Chapter 5, Problem 5.171QP

In 2012, Felix Baumgartner jumped from a balloon roughly 24 mi above Earth, breaking the record for the highest skydive. He reached speeds of more than 700 miles per hour and became the first skydiver to exceed the speed of sound during free fall. The helium-filled plastic balloon used to carry Baumgartner to the edge of space was designed to expand to 8.5 × 10⁸ L in order to accommodate the low pressures at the altitude required to break the record. (a) Calculate the mass of helium in the balloon from the conditions at the time of the jump (8.5 × 10⁸ L, −67.8°C, 0.027 mmHg). (b) Determine the volume of the helium in the balloon just before it was released, assuming a pressure of 1.0 atm and a temperature of 23°C.

Interpretation Introduction

Interpretation:

The mass of the helium in the balloon has 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.

$\text{PV = nRT}$

Answer to Problem 5.171QP

The mass of the helium in the balloon is $\text{7200}\text{g}\text{He}\text{.}$

Explanation of Solution

Given:

Pressure is $\text{0}\text{.027}\text{mmHg}$

Volume is $\text{8}\text{.5}\text{×}{\text{10}}^{\text{8}}\text{L}$

The mass of the helium in the balloon can be calculated as

$\begin{array}{l}\text{ n}\text{= }\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.027}\text{mmHg}\left(\frac{\text{1}\cancel{\text{atm}}}{\text{760}\cancel{\text{mmHg}}}\right)\text{8}\text{.5}\text{×}{\text{10}}^{\text{8}}\cancel{\text{L}}}{\text{0}\text{.0821}\frac{\cancel{\text{L}}\text{.}\cancel{\text{atm}}}{\cancel{\text{K}}\text{.}\text{mol}}\text{(-67}\text{.8}\text{+}\text{273}\text{.15)}\cancel{\text{K}}}\\ \text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{He}\end{array}$

Converting moles to grams to gives

$\text{1800}\cancel{\text{mol}}\text{He}\left(\frac{\text{4}\text{.003}\text{g}\text{He}}{\text{1}\cancel{\text{mol}}\text{He}}\right)\text{=}\text{7200}\text{g}\text{He}$

Hence, the mass of the helium in the balloon is $\text{7200}\text{g}\text{He}\text{.}$

Interpretation Introduction

Interpretation:

The volume of the helium in the balloon can 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.

$\text{PV = nRT}$

Answer to Problem 5.171QP

The volume of the helium in the balloon is $\text{4}\text{.4}\text{×}{\text{10}}^{\text{4}}\text{L}\text{.}$

Explanation of Solution

Given data:

Pressure is $\text{0}\text{.027}\text{mmHg}$

Volume is $\text{8}\text{.5}\text{×}{\text{10}}^{\text{8}}\text{L}$

The volume of the helium in the balloon has to be calculated as

Rearranging $\text{PV = nRT}$ to place the variables $\text{(P,V,T)}$ on the left and the constants $(R,n)$ on the right gives

$\begin{array}{l}\frac{\text{PV}}{\text{T}}\text{ =}\text{nR}\\ \text{=}\text{constant}\text{Þ}\frac{{\text{P}}_{\text{jump}}{\text{V}}_{\text{jump}}}{{\text{T}}_{\text{jump}}}\text{=}\frac{{\text{P}}_{\text{surface}}{\text{V}}_{\text{surface}}}{{\text{T}}_{\text{surface}}}\end{array}$

Plugging in the known values

$\begin{array}{l}\text{=}\frac{\text{(0}\text{.027}\text{mmHg)}\left(\text{8}\text{.5}\text{×}{\text{10}}^{\text{8}}\text{L}\right)}{\text{(-67}\text{.8}\text{+}\text{273}\text{.15}\text{)}\text{K}}\\ \\ \text{=}\frac{\text{1}\text{atm}\left(\frac{\text{760}\text{mm}\text{Hg}}{\text{atm}}\right){\text{V}}_{\text{surface}}}{\text{(23+}\text{273}\text{.15)}\text{K}}\end{array}$

And solving for ${\text{V}}_{\text{surface}}$ gives

$\begin{array}{l}{\text{V}}_{\text{surface}}\text{ =}\frac{\cancel{\left(\text{0}\text{.027}\text{mmHg}\right)}\text{(296}\cancel{\text{K}}\text{)}\text{(8}\text{.5}\text{×}{\text{10}}^{\text{8}}\text{L)}}{\text{(760}\cancel{\text{mmHg}}\text{)}\text{(205}\text{.3}\cancel{\text{K}}\text{)}}\\ \text{=}\text{4}\text{.4}\text{×}{\text{10}}^{\text{4}}\text{L}\end{array}$

Hence, the volume of the helium in the balloon is $\text{4}\text{.4}\text{×}{\text{10}}^{\text{4}}\text{L}\text{.}$