Question

A certain substance has a heat of vaporization of 66.77 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 329 K?

Answer 1

The vapour pressure will be 7.00 times higher at 358 K.

Explanation:

Chemists often use the Clausius-Clapeyron equation to determine the vapour pressure at different temperatures:

`color(blue)(bar(ul(|color(white)(a/a) ln(p_2/p_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "`

where

`p_1` and `p_2` are the vapour pressures at temperatures
`T_1` and `T_2`

`Δ_"vap"H` = the enthalpy of vaporization of the liquid

`Rcolor(white)(mmll)` = the Universal Gas Constant

In your problem,

`p_1 = p_1; color(white)(mm)T_1 = "329 K"`

`p_2 = 7.00p_1; T_2 = ?`

`Δ_text(vap)H = "66.77 kJ·mol"`

`R = "8.314 J·K"^"-1""mol"^"-1"`

`ln((7.00color(red)(cancel(color(black)(p_1))))/(color(red)(cancel(color(black)(p_1))))) = ("66 700" color(red)(cancel(color(black)("J·mol"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))(1/"329 K" - 1/T_2)`

`1.946 = "8023 K" × (1/"329 K" -1/T_2) = 24.38 - "8023 K"/T_2`

`"8023 K"/T_2 = 22.44`

`T_2 = "8023 K"/22.44 = "358 K"`