I am given three pairs of (pressure $p$, molar volume $V_m$) measurements at constant known temperature $T$: (0.750000, 29.8649), (0.500000, 44.8090), (0.250000, 89.6384). The goal is to determine the ideal gas constant $R$ plotting $p$ vs $1/V_m$ or $V_m$ vs $1/p$ and using the ideal gas law $pV_m = RT$. However, these plots produce best fit lines with slightly different slopes. I used Microsoft Excel to do this. The resulting equations of the best fit lines and $R^2$ values are:
$$y = 22.3932159243x + 0.0002057994 \qquad R^2 = 0.9999999746$$
for the $p$ vs $1/V_m$ plot, and
$$y = 22.4149788462x + 0.0214038462 \qquad R^2 = 0.9999999998$$
for the $V_m$ vs $1/p$ plot.
I am completely confused by this and would appreciate if someone could shed some light on this "paradox".
Best Answer
This is perfectly normal in the least square sense.
Suppose that the data are $(x,y)$. If you fit $y=a + b x$ you minimize the vertical distance on $y$ and vice-versa. So, except if there is absolutely no noise, the results of the regression cannot be the same.
In fact, if I had this problem, what I should do is to write $$P\,V=c$$ and minimize with respect to $c$ the real sum of squares $$SSQ=\sum_{i=1}^n (P_i\, V_i -c)^2$$ which will just give $$c=\frac 1 n \sum_{i=1}^n P_i\, V_i$$ With your data $c=\frac{2688511}{120000}=22.4043$ which is neither $22.3932$ nor $22.4150$ but something in the middle.