Question 840064
The data shows that pressure is a linear function of temperature.
 
PLOTTING THE DATA POINTS to get a graph, you realize that pressure is a linear function of temperature. Your teacher may expect you to plot the data, and "eyeball" a line that goes through all the points.
{{{drawing(600,300,-300,300,-0.5,2.5,
grid(0),
red(circle(-136,0.5,3)),red(circle(-25,0.91,3)),
red(circle(0,1,3)),red(circle(25,1.09,3)),
red(circle(100,1.37,3)),red(circle(273,2,3)),
green(line(-273,0,300,2.1))
)}}} From the graph you realize that the points are almost perfectly aligned, so it is a linear function.
From the line you may be able to estimate that the point with {{{y=0}}} (pressure equal zero atmosphere) corresponds to {{{x=-273}}} ( {{{-273^o}}}{{{C}}} ).
You do not need to read that from the graph. Knowing that
as the temperature increases from {{{0^o}}}{{{C}}} to {{{273^o}}}{{{C}}} ,
the pressure increases by {{{"1.00 atm"}}} ,from {{{"1.00 atm"}}} to {{{"2.00 atm"}}} ,
you would figure out that
as the temperature decreases from {{{0^o}}}{{{C}}} to {{{-273^o}}}{{{C}}} ,
the pressure will decrease by {{{"1.00 atm"}}} ,from {{{"1.00 atm"}}} to {{{"0.00 atm"}}} .
It could also be that your teacher expects you to calculate the slope, using two points, and write the linear function, to find the x intercept.
Using the points for {{{0^o}}}{{{C}}} to {{{-273^o}}}{{{C}}} ,
{{{slope=(1.37-1.00)/100=0.37}}}.
We know the y-intercept is {{{y=1.00}}}, since the {{{y}}} (pressure) for {{{x=0^o}}}{{{C}}} is {{{"1.00 atm"}}} , so the linear function is
{{{y=0.37x+"1.00"}}} or {{{pressure=0.37temperature+"1.00"}}} .
Hopefully you are not expected to go into statistic calculations.
 
QUICK ANSWER, WITHOUT PLOTTING, AND WITHOUT A CALCULATOR:
You see that as the temperature increases by {{{25^o}}}{{{C}}} degrees
(from {{{-25^o}}}{{{C}}} to {{{0^o}}}{{{C}}} or from {{{0^o}}}{{{C}}} to {{{25^oC}}} ),
the pressure increases by {{{9atm}}} ,
and for a temperature increases 4 times as large
( {{{100^o}}}{{{C}}} increase, from {{{0^o}}}{{{C}}} to {{{100^o}}}{{{C}}} ),
the pressure increases by about 4 times as much, {{{37atm}}} .
The rate of increase (the slope of the line) seems to be about {{{0.37}}} atmosphere per degree.
You also see that the increase in pressure for a temperature increase of {{{273^o}}}{{{C}}} is {{{"1.00 atm"}}} ,
and for half as large a change in temperature (from {{{-136^o}}}{{{C}}} to {{{0^o}}}{{{C}}} ),
the pressure increases by half as much, {{{0.50atm}}} .
You could calculate the slope from {{{0^o}}}{{{C}}} to {{{273^o}}}{{{C}}} to verify that it is also about {{{0.37}}} , but that takes time.
nowing that
as the temperature increases from {{{0^o}}}{{{C}}} to {{{273^o}}}{{{C}}} ,
the pressure increases by {{{"1.00 atm"}}} ,from {{{"1.00 atm"}}} to {{{"2.00 atm"}}} ,
you would figure out that
as the temperature decreases from {{{0^o}}}{{{C}}} to {{{-273^o}}}{{{C}}} ,
the pressure will decrease by {{{"1.00 atm"}}} ,from {{{"1.00 atm"}}} to {{{"0.00 atm"}}} ,
without ever using a calculator or pencil and paper.
 
NOTE:
As a non-famous chemist of the 20th-21st centuries, I would not extrapolate that far out of the data range to draw a conclusion.
Besides, they taught me that in the 9th grade, in physics and chemistry classes.
A physicist in the 1700's would draw a conclusion from such limited data as he could obtain at the time. If he was a famous and respected scientist, people would eventually believe him, and his conclusion would become a law of physics named after him that would be included in physics and chemistry textbooks.