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.  • What values for $x_1$, $x_2$, $y_1$ and $y_2$ did you get from the diagram? Sep 24 '16 at 22:35
• I used the values .09531-6.25570/.005-.00333= -6.16030/0.00167= -3689
– Karl
Sep 25 '16 at 1:38
• Then you have done it right and your result is probably within the error bounds. You can use linear regression if you want to treat the problem mathematically more rigorously. Sep 25 '16 at 8:05
• Thank you both for your input. I can not tell you how much of a relief it is to get some clarification on this problem. As I move on I will get a better understanding of what you mean by 'error bounds'. Linear regression and programming will have to wait, but I will remember your remarks.
– Karl
Sep 25 '16 at 13:02
• I still have some questions about this kind of graph. First, on the x-axis, why do we begin with the inverse of 300K? This creates a negative slope. Why don't we plot the inverse of 200K as our first point? Second, and this may be beyond my mathematical understanding, what is it about the natural log that allows us to make a linear graph of pressure as a function of temperature? Why don't we see disproportional jumps in the natural log of pressure in relation to temperature ultimately producing a curve?
– Karl
Sep 25 '16 at 13:24

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 <- 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$ln_vp <- log(my_data$vapor_pressure)
my_data$inverse_T <- 1/(my_data$temperature)
linear_fit <- 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%.