Can I get some help please? Thanks!

Using the slope method

The change in pressure per degree C
`20^oC - 10^oC = 10^oC`
`750 - 726 = 24/10 = ( 2.4 mm Hg)/1^oC`

The change in pressure per degree C second interval

# 40^o C - 20^oC = 20^oC

`800 -750 = 50/20 = ( 2.5mmHg)/1.0^oC`

The change in pressure per degree C third interval

`880 - 800 = 80/30 = (2.6 mmHg)/1.0^oC`

The change in pressure per degree C fourth interval

960- 880 = 80/30 = (2.6mm Hg)/1^oC #

averaging the change in pressure for the four intervals

`(2.4 + 2.5 + 2.6 + 2.6)/4 = 2.525`

The change in temperature 150-100 = 50 so

`50^oC xx( 2.525mmHg)/1^0C = (126.25 mmHg)`

Adding this change in pressure to the value of 960 gives

960 + 126 = 1086 mmHg.

this is an approximate value as the the intervals do not give a perfect straight line slope.

I would use Excel or a graphing calculator to plot a graph of `p` vs. `t` and get an equation for the line of best fit.

I don’t have a graphing calculator, so I plotted the data in Excel.

Excel told me that the equation of the best-fit line was

`y = 2.608x + 698.01` or

`p = 2.608t + 698.01`

Substituting `t = "150 °C"`, I get

`p = (2.608 × 150 + 698.01)color(white)(l)"mmHg" = "(391.2 + 698.01) mmHg" = "1089 mmHg"`