1. The vapor pressure, P of a certain liquid was measured at two temperatures, T. from this data:

T(K) and P(kpa)
275 3.57
825 6.57

for the enthalpy how would this be plotted on the graph

point 1:
point 2:

2. What is the rise, run, and slope

3. What is the enthalpy of vaporation of this liquid (in j/mol)?

http://www2.ohlone.edu/people/jklent/labs/101A_labs/Clausius-Clapeyron.pdf

http://www.science.uwaterloo.ca/~cchieh/cact/c123/clausius.html

To plot the points on the graph:

1. Assign the temperature (T) as the x-axis and the vapor pressure (P) as the y-axis.
2. Plot the first point (275 K, 3.57 kPa) on the graph.
3. Plot the second point (825 K, 6.57 kPa) on the graph.
4. Connect these two points with a straight line to represent the relationship between temperature and vapor pressure.

The rise (vertical change), run (horizontal change), and slope of the line can be determined as follows:

1. Rise: Subtract the y-values of the two points (6.57 kPa - 3.57 kPa = 3.0 kPa).
2. Run: Subtract the x-values of the two points (825 K - 275 K = 550 K).
3. Slope: Divide the rise by the run (3.0 kPa / 550 K ≈ 0.0055 kPa/K).

The enthalpy of vaporization can be determined using the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
T1 and T2 are the corresponding temperatures in Kelvin.
ΔHvap is the enthalpy of vaporization in joules per mole.
R is the ideal gas constant (8.314 J/(mol·K)).

Using the data provided:
P1 = 3.57 kPa; T1 = 275 K
P2 = 6.57 kPa; T2 = 825 K

Rearranging the equation gives:
ΔHvap = -(ln(P2/P1) * R) / (1/T2 - 1/T1)

Calculating:
ΔHvap = -(ln(6.57/3.57) * 8.314) / (1/825 - 1/275)
≈ -(0.438 * 8.314) / (0.0012121 - 0.0036364)
≈ -3.633 J/mol

Therefore, the enthalpy of vaporization of this liquid is approximately -3.633 J/mol.