Question

The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At \(15^\circ {\rm{C}}\) the pressure is \(101.3\)kPa at sea level and \(87.14\) kPa at \(h = 1000{\rm{ m}}\).

(a) What is the pressure at an altitude of 3000 m?

(b) What is the pressure at the top of Mount McKinley, at an altitude of 6187 m?

Short answer

(a) The pressure at an altitude of 3000 m is \(64.5\;{\rm{kPa}}\).

(b) The pressure at an altitude of 6187 m is \(39.9\;{\rm{kPa}}\).

Step-by-step solution

(a) Step 1: Write the function for change in temperature

Assume\(P\left( h \right)\) be the pressure at altitude \(h\).

According to the given condition, \(\frac{{dP}}{{dh}} = kP\) and \(P\left( 0 \right) = 101.3\).

This implies that,

\(\begin{aligned}P\left( h \right) &= P\left( 0 \right){e^{kh}}\\ &= 101.3{e^{kh}}\end{aligned}\)

Determine the value of \(k\)

Substitute \(1000\) for \(h\) in the equation \(P\left( h \right) = 101.3{e^{kh}}\) and simplify it.

\(\begin{aligned}P\left( {1000} \right) &= 101.3{e^{k\left( {1000} \right)}}\\87.14 &= 101.3{e^{1000k}}\\{e^{1000k}} &= \frac{{87.14}}{{101.3}}\\k &= \frac{1}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)\end{aligned}\)

This implies that, \(P\left( h \right) = 101.3{e^{\frac{h}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\).

Determine the pressure at an altitude of 3000 m

Substitute \(3000\) for \(h\) into the equation \(P\left( h \right) = 101.3{e^{\frac{h}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\) and simplify it.

\(\begin{aligned}P\left( {3000} \right) &= 101.3{e^{\frac{{3000}}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\\ &= 101.3{e^{3{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\\ &\approx 64.5\;{\rm{kPa}}\end{aligned}\)

Thus, the pressure at an altitude of 3000 m is \(64.5\;{\rm{kPa}}\).

(b) Step 4: Determine the pressure at an altitude of 6187 m

Substitute \(6187\) for \(h\) into the equation \(P\left( h \right) = 101.3{e^{\frac{h}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\) and simplify it.

\(\begin{aligned}P\left( {3000} \right) &= 101.3{e^{\frac{{6187}}{{1000}}{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\\ &= 101.3{e^{6.187{\rm{ln}}\left( {\frac{{87.14}}{{101.3}}} \right)}}\\ &\approx 39.9\;{\rm{kPa}}\end{aligned}\)

Thus, the pressure at an altitude of 6187 m is \(39.9\;{\rm{kPa}}\).