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Chapter 10: Problem 22

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into \((\mathbf{a})\) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of Hg.

Short Answer

Expert verified
The pressure of Hurricane Wilma in different units is approximately: \(\mathbf{a})\) 0.8707 atmospheres, \(\mathbf{b})\) 893.42 Torr, and \(\mathbf{c})\) 26.0571 inches of Hg.

Step by step solution

01

Convert from millibars to atmospheres

To convert the pressure reading of Hurricane Wilma from millibars to atmospheres, we can use the conversion factor: 1 atm = 1013.25 mbar. This means that we need to divide the given pressure (882 mbar) by the conversion factor. \(\mathrm{atm} = \frac{882\,\mathrm{mbar}}{1013.25\,\mathrm{mbar}}\) Now, we can calculate the pressure in atmospheres.
02

Calculate the pressure in atmospheres

Now we can calculate the pressure using the formula from Step 1: \(\mathrm{atm} = \frac{882}{1013.25}\) \(\mathrm{atm} ≈ 0.8707\) So, the pressure of Hurricane Wilma in atmospheres is approximately 0.8707 atm.
03

Convert from millibars to torr

Now we will convert the 882 mbar pressure into torr using the conversion factor: 1 Torr = 0.98692 mbar. To accomplish this, we must divide the given pressure value by the conversion factor: \(\mathrm{Torr} = \frac{882\,\mathrm{mbar}}{0.98692\,\mathrm{mbar}}\) Now, we can calculate the pressure in torr.
04

Calculate the pressure in torr

Now we can calculate the pressure using the formula from Step 3: \(\mathrm{Torr} = \frac{882}{0.98692}\) \(\mathrm{Torr} ≈ 893.42\) So, the pressure of Hurricane Wilma in torr is approximately 893.42 Torr.
05

Convert from millibars to inches of Hg

Finally, we will convert the 882 mbar pressure into inches of Hg using the conversion factor: 1 in Hg = 33.8638 mbar. To do this, we must divide the given pressure value by the conversion factor: \(\mathrm{in\, Hg} = \frac{882\,\mathrm{mbar}}{33.8638\,\mathrm{mbar}}\) Now, we can calculate the pressure in inches of Hg.
06

Calculate the pressure in inches of Hg

Now we can calculate the pressure using the formula from Step 5: \(\mathrm{in\, Hg} = \frac{882}{33.8638}\) \(\mathrm{in\, Hg} ≈ 26.0571\) So, the pressure of Hurricane Wilma in inches of Hg is approximately 26.0571 in Hg. After solving all the required conversions, we found that the pressure of Hurricane Wilma was: - \(\mathbf{a})\) 0.8707 atmospheres - \(\mathbf{b})\) 893.42 Torr - \(\mathbf{c})\) 26.0571 inches of Hg.

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Most popular questions from this chapter

(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{K} : \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{HBr}\) . (b) Calculate the rms speeds of CO and \(\mathrm{Cl}_{2}\) molecules at 300 \(\mathrm{K}\) . (c) Calculate the most probable speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at 300 \(\mathrm{K}\) .

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To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) light-bulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

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Consider the following gases, all at STP: Ne, SF \(_{6}, \mathrm{N}_{2}, \mathrm{CH}_{4}\) . (a) Which gas is most likely to depart from the assumption of the kinetic-molecular theory that says there are no attractive or repulsive forces between molecules? (b) Which one is closest to an ideal gas in its behavior? (c) Which one has the highest root-mean-square molecular speed at a given temperature? (d) Which one has the highest total molecular volume relative to the space occupied by the gas? (e) Which has the highest average kinetic-molecular energy? (f) Which one would effuse more rapidly than \(\mathrm{N}_{2} ?\) (g) Which one would have the largest van der Waals \(b\) parameter?
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