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Chapter 5: Problem 150

A relation known as the barometric formula is useful for estimating the change in atmospheric pressure with altitude. The formula is given by \(P=P_{0} e^{-g .1 h / R T},\) where \(P\) and \(P_{0}\) are the pressures at height \(h\) and sea level, respectively; \(g\) is the acceleration due to gravity \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) ; \mathscr{M}\) is the average molar mass of air \((29.0 \mathrm{~g} / \mathrm{mol}) ;\) and \(R\) is the gas constant. Calculate the atmospheric pressure in atm at a height of \(5.0 \mathrm{~km}\), assuming the temperature is constant at \(5^{\circ} \mathrm{C}\) and \(P_{0}=1.0 \mathrm{~atm} .\)

Short Answer

Expert verified
The atmospheric pressure at a height of 5km is approximately 0.5 atm.

Step by step solution

01

Convert Units

First, perform the necessary conversions. Convert height from kilometers to meters and temperature from Celsius to Kelvin. For the height, since \(1 km = 1000 m\), therefore \(5.0 km = 5.0 * 1000 m = 5000m\). For the temperature, T in Kelvin \(= 5^{\circ} \mathrm{C} + 273 = 278K \).
02

Substitute Values into Equation

Next, substitute the given values and the converted values into the formula. In the barometric pressure formula \(P=P_{0} e^{-g .1 h / R T}\), replace \(P_{0}\) with 1.0 atm (the pressure at sea level), \(g\) with 9.8 \(m/s^2\), \(h\) with 5000 m, \(R\) with 0.0821 \(atm/mol*K\) and \(T\) with 278 K.
03

Solve for P

At this point, it's a matter of calculation. After substituting all values, the equation becomes \(P = 1.0 * e^(-9.8 * 5000 / (0.0821*278))\). Simplify this to find the value of P (the atmospheric pressure at a certain height).
04

Calculation and Result

Evaluate the equation to obtain the result: solved value for P. By doing so, one can find the atmospheric pressure at a height of 5km.

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Most popular questions from this chapter

What does the Maxwell speed distribution curve tell us? Does Maxwell's theory work for a sample of 200 molecules? Explain.

A gas at \(772 \mathrm{mmHg}\) and \(35.0^{\circ} \mathrm{C}\) occupies a volume of \(6.85 \mathrm{~L}\). Calculate its volume at STP.

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(a) Show that the pressure exerted by a fluid \(P\) (in pascals) is given by \(P=h d g,\) where \(h\) is the column of the fluid in meters, \(d\) is the density in \(\mathrm{kg} / \mathrm{m}^{3},\) and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (Hint: See Appendix 1.) (b) The volume of an air bubble that starts at the bottom of a lake at \(5.24^{\circ} \mathrm{C}\) increases by a factor of 6 as it rises to the surface of water where the temperature is \(18.73^{\circ} \mathrm{C}\) and the air pressure is 0.973 atm. The density of the lake water is \(1.02 \mathrm{~g} / \mathrm{cm}^{3}\). Use the equation in (a) to determine the depth of the lake in meters.
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