When a student plotted ln [vapor pressure of a gas] vs. inverse Kelvin temperature, she obtained a straight line with a slope equal to -8.000 x 10^3 K.
According to the Clausius Clapeyron equation, Hvap is ? KJ/mol
http://www.tau.ac.il/~phchlab/experiments/iodine/clauclap.html
This is the equation ln(P2/P1) = -DHvap/R * (1/T2- 1/T1)
but how do i get the pressure values and tempature values?
To determine Hvap, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its enthalpy of vaporization (Hvap). The equation is given as:
ln(P2/P1) = -ΔHvap/R * [1/T2 - 1/T1]
In this equation:
- P1 and P2 are the vapor pressures at temperatures T1 and T2 in Kelvin, respectively.
- ΔHvap is the enthalpy of vaporization in Joules per mole (J/mol).
- R is the ideal gas constant, which is equal to 8.314 J/(mol∙K).
In this case, the student plotted ln(vapor pressure) on the y-axis and 1/Temperature on the x-axis and obtained a straight line with a slope of -8.000 x 10^3 K.
The slope of the line represents -ΔHvap/R, so we can rearrange the equation to solve for ΔHvap:
-8.000 x 10^3 K = -ΔHvap/8.314 J/(mol∙K)
To find ΔHvap in kilojoules per mole (kJ/mol), we need to convert the units:
-8.000 x 10^3 K * 8.314 J/(mol∙K) = ΔHvap
Calculating this expression will give us the magnitude of ΔHvap.