- #1

Peter G.

- 442

- 0

I have researched the equation that links the two and I know that to have a straight line graph I need to plot:

T

^{2}against 1/K

However, I do not know why; how I get to that conclusion.

I have read about log-log graphs and played around with the concept to see if it would yield the plot I mentioned above. I tried plotting for the log of T against the log of K and I did get a straight line plot. I checked if my gradient was equal to -0.5 and my y-intercept was equal to (log 2∏-0.5 log m) and they were, but I still do not understand why I have to plot T

^{2}against 1/K.

Can anyone please shed some light onto this please?

Thanks in advance!