The vapor pressure, P, of a certain liquid was measured at two temperatures, T.
Represented in a table:
T (in Kelvin) ------- P(in kPa)
[325] --------------- [4.00]
[775] --------------- [8.59]
What is the enthalpy of vaporization for this liquid? (Delta H vap) in kJ/mol?
If you were going to graphically determine the enthalpy of vaporization (delta H vap) for this liquid, what points would you plot?
x1 =
x2 =
y1 =
y2 =
Determine the rise, run, and slope of these points.
much help would be greatly appreciated! thank you
Use the Clausius-Clapeyron equation. Here is a site that shows the plot.
http://www.science.uwaterloo.ca/~cchieh/cact/c123/clausius.html
To determine the enthalpy of vaporization (ΔHvap) for the liquid, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where P2 and P1 are the vapor pressures at temperatures T2 and T1 respectively, ΔHvap is the enthalpy of vaporization, R is the gas constant (8.314 J/(mol·K)), and T2 and T1 are the temperatures in Kelvin.
In this case, we have two sets of data:
T1 = 325 K, P1 = 4.00 kPa
T2 = 775 K, P2 = 8.59 kPa
To find ΔHvap, we need to rearrange the equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Let's plug in the values and solve for ΔHvap:
ln(8.59/4.00) = (-ΔHvap/8.314) * (1/775 - 1/325)
ln(2.15) = (-ΔHvap/8.314) * (0.001290 - 0.003077)
Now, let's calculate ΔHvap by rearranging the equation:
ΔHvap = -(ln(2.15) * 8.314) / (0.001290 - 0.003077)
ΔHvap ≈ 40.93 kJ/mol
So, the enthalpy of vaporization for this liquid is approximately 40.93 kJ/mol.
-------------
To graphically determine the enthalpy of vaporization (ΔHvap), we need to plot the points on a graph. In this case, we have two data points:
x1 = T1 = 325 K
x2 = T2 = 775 K
y1 = P1 = 4.00 kPa
y2 = P2 = 8.59 kPa
Now, let's calculate the rise, run, and slope of these points.
Rise (y2 - y1) = 8.59 - 4.00 ≈ 4.59 kPa
Run (x2 - x1) = 775 - 325 = 450 K
Slope (Δy / Δx) = Rise / Run = 4.59 / 450 ≈ 0.0102 kPa/K
Therefore, the rise is approximately 4.59 kPa, the run is 450 K, and the slope is approximately 0.0102 kPa/K. These values represent the change in vapor pressure with respect to temperature, which can be used to determine the enthalpy of vaporization.
To determine the enthalpy of vaporization (ΔHvap), we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where P1 and P2 are the vapor pressures at temperatures T1 and T2, ΔHvap is the enthalpy of vaporization, R is the gas constant (8.314 J/mol·K), and T1 and T2 are the temperatures in Kelvin.
Using the given data:
T1 = 325 K
P1 = 4.00 kPa
T2 = 775 K
P2 = 8.59 kPa
We can rewrite the equation in terms of natural logarithms:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Now we can substitute the values:
ln(8.59/4.00) = -ΔHvap/(8.314 J/mol·K) * (1/775 K - 1/325 K)
Calculating the natural logarithm:
ln(2.1475) = -ΔHvap/(8.314 J/mol·K) * (0.0012903 K^-1)
Rearranging the equation to solve for ΔHvap:
ΔHvap = -8.314 J/mol·K * ln(2.1475) / (0.0012903 K^-1)
Now we just need to convert the units to kJ/mol by dividing by 1000:
ΔHvap ≈ (-8.314 J/mol·K * ln(2.1475) / (0.0012903 K^-1)) / 1000
ΔHvap ≈ -40.657 kJ/mol
Therefore, the enthalpy of vaporization for this liquid is approximately -40.657 kJ/mol.
For the graphical determination of ΔHvap, we need to plot the points (T1, ln(P1)) and (T2, ln(P2)):
x1 = 325 K
x2 = 775 K
y1 = ln(4.00 kPa)
y2 = ln(8.59 kPa)
Now let's calculate the rise, run, and slope:
rise = y2 - y1 = ln(8.59 kPa) - ln(4.00 kPa)
run = x2 - x1 = 775 K - 325 K
slope = rise / run = (ln(8.59 kPa) - ln(4.00 kPa)) / (775 K - 325 K)
I hope this helps! Let me know if you have any further questions.