Chapter 5, Problem 5.133QP

Atop Mt. Everest, the atmospheric pressure is 210 mmHg and the air density is 0.426 kg/m³. (a) Calculate the air temperature, given that the molar mass of air is 29.0 g/mol. (b) Assuming no change in air composition, calculate the percent decrease in oxygen gas from sea level to the top of Mt. Everest.

Interpretation Introduction

Interpretation:

The air temperature and the percent decrease in oxygen gas from sea level have to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

Mt. Everest to the moles of ${\text{O}}_{\text{2}}$ at sea level is calculated by given formula

  $\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)}}\text{=}\frac{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{\text{RT}}}{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)V}}{\text{RT}}}$

Answer to Problem 5.133QP

The air temperature $\text{229}\text{K}\text{=}\text{-44°C}$.

Explanation of Solution

To convert density to units of g/L

  $\frac{\text{0}\text{.426}\text{kg}}{\text{1}{\text{m}}^{\text{3}}}\text{×}\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\text{×}{\left(\frac{\text{0}\text{.01}\text{m}}{\text{1}\text{cm}}\right)}^{\text{3}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{1}\text{L}}\text{=}\text{0}\text{.426}\text{g/L}$

Let’s assume a volume of 1.00 L of air. This air sample will have a mass of 0.426 g.

We can substitute into the ideal gas equation to calculate the air temperature.

  $\text{0}\text{.426}\text{g}\text{air}\text{×}\frac{\text{1}\text{mol}\text{air}}{\text{29}\text{.0}\text{g}\text{air}}\text{=}\text{0}\text{.0147}\text{mol}\text{air}$

    $\text{T=}\frac{\text{PV}}{\text{nR}}\text{=}\frac{\left(\text{210}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\text{(1}\text{.00L)}}{\text{(0}\text{.0147 mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)}\text{=}\text{229}\text{K}\text{=}\text{-44°C}$

Theair temperature iscalculated by plugging in the values the moles of air, volume and pressure.  The air temperature was found to be $\text{229}\text{K}\text{=}\text{-44°C}$.

Interpretation Introduction

Interpretation:

The air temperature and the percent decrease in oxygen gas from sea level have to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

Mt. Everest to the moles of ${\text{O}}_{\text{2}}$ at sea level is calculated by given formula

  $\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)}}\text{=}\frac{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{\text{RT}}}{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)V}}{\text{RT}}}$

Answer to Problem 5.133QP

As a result, the percent decrease in oxygen gas from sea level to the top of Mt. Everest is $\text{72}\text{.4\%}\text{.}$

Explanation of Solution

In order to determine the percent decrease in oxygen gas, let’s compare moles of ${\text{O}}_{\text{2}}$ at the top of Mt. Everest to the moles of ${\text{O}}_{\text{2}}$ at sea level.

  $\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)}}\text{=}\frac{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{\text{RT}}}{\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)V}}{\text{RT}}}$

  $\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)}}\text{=}\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(Mt}\text{.Everest)}}{{\text{P}}_{{\text{O}}_{\text{2}}}\text{(sea}\text{level)}}\text{=}\frac{\text{210}\text{mmHg}}{\text{760}\text{mmHg}}\text{=0}\text{.276}$

The above calculation shows that there is only $\text{27}\text{.6\%}$ as much oxygen at the top of Mt. Everest compared to sea level.  As a result, the percent decrease in oxygen gas from sea level to the top of Mt. Everest is $\text{72}\text{.4\%}\text{.}$