An individual molecule does not have a melting point or boiling point.
It is the bulk substance, which consists of many molecules, that has these physical properties.
Pressure and melting point
We can use the Clausius equation to calculate the effect of pressure on the melting point.
One form of the Clausius equation is
#color(blue)(bar(ul(|color(white)(a/a)(dp)/(dT) = (Δ_(fus)text(H))/(TΔv_text(fus))color(white)(a/a)|)))" "#
where
#Δ_text(fus)H color(white)(ll)=# the enthalpy of fusion of ice
#T color(white)(mmm)=# the Kelvin temperature
#Δv_text(fus) color(white)(ml)=# the change in specific volume for the phase change
We can integrate the Clausius equation:
#"int_(p₁)^(p₂)dp = (Δ_text(fus)H)/(Δv_text(fus))int_(T₁)^(T₂)(dT)/T#
#p_2 - p_1 = "6 bar- 1 bar = 5 bar" = (Δ_text(fus)H)/(Δv_text(fus))ln(T_2/T_1)#
#Δ_text(fus)H = "333.55 J·g"^"-1"#
#v_text(ice) = 1/("0.916 75 g·cm"^"-3") = "1.0908 cm"^3·"g"^"-1"#
#v_text(liq) = 1/("0.999 84 g·cm"^"-3") = "1.0001 cm"^3·"g"^"-1"#
#Δv_text(fus) = v_text(liq) - v_text(ice) = "(1.0001 - 1.0908) cm"^"-3""g"^"-1" = "-0.0907 cm"^3"g"^"-1"#
#"5 bar" = "-"("333.55 J"·color(red)(cancel(color(black)("g"^"-1"))))/("0.0907 cm"^3color(red)(cancel(color(black)("g"^"-1"))))ln(T_2/T_1)#
#ln(T_2/T_1) = "-"(5 color(red)(cancel(color(black)("bar"))) × 0.0907 color(red)(cancel(color(black)("cm"^3))))/(333.55 color(red)(cancel(color(black)("J")))) × (1 color(red)(cancel(color(black)("J"))))/(1 color(red)(cancel(color(black)("Pa")))·color(red)(cancel(color(black)("m"^3)))) × ( 10^5 color(red)(cancel(color(black)("Pa"))))/(1 color(red)(cancel(color(black)("bar")))) × (1 color(red)(cancel(color(black)("m"^3))))/(10^6 color(red)(cancel(color(black)("cm"^3)))) = "-0.000 1360"#
#T_2/T_1 = e^"-0.000 1360" = "0.999 86"#
#T_2 = "273.15 K × 0.999 86 = 273.112 K = -0.037 °C"#
Pressure and boiling point
Chemists often use the Clausius-Clapeyron equation to calculate the vapour pressure of a liquid 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
For example. the boiling point of water is 99.6 °C at 1 bar and its #Δ_text(vap)H = 40.66 "kJ·mol"^"-1"#.
We can calculate the boiling point at 6 bar.
#p_1 = "1 bar"; T_1 = "99.6 °C = 372.75 K"#
#p_2 = "6 bar"; T_2 = ?#
#Δ_text(vap)H = "40.66 kJ·mol"^"-1"#
#R = "8.314 J·K"^"-1""mol"^"-1"#
#ln((6 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar"))))) = ("40 660" 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/"372.75 K" - 1/T_2)#
#1.792 = 4891/372.75 - "4891 K"/T_2 = 13.12 - "4891 K"/T_2"#
#"4891 K"/T_2 = "13.12 - 1.792 = 11.33"#
#T_2 = "4891 K"/11.33 = "431.7 K" = "159 °C"#
At 6 bar, the boiling point of water is 159 °C.
