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Chapter 9: Problem 83

Near sea level, how high a hill must you ascend for the reading of a barometer you are carrying to drop by 1.0 cm Hg? Assume the temperature remains at \(20^{\circ} \mathrm{C}\) as you climb. The reading of a barometer on an average day at sea level is $76.0 \mathrm{cm}$ Hg. (W) tutorial: gauge)

Short Answer

Expert verified
Answer: The person must climb approximately 111.31 meters.

Step by step solution

01

Recall the Barometric Formula

The barometric formula relates pressure, altitude, and temperature as follows: \(P(h) = P_0 \cdot e^{-\frac{mg}{kT}h}\) where: - \(P(h)\) is the pressure at altitude \(h\) - \(P_0\) is the pressure at sea level - \(m\) is the molar mass of air (approximately \(29 \mathrm{g/mol}\)) - \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)) - \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} \mathrm{J/K/mol}\)) - \(T\) is the temperature in Kelvin - \(h\) is the altitude in meters We want to find the altitude (h) when the pressure drops by 1 cm Hg.
02

Convert given values to SI units

Before we plug in the values, we need to convert them to SI units. Convert the initial reading of 76 cm Hg to Pascals and the temperature to Kelvin: - Pressure at sea level: \(76.0 \mathrm{cm Hg} 脳 133.3 \mathrm{Pa/cm Hg} = 10132 \mathrm{Pa}\) - Temperature: \(20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\)
03

Calculate the change in pressure

The pressure drop we want to achieve is 1.0 cm Hg. Convert this value to Pascals: - Pressure drop: \(1.0 \mathrm{cm Hg} 脳 133.3 \mathrm{Pa/cm Hg} = 133.3 \mathrm{Pa}\) Now, we can calculate the final pressure \(P(h)\) after the pressure drop: - Final pressure: \(P(h) = P_0 - \Delta P = 10132 \mathrm{Pa} - 133.3 \mathrm{Pa} = 9998.7 \mathrm{Pa}\)
04

Solve for altitude h

Plug in the values into the barometric formula and solve for \(h\): \(9998.7 \mathrm{Pa} = 10132 \mathrm{Pa} \cdot e^{-\frac{(29 \times 10^{-3}\mathrm{kg/mol})(9.81\mathrm{m/s^2})}{(1.38 \times 10^{-23}\mathrm{J/K/mol})(293.15\mathrm{K})}h}\) Next, divide both sides by \(10132 \mathrm{Pa}\) and take the natural logarithm: \(ln(\frac{9998.7}{10132}) = -\frac{(29 \times 10^{-3}\mathrm{kg/mol})(9.81\mathrm{m/s^2})}{(1.38 \times 10^{-23}\mathrm{J/K/mol})(293.15\mathrm{K})}h\) Now, solve for \(h\): \(h = \frac{ln(\frac{9998.7}{10132})}{-\frac{(29 \times 10^{-3}\mathrm{kg/mol})(9.81\mathrm{m/s^2})}{(1.38 \times 10^{-23}\mathrm{J/K/mol})(293.15\mathrm{K})}}\) \(h 鈮� 111.31 \mathrm{m}\)
05

Interpret the result

The hill must be approximately 111.31 meters high (to the nearest meter) for the reading of the barometer to drop by 1.0 cm Hg near sea level, while the temperature remains at 20掳C.

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