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.
To find Hvap (the heat of vaporization) in kJ/mol using the Clausius Clapeyron equation, we have the following information:
- The student plotted ln [vapor pressure of a gas] vs. inverse Kelvin temperature and obtained a straight line with a slope equal to -9500 K.
- The vapor pressure is proportional to exp (-Hv/RT), with a constant of proportionality C.
- When one plots ln Pv vs 1/T, the relationship is ln (Pv) = ln V - Hv/RT, and the slope is -Hv/R.
Given that -Hv/R = -9500 K, we can solve for Hv:
Hv = -9500 K * R
The molar gas constant R is given as 1.987 cal/mole K. To convert it to kJ/mol K, we divide it by 4.184 (since 1 cal is approximately 4.184 J).
R = 1.987 cal/mol K / 4.184 = 0.476 kJ/mol K
Now, substitute this value into the equation to find Hv:
Hv = -9500 K * 0.476 kJ/mol K
Hv = -4,532 kJ/mol
Therefore, Hvap (the heat of vaporization) is equal to -4,532 kJ/mol.