Asked by yayhawknelson
When a student plotted ln [vapor pressure of a gas] vs. inverse Kelvin temperature, she obtained a straight line with a slope equal to -9500 K.
According to the Clausius Clapeyron equation, Hvap is ?
kJ/mol.
The vapor pressure is proportional to
exp (-Hv/RT), with a constant of proprtionality I'll call C
When one plots ln Pv vs 1/T, the relationship is
ln (Pv) = ln V -Hv/RT
and the slope is -Hv/R.
If -Hv/R = -9500 K,
Therefore Hv = 9500 R,
where R is the molar gas constant, 1.987 cal/mole K
Hv is called the heat of vaporization, not the heat of vapor pressure.
My symbol Pv in my previous post represents the vapor pressure.
According to the Clausius Clapeyron equation, Hvap is ?
kJ/mol.
The vapor pressure is proportional to
exp (-Hv/RT), with a constant of proprtionality I'll call C
When one plots ln Pv vs 1/T, the relationship is
ln (Pv) = ln V -Hv/RT
and the slope is -Hv/R.
If -Hv/R = -9500 K,
Therefore Hv = 9500 R,
where R is the molar gas constant, 1.987 cal/mole K
Hv is called the heat of vaporization, not the heat of vapor pressure.
My symbol Pv in my previous post represents the vapor pressure.
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