₁ value at same T₁ value.

T₁ = 77 °C = ( 77 + 273) K = 350 K

T₂ = 85 °C = ( 85 + 273) K = 358 K

We know that :   P₂ / P₁ = 2

P₁ and P₂ : vapor pressures at temperatures T₁ and T₂

Using the Clausius- Clapeyron equation, we can calculate the enthalpy of vaporization of the liquid:

ln P₂/P₁  =  - ΔHvap / R × (1/T₂ - 1/T₁)

ΔHvap = - R ln (P₂/P₁) / (1/T₂ - 1/T₁)            R = 8,314 J /mol . K

ΔHvap = - 8,314 J /mol . K × (ln 2) / ( 1/358 - 1/350) = 96047 J/mol

Now, we can calculate T₂ if T₁ = 77 °C and  P₂/P₁ = 7 :

T₁ = 77 °C = 350 K

ln P₂/P₁  =  - ΔHvap / R × (1/T₂ - 1/T₁)

ln 7  =  - (96047 J/mol / 8,314 J /mol . K) × (1 /T₂ - 1 /350 K)

1,946 = - 11552,4 × (1 /T₂ - 1 /350)         ⇒   T₂  =  372 K