The atmospheric pressure is the pressure exerted by the air in the earth’s atmosphere - Leaving Cert Mathematics - Question 9 - 2012

To find the percentage error for Hannah’s model, we can use the formula:

ext{Percentage Error} = rac{| ext{Estimated Value} - ext{Actual Value} |}{ ext{Actual Value}} imes 100

Calculating these errors for each row leads to:

• For altitude 0 km: rac{|101.3 - 101.3|}{101.3} imes 100 = 0\\%
• For altitude 1 km: rac{|89.4 - 89.9|}{89.9} imes 100 \\approx 0.56\%
• For altitude 2 km: rac{|79.0 - 79.5|}{79.5} imes 100 \\approx 0.63\%
• For altitude 3 km: rac{|69.7 - 70.1|}{70.1} imes 100 \\approx 0.57\%
• For altitude 4 km: rac{|61.6 - 61.6|}{61.6} imes 100 = 0\%
• For altitude 5 km: rac{|54.4 - 54.0|}{54.0} imes 100 \\approx 0.74\%

Taking the maximum error, Hannah’s model is accurate to within approximately 1%.

Let’s take an altitude of 2 km for verification:

Using Thomas’s model:

$p = 101.3 imes e^{-0.12484 imes 2}$

Calculating:

$p = 101.3 imes e^{-0.24968} \\approx 101.3 imes 0.7788 \\approx 78.87 kPa$

Using Hannah’s model:

Pressure at 2 km = 79.5 kPa

Difference = $|78.87 - 79.5| \\approx 0.63 kPa$, which is less than 0.01 kPa.

Thomas might have derived the constant 0.1244 through various methods:

• He assumed the pressure followed the exponential form $p = 101.3 e^{-kh}$ and used observations to estimate $k$.
• By plotting values of pressure against altitude and fitting a best-fit line to find the slope, he could have calculated $k$.

Hannah’s model produces pressure values at discrete (specific whole number) altitudes, while Thomas’s model provides pressure values at any real value of altitude, allowing for smoother variations.

A continuous model allows for calculations at any altitude, providing flexibility to analyze pressure at non-integer altitudes, which is useful for more precise scientific applications.

Using Thomas's model:

p(8848) = 101.3 imes e^{-0.12484 imes 8.848} \\approx 101.3 imes e^{-1.106}$ \\approx 33.7 kPa

Given that at sea level: $p(0)=101.3 kPa$ We want: p(h) = rac{101.3}{2} = 50.65 kPa Using the formula: $50.65 = 101.3 e^{-0.12484h}$ Solving for $h$ gives:

Using the pressure model, the change in pressure per floor is approximately: