The equation for the best fit line was y = -4058.7x + 16.10. Using the equation of the best fit line, calculate the heat of vaporization (delta H) and the NORMAL BOILING POINT (in Celsius) of this volatile liquid.
Calculated:
dHvap = 33.7kJ/mol
----> I'm really confused on how to find the normal boiling point. I know that the pressure at normal boiling point is 101.3kPa. But I don't know where to go from there.
Use the Clausius-Clapeyron equation. You know dHvap. Use P1 = 101.325 with T1 = unknown normal boiling point. P2 and T2 I assume are in the data/plot.
Okay thank you, since they did not give me the data plots, I'm guessing I should sketch the graph on my own.
To find the normal boiling point using the equation of the best fit line, we need to understand the relationship between temperature (in Celsius) and pressure (in kPa) at different points along the best fit line.
The equation of the best fit line, y = -4058.7x + 16.10, represents the relationship between temperature (y) and pressure (x). However, in this case, we want to find the temperature at which the pressure is 101.3 kPa, which represents the normal boiling point.
To do this, we can substitute the pressure value (101.3 kPa) into the equation and solve for temperature (y).
101.3 = -4058.7x + 16.10
Now, let's solve for x:
-4058.7x = 101.3 - 16.10
-4058.7x = 85.20
Dividing both sides of the equation by -4058.7:
x = -85.20 / -4058.7
x ≈ 0.021 kPa
Therefore, the temperature (y) corresponding to a pressure of 101.3 kPa, which represents the normal boiling point, is approximately 0.021.
Thus, the normal boiling point of the volatile liquid is approximately 0.021 degrees Celsius.