Chapter 5, Problem 155CP

Interpretation Introduction

Interpretation: For the given conditions, total mass that can be lifted by a balloon should be determined.

Concept introduction:

• Total mass that can be lifted by a balloon is the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon
• Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

R= $\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}$, gas constant

• Volume of sphere  $\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}$

  r = radius of sphere

  $Radius=\frac{Diameter}{2}$

Answer to Problem 155CP

Answer

• Total mass that can be lifted by a balloon, when the average molar mass of air $\text{29}\text{.0g/mol}$ = $\text{1}\text{.01}\times {\text{10}}^{4}g$
• Total mass that can be lifted by a balloon, when the balloon is filled with helium is

  $6.65\times {\text{10}}^{4}g$

• Total mass that can be lifted by a balloon if it were on the ground in dever, colarado is $8.7\times {\text{10}}^{3}g$

Explanation of Solution

Explanation

To determine: Mass of hot air when the average molar mass of air $\text{29}\text{.0g/mol}$

Mass of hot air = $6.70\times {10}^{4}g$

Volume of sphere  $\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}$

r = radius of sphere

Here in the case of balloon,

$\begin{array}{l}\text{Diameter}\text{=}\text{5}\text{.00}\text{m}\\ \text{Radius=}\frac{\text{Diameter}}{\text{2}}\\ \text{Radius=}\frac{\text{5}\text{.00}}{\text{2}}\text{=2}\text{.50}\text{m}\end{array}$

$\begin{array}{l}\text{Volume}\text{of}\text{hot}\text{air}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \text{=}\frac{\text{4}}{\text{3}}\text{π}{\left(\text{2}\text{.50}\text{m}\right)}^{\text{3}}\text{=65}\text{.4}{\text{m}}^{\text{3}}\end{array}$

$\begin{array}{l}\text{Since,1}{\text{cm}}^{\text{3}}\text{=}\text{1000L}\\ \text{65}\text{.4}{\text{m}}^{\text{3}}\text{×}\left(\frac{\text{10}\text{dm}}{\text{m}}\right)\text{×}\frac{\text{1L}}{{\text{dm}}^{\text{3}}}\text{=}\text{6}{\text{.54×10}}^{\text{4}}\text{L}\end{array}$

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Here, from the given data,

$\begin{array}{l}\text{P}\text{=745}\text{torr}\text{=}\text{0}\text{.980}\text{atm}\text{Since,}\text{1atm=760}\text{torr}\\ \text{745}\text{torr=}\frac{\text{745}\text{torr}}{\text{760}\text{torr}}\\ \text{=0}\text{.980}\text{atm}\\ \text{V}\text{=}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}\\ \text{T=65°C}\text{=}\text{338K}\text{Since,K}\text{=°C}\text{+}\text{273}\\ \text{=}\left(\text{65°C+273}\right)\text{K}\\ \text{=338}\text{K}\\ \text{R=}\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\\ \text{n=}\frac{\text{P}\text{V}}{\text{RT}}\text{=}\frac{\text{0}\text{.980}\text{atm×}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}}{\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\text{×}\text{338}\text{K}}\\ \text{=2}\text{.31}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{air}\\ \text{Mass}\text{of}\text{hot}\text{air=number}\text{of}\text{moles}\text{of}\text{air}\text{×}\text{molar}\text{mass}\text{of}\text{hot}\text{air}\\ \text{Molar}\text{mass}\text{of}\text{hot}\text{air}\text{=}\text{29}\text{.0g/mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{air=}\text{2}\text{.31}\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{Mass}\text{of}\text{hot}\text{air}\text{=}\text{2}\text{.31}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{×}\text{29}\text{.0g/mol}\\ \text{=}\text{6}\text{.70}{\text{×10}}^{\text{4}}\text{g}\\ \\ \end{array}$

To determine: Mass of air displaced when the average molar mass of air $\text{29}\text{.0g/mol}$

Mass of air displaced $=7.71\times 10^{4}g$

Mass of Helium

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Here, from the given data,

$\begin{array}{l}\text{P}\text{=745}\text{torr}\text{=}\text{0}\text{.980}\text{atm}\text{Since,}\text{1atm=760}\text{torr}\\ \text{745}\text{torr=}\frac{\text{745}\text{torr}}{\text{760}\text{torr}}\\ \text{=0}\text{.980}\text{atm}\\ \text{V}\text{=}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}\\ \text{T=21°C}\text{=}\text{294K}\text{Since,K}\text{=°C}\text{+}\text{273}\\ \text{=}\left(\text{21°C+273}\right)\text{K}\\ \text{=294}\text{K}\\ \text{R=}\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\\ \text{n=}\frac{\text{P}\text{V}}{\text{RT}}\text{=}\frac{\text{0}\text{.980}\text{atm×}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}}{\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\text{×}\text{294}\text{K}}\\ \text{=2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{air}\\ \text{Mass}\text{of}\text{hot}\text{air dispalced=number}\text{of}\text{moles}\text{of}\text{air}\text{×}\text{molar}\text{mass}\text{of}\text{hot}\text{air}\\ \text{Molar}\text{mass}\text{of}\text{hot}\text{air}\text{=}\text{29}\text{.0g/mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{air=}\text{2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{Mass}\text{of}\text{air}\text{dispalced}\text{=}\text{2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{×}\text{29}\text{.0g/mol}\\ \text{=}\text{7}\text{.71}{\text{×10}}^{\text{4}}\text{g}\\ \\ \end{array}$

To determine: Total mass that can be lifted by a balloon, when the average molar mass of air $\text{29}\text{.0g/mol}$

Total mass that can be lifted by a balloon, when the average molar mass of air $\text{29}\text{.0g/mol}$ = $\text{1}\text{.01}\times {\text{10}}^{4}g$

Total mass that can be lifted by a balloon is the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon.

$\begin{array}{l}\text{Total mass that can be lifted by a balloon=Mass}\text{of}\text{air}\text{diaplaced-Mass }\text{of}\text{hot}\text{air}\\ \text{=}\text{7}\text{.71}\text{×10}{}^{\text{4}}\text{g}\text{-}\text{6}\text{.70}{\text{×10}}^{\text{4}}\text{g}\\ \text{=}\text{1}\text{.01}{\text{×10}}^{\text{4}}\text{g}\\ \end{array}$

To determine: Total mass that can be lifted by a balloon, when the balloon is filled with helium.

Total mass that can be lifted by a balloon, when the balloon is filled with helium is $6.65\times {\text{10}}^{4}g$

• Mass of air displaced is same, that is $\text{7}\text{.71}\text{×10}{}^{\text{4}}\text{g}$

$Massofhelium=numberofmolesofhelium\times molarmassofhelium$

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Here, from the given data,

$\begin{array}{l}\text{P}\text{=745}\text{torr}\text{=}\text{0}\text{.980}\text{atm}\text{Since,}\text{1atm=760}\text{torr}\\ \text{745}\text{torr=}\frac{\text{745}\text{torr}}{\text{760}\text{torr}}\\ \text{=0}\text{.980}\text{atm}\\ \text{V}\text{=}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}\\ \text{T=21°C}\text{=}\text{294K}\text{Since,K}\text{=°C}\text{+}\text{273}\\ \text{=}\left(\text{21°C+273}\right)\text{K}\\ \text{=294}\text{K}\\ \text{R=}\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\end{array}$

$\begin{array}{l}\text{n=}\frac{\text{P}\text{V}}{\text{RT}}\text{=}\frac{\text{0}\text{.980}\text{atm×}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}}{\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\text{×}\text{294}\text{K}}\\ \text{=2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{air}\\ \text{Mass}\text{of}\text{helium=number}\text{of}\text{moles}\text{of}\text{helium}\text{×}\text{molar}\text{mass}\text{of helium}\\ \text{Molar}\text{mass}\text{of}helium\text{=}4.003\text{g/mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{air=}\text{2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{Mass}\text{of}helium\text{=}\text{2}\text{.66}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{×}4.003\text{g/mol}\\ \text{=}1.06{\text{×10}}^{\text{4}}\text{g}\end{array}$

• Mass of He= $\text{1}{\text{.06×10}}^{\text{4}}\text{g}$

$\begin{array}{l}\text{Total mass that can be lifted by a balloon=Mass}\text{of}\text{air}\text{diaplaced-Mass }\text{of}\text{helium}\\ \text{=}\text{7}\text{.71}\text{×10}{}^{\text{4}}\text{g}\text{-}\text{1}{\text{.06×10}}^{\text{4}}\text{g}\\ \text{=}\text{6}{\text{.65×10}}^{\text{4}}\text{g}\\ \end{array}$

To determine: Mass of hot air when if it were on the ground in Denver, Colorado

Mass of hot air $=5.66\times 10^{4}g$

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Here, from the given data,

$\begin{array}{l}\text{P}\text{=630torr}\text{=}\text{0}\text{.828atm}\text{Since,}\text{1atm=760}\text{torr}\\ 630\text{torr=}\frac{\text{630}\text{torr}}{\text{760}\text{torr}}\\ \text{=0}\text{.828}\text{atm}\\ \text{V}\text{=}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}\\ \text{T=65°C}\text{=}\text{338K}\text{Since,K}\text{=°C}\text{+}\text{273}\\ \text{=}\left(\text{65°C+273}\right)\text{K}\\ \text{=338}\text{K}\\ \text{R=}\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\\ \text{n=}\frac{\text{P}\text{V}}{\text{RT}}\text{=}\frac{\text{0}\text{.828}\text{atm×}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}}{\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\text{×}\text{338}\text{K}}\\ \text{=1}\text{.95×}{\text{10}}^{\text{3}}\text{mol}\text{air}\\ \text{Mass}\text{of}\text{hot}\text{air=number}\text{of}\text{moles}\text{of}\text{air}\text{×}\text{molar}\text{mass}\text{of}\text{hot}\text{air}\\ \text{Molar}\text{mass}\text{of}\text{hot}\text{air}\text{=}\text{29}\text{.0g/mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{air=}1.95\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{Mass}\text{of}\text{hot}\text{air}\text{=}1.95\text{×}{\text{10}}^{\text{3}}\text{mol}\text{×}\text{29}\text{.0g/mol}\\ \text{=}5.66{\text{×10}}^{\text{4}}\text{g}\\ \\ \end{array}$

To determine: Mass of air displaced when if it were on the ground in Denver, Colorado

Mass of air displaced= $=6.53\times 10^{4}g$

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Here, from the given data,

$\begin{array}{l}\text{P}\text{=630torr}\text{=}\text{0}\text{.828atm}\text{Since,}\text{1atm=760}\text{torr}\\ \text{630}\text{torr=}\frac{\text{630}\text{torr}}{\text{760}\text{torr}}\\ \text{=0}\text{.828}\text{atm}\\ \text{V}\text{=}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}\\ \text{T=21°C}\text{=}\text{294K}\text{Since,K}\text{=°C}\text{+}\text{273}\\ \text{=}\left(\text{21°C+273}\right)\text{K}\\ \text{=294}\text{K}\\ \text{R=}\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\\ \text{n=}\frac{\text{P}\text{V}}{\text{RT}}\text{=}\frac{\text{0}\text{.828}\text{atm×}\text{6}\text{.54}{\text{×10}}^{\text{4}}\text{L}}{\frac{\text{0}\text{.08206}\text{Latm}}{\text{K}\text{mol}}\text{×}\text{294}\text{K}}\\ \text{=2}\text{.25×}{\text{10}}^{\text{3}}\text{mol}\text{air}\\ \text{Mass}\text{of}\text{hot}\text{air=number}\text{of}\text{moles}\text{of}\text{air}\text{×}\text{molar}\text{mass}\text{of}\text{hot}\text{air}\\ \text{Molar}\text{mass}\text{of}\text{hot}\text{air}\text{=}\text{29}\text{.0g/mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{air=}\text{2}\text{.25}\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{Mass}\text{of}\text{hot}\text{air}\text{=}\text{2}\text{.25}\text{×}{\text{10}}^{\text{3}}\text{mol}\text{×}\text{29}\text{.0g/mol}\\ \text{=}\text{6}\text{.53}{\text{×10}}^{\text{4}}\text{g}\text{of air}\text{displaced}\text{.}\\ \\ \end{array}$

To determine: Total mass that can be lifted by a balloon, if it were on the ground in Denver, Colorado

Total mass that can be lifted by a balloon, , if it were on the ground in Denver, Colorado

 = $8.7\times {\text{10}}^{3}g$

Total mass that can be lifted by a balloon is the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon.

$\begin{array}{l}\text{Total mass that can be lifted by a balloon=Mass}\text{of}\text{air}\text{diaplaced-Mass }\text{of}\text{hotair}\\ \text{=}\text{6}\text{.53×10}{}^{\text{4}}\text{g}\text{-}\text{5}{\text{.66×10}}^{\text{4}}\text{g}\\ \text{=}\text{8}{\text{.7×10}}^{\text{3}}\text{g}\\ \end{array}$

Conclusion

Conclusion

For the given conditions, total mass that can be lifted by a balloon is determined by calculating difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon