Clausius-Clapeyron - Slope of the best-fit-line I'm using the equation $\frac{y_2-y_1}{x_2-x_1}$ to find the slope of the best fit line for this problem but I can't get the figure shown in the book (-3773x K).  I've been at this a while and I really need to know so I can move on.  What am I doing wrong?  I've looked at half a dozen general chemistry texts and none of them make this step explicit.
Using your formula, $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$, and eyeballing the most extreme points on the chart, $\mathrm{slope} = \frac{0-6.2}{0.005 - 0.0033}$, I get a slope of 3647 K, pretty close to the "official" answer. 

If you want to get more fancy, you could use a software program to fit this data.

my_data &lt;- data.frame(vapor_pressure = c(1.1, 6, 28, 95, 293, 521),
                      temperature = c(200, 220, 240, 260, 280, 300))

# linear regression on transformed variables
my_data$ln_vp &lt;- log(my_data$vapor_pressure)
my_data$inverse_T &lt;- 1/(my_data$temperature)
linear_fit &lt;- lm(data = my_data, formula = ln_vp ~ inverse_T)
linear_fit$coefficients['inverse_T']




inverse_T: -3787.46829114867


This answer is even closer to the "official" answer.  The remaining difference is probably due to rounding errors or differences in fitting algorithms used.  It only affects the answer by 0.3%.