As #y=mx+c#, and we know that #p=y# and #x=(1/H)#, then #b# must be the gradient of the line.
We can use the gradient formula, if we use 2 points from the graph:
#(y_2-y_1)/(x_2-x_1)=m#
I will choose the points #4, 2.0 times 10^5#=#x_2,y_2#
and
#2, 1.0 times 10^5#=#x_1, y_1#
Plug everything in:
#((2.0 times 10^5)-(1.0 times 10^5))/(4-2)=(10 000)/2=50000= 5.0 times 10^4#- which is within the acceptable range.
When it comes to the unit of #b#:
#y# has a unit of Pascals, #Pa=F/A=Nm^-2=(kgms^-2)/(m^2)=(kgm^-1s^-2)#
while #x# has a unit of #m^-1#, so we must multiply it by #kgs^-2# to get to the correct unit. as #y=mx#
So the unit for #b# is #kgs^-2#- which could be written as #Pa times m#. (Which is still equivalent to #kgs^-2#).
