Chapter 5, Problem 120AE

a)

Interpretation Introduction

Interpretation:Mass of that hot air balloon can liftis to be determined.

Concept introduction: Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Explanation of Solution

Expression for volume of sphere is as follows:

  $V=\frac{4}{3}\text{π}{\left(\frac{\text{diameter}}{2}\right)}^3$

Value of diameter is $\text{5}\text{.00 m}$ .

Value of $\text{π}$ is 3.14.

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{4}{3}\left(3.14\right){\left(\frac{\text{5}\text{.00 m}}{2}\right)}^3\\ =65.4{\text{ m}}^3\end{array}$

Conversion of $65°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =65°\text{C}+273\\ =338\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air.

  $n=\frac{PV}{RT}$

Value of $P$ is $745\text{ torr}$ .

Value of $T$ is $338\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(745\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(338\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.31\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.31\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =6.70\times {10}^4\text{ g}\end{array}$

Conversion of $21°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =21°\text{C}+273\\ =294\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air displaced.

  $n=\frac{PV}{RT}$

Value of $P$ is $745\text{ torr}$ .

Value of $T$ is $294\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(745\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(294\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.66\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.66\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =7.71\times {10}^4\text{ g}\end{array}$

Total mass that can lifted by balloon can be calculated as follows:

  $\begin{array}{c}\text{Total mass}=\text{Mass of air displaced}-\text{Mass of hot air}\\ =\text{7}\text{.71}\times {\text{10}}^4\text{ g}-6.70\times {10}^4\text{ g}\\ =1.01\times {10}^4\text{ g}\end{array}$

Hence total mass that can be lifted by balloon is $1.01\times {10}^4\text{ g}$ .

b)

Interpretation Introduction

Interpretation:Total mass that balloon can lift if it filled with helium is to be determined.

Concept introduction: Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

  $P$ is pressure of the gas.

  $V$ is volume of gas.

  $n$ denotes moles of gas.

  $R$ is gas constant.

  $T$ is temperature of gas.

Explanation of Solution

Expression for volume of sphere is as follows:

  $V=\frac{4}{3}\text{π}{\left(\frac{\text{diameter}}{2}\right)}^3$

Value of diameter is $\text{5}\text{.00 m}$ .

Value of $\text{π}$ is 3.14.

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{4}{3}\left(3.14\right){\left(\frac{\text{5}\text{.00 m}}{2}\right)}^3\\ =65.4{\text{ m}}^3\end{array}$

Conversion of $21°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =21°\text{C}+273\\ =294\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for helium.

  $n=\frac{PV}{RT}$

Value of $P$ is $745\text{ torr}$ .

Value of $T$ is $294\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(745\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(294\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.66\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.31\times {10}^3\text{ mol}\right)\left(4.00\text{ g/mol}\right)\\ =9.24\times {10}^3\text{ g}\end{array}$

Hence total mass that can be lifted by balloon is $9.24\times {10}^3\text{ g}$ .

c)

Interpretation Introduction

Interpretation: Mass of that hot air balloon can lift at $630\text{ torr}$ is to be determined.

Concept introduction: Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

  $P$ is pressure of the gas.

  $V$ is volume of gas.

  $n$ denotes moles of gas.

  $R$ is gas constant.

  $T$ is temperature of gas.

Explanation of Solution

Expression for volume of sphere is as follows:

  $V=\frac{4}{3}\text{π}{\left(\frac{\text{diameter}}{2}\right)}^3$

Value of diameter is $\text{5}\text{.00 m}$ .

Value of $\text{π}$ is 3.14.

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{4}{3}\left(3.14\right){\left(\frac{\text{5}\text{.00 m}}{2}\right)}^3\\ =65.4{\text{ m}}^3\end{array}$

Conversion of $65°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =65°\text{C}+273\\ =338\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air.

  $n=\frac{PV}{RT}$

Value of $P$ is $630\text{ torr}$ .

Value of $T$ is $338\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(630\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(338\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =1.95\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(1.95\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =5.66\times {10}^4\text{ g}\end{array}$

Conversion of $21°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =21°\text{C}+273\\ =294\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air displaced.

  $n=\frac{PV}{RT}$

Value of $P$ is $630\text{ torr}$ .

Value of $T$ is $294\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(630\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(294\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.25\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.25\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =6.52\times {10}^4\text{ g}\end{array}$

Total mass that can lifted by balloon can be calculated as follows:

  $\begin{array}{c}\text{Total mass}=\text{Mass of air displaced}-\text{Mass of hot air}\\ =6.52\times {10}^4\text{ g}-5.66\times {10}^4\text{ g}\\ =8.60\times {10}^3\text{ g}\end{array}$

Hence total mass that can be lifted by balloon is $8.60\times {10}^3\text{ g}$ .

d)

Interpretation Introduction

Interpretation: Mass of that hot air balloon can lift at temperature $-8°\text{C}$ is to be determined.

Concept introduction: Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Explanation of Solution

Expression for volume of sphere is as follows:

  $V=\frac{4}{3}\text{π}{\left(\frac{\text{diameter}}{2}\right)}^3$

Value of diameter is $\text{5}\text{.00 m}$ .

Value of $\text{π}$ is 3.14.

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{4}{3}\left(3.14\right){\left(\frac{\text{5}\text{.00 m}}{2}\right)}^3\\ =65.4{\text{ m}}^3\end{array}$

Conversion of $65°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =65°\text{C}+273\\ =338\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air.

  $n=\frac{PV}{RT}$

Value of $P$ is $745\text{ torr}$ .

Value of $T$ is $338\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(745\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(338\text{K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.31\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.31\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =6.70\times {10}^4\text{ g}\end{array}$

Conversion of $-8°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =-8°\text{C}+273\\ =265\text{ K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for air displaced.

  $n=\frac{PV}{RT}$

Value of $P$ is $745\text{ torr}$ .

Value of $T$ is $265\text{ K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $65.4{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\left(\frac{\left(745\text{ torr}\right)\left(65.4{\text{ m}}^3\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(265\text{ K}\right)}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.95\times {10}^3\text{ mol}\end{array}$

Mass of hot air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(2.95\times {10}^3\text{ mol}\right)\left(29.0\text{ g/mol}\right)\\ =8.56\times {10}^4\text{ g}\end{array}$

Total mass that can lifted by balloon can be calculated as follows:

  $\begin{array}{c}\text{Total mass}=\text{Mass of air displaced}-\text{Mass of hot air}\\ =8.56\times {10}^4\text{ g}-6.70\times {10}^4\text{ g}\\ =1.86\times {10}^4\text{ g}\end{array}$

Hence total mass that can be lifted by balloon is $1.86\times {10}^4\text{ g}$ .