Take a look at the two-column data.

 T (oC) P (kPa) 0 0.612 10 1.227 20 2.536 30 4.242 40 7.37 50 12.33 60 19.9 70 31.15 75.7 46.12 89.7 70.1 100 101.32 120 198.5 200 1554.3

If we look closely at the data and pay attention to where the uncertainties are, we can conclude that the temperature readings have huge uncertainties. Are they real uncertainties, or are they sloppily recorded without consideration of significant figures? We could ask the person who took the data for clarification, but let’s assume we don’t know who took the data, and we have to proceed with our graphing exercise because it is due next Friday!

Okay, the due date is near and we can’t wait; we will just have to assume that the data was collected with a temperature measuring device that is not very good. According to significant figure rules, the uncertainties in the temperature readings are as follows:

 Temperature (oC) Uncertainty in temperature readings Range in temperature (oC) 0 1 sig fig; ± 0.5o -0.5 – 0.5 10 1 sig fig; ± 5oC 5 -15 20 1 sig fig; ± 5oC 15 -25 30 1 sig fig; ± 5oC 25 -35 40 1 sig fig; ± 5o 35 -45 50 1 sig fig; ± 5oC 45 -55 60 1 sig fig; ± 5oC 55 -65 70 1 sig fig; ± 5oC 65 -75 75.7 3 sig figs; ± 0.05oC 75.65 -75.75 89.7 3 sig figs; ± 0.05oC 89.65 -89.75 100 1 sig fig; ± 50oC 50 -150 120 1 sig fig; ± 5oC 115 -125 200 1 sig fig; ± 50oC 150 -250

We keep these uncertainties in mind and go to Excel to plot some graphs!

Graph 1: Use the data as is and plot “ln Pressure Vs. 1/Temperature”. Here is the graph.

Graph 2: Use the upper error limit on temperature and plot “ln Pressure Vs. 1/Temperature”. Here is the graph.

Graph 3: Use the lower error limit on temperature and plot “ln Pressure Vs. 1/Temperature”. Here is the graph.

Graph 2 and 3 illustrate the worst case scenario where maximum error could be observed. The trend lines for the three graphs are summarized below.

Graph 1 : Trendline equation is y = -5085x + 18.23 (Best line)
Graph 2 : Trendline equation is y = -4501x + 16.16
Graph 3 : Trendline equation is y = -5353x + 19.6

This gives us a +/- uncertaintiesfor the slope (approximately +300 and -600) and intercept (approximately +2 and -3):

• The uncertainty in the slope is in the hundreds digit, therefore the slope would be -5100 K, with 2 significant figures.
• The uncertainty in the intercept is in the ones digit, therefore the intercept would be 18, with 2 significant figures.

From the slope, we determine ΔHvap of water.

slope = ΔHvap / R where R is the gas constant, 8.314 J/moleK.

ΔHvap = (5100)(8.314) = 42 kJ/mole (2 significant figures)

The above ΔHvap is calculated assuming maximum errors in the temperature measurements.

Since Graph 1 shows a very good fit of the data to a straight line ( R2= 0.998), perhaps the temperature measurements were more exact than how the data was recorded. Let’s determine the statistical error of the data using Excel’s LINEST function … (see next post)