HW 8
1. 10 points The barometric equation P = 𝑃𝑃0 𝑒𝑒 −𝑔𝑔𝑔𝑔ℎ/𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 and Clausius-Clapeyron
𝑑𝑑𝑑𝑑 𝑃𝑃∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
equation = can be used to estimate the boiling point of a liquid at a higher
𝑑𝑑𝑑𝑑 𝑅𝑅𝑇𝑇 2
altitude. Use these equations to derive a single equation to make the calculation. Use
this equation to solve the boiling point of water 2miles above sea level (M=0.0289
kg/mol, T = 300K, and assume the enthalpy of vaporization of water is 44.0 kJ/mol
independent of temperature).
Solution
𝑔𝑔𝑔𝑔ℎ
−𝑅𝑅𝑇𝑇
From P = 𝑃𝑃0 𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎
𝑃𝑃 𝑔𝑔𝑔𝑔ℎ
−
= 𝑒𝑒 𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎
𝑃𝑃0
ln � � = −
𝑃𝑃
𝑃𝑃0
𝑔𝑔𝑔𝑔ℎ
𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎
1
𝑃𝑃∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
From
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
=
𝑅𝑅𝑇𝑇 2
2
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
𝑑𝑑𝑑𝑑
𝑃𝑃
=
𝑅𝑅
∙
𝑑𝑑𝑑𝑑
𝑇𝑇 2
2
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻 1
lnP� 𝑃𝑃𝑃𝑃0 = �− � |𝑇𝑇𝑇𝑇0
𝑅𝑅 𝑇𝑇
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
− )2
𝑃𝑃 1 1
ln � � = (
𝑃𝑃0 𝑅𝑅 𝑇𝑇0 𝑇𝑇
Combine these two equations,
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
− )2
𝑃𝑃 𝑔𝑔𝑔𝑔ℎ 1 1
ln � � = − = (
𝑃𝑃0 𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅 𝑇𝑇0 𝑇𝑇
1 1 𝑔𝑔𝑔𝑔ℎ
Thus = +
𝑇𝑇 𝑇𝑇0 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 ∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
And here 𝑇𝑇0 = 373.15 K, M = 0.0289 kg ∙ mol −1 , 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 = 300 K, h = 2mile =
3.219 × 10 3 m
g = 9.8 m ∙ s −2 , ∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻 = 44000𝐽𝐽 ∙ mol −1
Thus T=363K. 1
1. 10 points The barometric equation P = 𝑃𝑃0 𝑒𝑒 −𝑔𝑔𝑔𝑔ℎ/𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 and Clausius-Clapeyron
𝑑𝑑𝑑𝑑 𝑃𝑃∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
equation = can be used to estimate the boiling point of a liquid at a higher
𝑑𝑑𝑑𝑑 𝑅𝑅𝑇𝑇 2
altitude. Use these equations to derive a single equation to make the calculation. Use
this equation to solve the boiling point of water 2miles above sea level (M=0.0289
kg/mol, T = 300K, and assume the enthalpy of vaporization of water is 44.0 kJ/mol
independent of temperature).
Solution
𝑔𝑔𝑔𝑔ℎ
−𝑅𝑅𝑇𝑇
From P = 𝑃𝑃0 𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎
𝑃𝑃 𝑔𝑔𝑔𝑔ℎ
−
= 𝑒𝑒 𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎
𝑃𝑃0
ln � � = −
𝑃𝑃
𝑃𝑃0
𝑔𝑔𝑔𝑔ℎ
𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎
1
𝑃𝑃∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
From
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
=
𝑅𝑅𝑇𝑇 2
2
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
𝑑𝑑𝑑𝑑
𝑃𝑃
=
𝑅𝑅
∙
𝑑𝑑𝑑𝑑
𝑇𝑇 2
2
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻 1
lnP� 𝑃𝑃𝑃𝑃0 = �− � |𝑇𝑇𝑇𝑇0
𝑅𝑅 𝑇𝑇
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
− )2
𝑃𝑃 1 1
ln � � = (
𝑃𝑃0 𝑅𝑅 𝑇𝑇0 𝑇𝑇
Combine these two equations,
∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
− )2
𝑃𝑃 𝑔𝑔𝑔𝑔ℎ 1 1
ln � � = − = (
𝑃𝑃0 𝑅𝑅𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅 𝑇𝑇0 𝑇𝑇
1 1 𝑔𝑔𝑔𝑔ℎ
Thus = +
𝑇𝑇 𝑇𝑇0 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 ∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻
And here 𝑇𝑇0 = 373.15 K, M = 0.0289 kg ∙ mol −1 , 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 = 300 K, h = 2mile =
3.219 × 10 3 m
g = 9.8 m ∙ s −2 , ∆𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻 = 44000𝐽𝐽 ∙ mol −1
Thus T=363K. 1