Take a look at the two-column data.

T (^{o}C) |
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 ( ^{o}C) |
Uncertainty in temperature readings |
Range in temperature ( ^{o}C) |

0 | 1 sig fig; ± 0.5^{o} |
-0.5 – 0.5 |

10 | 1 sig fig; ± 5^{o}C |
5 -15 |

20 | 1 sig fig; ± 5^{o}C |
15 -25 |

30 | 1 sig fig; ± 5^{o}C |
25 -35 |

40 | 1 sig fig; ± 5^{o} |
35 -45 |

50 | 1 sig fig; ± 5^{o}C |
45 -55 |

60 | 1 sig fig; ± 5^{o}C |
55 -65 |

70 | 1 sig fig; ± 5^{o}C |
65 -75 |

75.7 | 3 sig figs; ± 0.05^{o}C |
75.65 -75.75 |

89.7 | 3 sig figs; ± 0.05^{o}C |
89.65 -89.75 |

100 | 1 sig fig; ± 50^{o}C |
50 -150 |

120 | 1 sig fig; ± 5^{o}C |
115 -125 |

200 | 1 sig fig; ± 50^{o}C |
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* = -5085*x* + 18.23 (Best line)

Graph 2 : Trendline equation is *y* = -4501*x* + 16.16

Graph 3 : Trendline equation is *y* = -5353*x* + 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 ΔH_{vap} of water.

slope = ΔH_{vap} / R where R is the gas constant, 8.314 J/moleK.

ΔH_{vap} = (5100)(8.314) = **42 kJ/mole (2 significant figures)**

The above ΔH_{vap} is calculated assuming maximum errors in the temperature measurements.

Since Graph 1 shows a very good fit of the data to a straight line ( R^{2}= 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)

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