Chemists often use the Clausius-Clapeyron equation to estimate the vapour pressures of pure liquids:
#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
#R# = the Universal Gas Constant
In your problem,
#P_1 = "100 mmHg"#; #T_1 = "34.9 °C" = "308.05 K"#
#P_2 = "?"#; #color(white)(mmmmm)T_2 = "55.5 °C" = "328.65 K"#
#Δ_"vap"H = "42.3 kJ/mol"#
#R = "0.008 314 kJ·K"^"-1""mol"^"-1"#
#ln(P_2/("100 mmHg")) = (42.3 color(red)(cancel(color(black)("kJ·mol"^"-1"))))/("0.008 314" color(red)(cancel(color(black)("kJ·K"^"-1""mol"^"-1"))))(1/(308.05 color(red)(cancel(color(black)("K")))) - 1/(328.65 color(red)(cancel(color(black)("K")))))#
#ln(P_2/("100 mmHg")) = 5088 × 2.035 × 10^"-4" = 1.035#
#P_2/("100 mmHg") = e^1.035 = 2.816#
#P_2 = "2.816 × 100 mmHg = 282 mmHg"#
The on-line calculator here predicts a value of 286 mmHg. Close enough!
