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Chapter 10: Problem 19

The typical atmospheric pressure on top of Mount Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to \((\mathbf{a})\) atm, \((\mathbf{b})\) \(\mathrm{mm} \mathrm{Hg}\),\((\mathbf{c})\) pascals,\((\mathbf{d})\) bars , \((\mathbf{e})\) psi.

Short Answer

Expert verified
The atmospheric pressure on top of Mount Everest is approximately \(0.3487 \,atm\), \(265 \,mmHg\), \(35342 \,Pa\), \(0.3534 \,bar\), and \(5.12 \,psi\).

Step by step solution

01

Convert to atmospheres (atm)

To convert the pressure to atmospheres (atm), we will use the conversion factor 1 atm = 760 torr: \(Pressure_{atm} = \frac{Pressure_{torr}}{760 \,torr} = \frac{265 \,torr}{760 \,torr} = 0.3487 \,atm\)
02

Convert to millimeters of mercury (mmHg)

Since 1 torr is equivalent to 1 mmHg, the pressure in mmHg is the same as the given pressure in torr: \(Pressure_{mmHg} = 265 \,mmHg\)
03

Convert to pascals (Pa)

To convert the pressure to pascals (Pa), we will first convert the pressure to atmospheres using the conversion factor 1 atm = 760 torr, and then use the conversion factor 1 atm = 101325 Pa: \(Pressure_{Pa} = Pressure_{atm} * 101325 \,Pa = 0.3487 \,atm * 101325 \,Pa = 35342 \,Pa\)
04

Convert to bars

To convert the pressure to bars, we will first convert the pressure to atmospheres using the conversion factor 1 atm = 760 torr, and then use the conversion factor 1 atm = 1.01325 bar: \(Pressure_{bar} = Pressure_{atm} * 1.01325 \,bar = 0.3487 \,atm * 1.01325 \,bar = 0.3534 \,bar\)
05

Convert to pounds per square inch (psi)

To convert the pressure to pounds per square inch (psi), we will first convert the pressure to atmospheres using the conversion factor 1 atm = 760 torr, and then use the conversion factor 1 atm = 14.696 psi: \(Pressure_{psi} = Pressure_{atm} * 14.696 \,psi = 0.3487 \,atm * 14.696 \,psi = 5.12 \,psi\) The atmospheric pressure on top of Mount Everest, in various units, is as follows: a) 0.3487 atm b) 265 mmHg c) 35342 Pa d) 0.3534 bar e) 5.12 psi

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Most popular questions from this chapter

If 5.15 gof \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a 75.0 - -mL tube filled with 760 torr of \(\mathrm{N}_{2}\) gas at \(32^{\circ} \mathrm{C},\) and the tube is heated to \(320^{\circ} \mathrm{C},\) the \(\mathrm{Ag}_{2} \mathrm{O}\) decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? \begin{equation}\begin{array}{l}{\text { (a) Helium is a monatomic gas, whereas nearly all the molecules }} \\ {\text { that make up air, such as nitrogen and oxygen, are }} \\ {\text { diatomic. }} \\ {\text { (b) The average speed of helium atoms is greater than the }} \\ {\text { average speed of air molecules, and the greater speed }} \\ {\text { of collisions with the balloon walls propels the balloon }} \\ {\text { upward. }}\\\\{\text { (c) Because the helium atoms are of lower mass than the average }} \\ {\text { air molecule, the helium gas is less dense than air. }} \\\ {\text { The mass of the balloon is thus less than the mass of the }} \\\ {\text { air displaced by its volume. }}\\\\{\text { (d) Because helium has a lower molar mass the average }} \\ {\text { air molecule, the helium atoms are in faster motion. This }} \\ {\text { means that the temperature of the helium is greater than }} \\ {\text { the air temperature. Hot gases tend to rise. }}\end{array} \end{equation}

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) . (b) Calculate the molar mass of a vapor that has a density of 7.135 \(\mathrm{g} / \mathrm{L}\) at \(12^{\circ} \mathrm{C}\) and 743 torr.

A sample of 3.00 \(\mathrm{g}\) of \(\mathrm{SO}_{2}(g)\) originally in a 5.00 -L vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C} .\) A sample of 2.35 \(\mathrm{g}\) of \(\mathrm{N}_{2}(g)\) originally in a \(2.50-\mathrm{L}\) vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same 10.0 -L vessel. (a) What is the partial pressure of \(S O_{2}(g)\) in the larger container? (b) What is the partial pressure of \(N_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if 1.50 mol has a pressure of 1.25 atm at a temperature of \(-6^{\circ} \mathrm{C} ; \mathbf{b}\) ) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies 478 \(\mathrm{mL}\) at 750 torr; (c) the pressure, in atmospheres, if 0.00245 \(\mathrm{mol}\) occupies 413 \(\mathrm{mL}\) at \(138^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if 126.5 \(\mathrm{L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of 11.25 \(\mathrm{kPa}\) .
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