Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

Explanation of Solution

Given:

Load, $F=110\text{N}$

weight of the balloon’s envelope, $w=50.0\text{}\text{N}$

Volume of the balloon when it is fully inflated, $V=32.0{\text{m}}^{\text{3}}$

Temperature of air, $T=0°\text{C = 273 K}$

Atmospheric pressure, $P=1\text{atm}\text{=1}{\text{.01×10}}^{\text{5}}\text{Pa}$

Net upward force on balloon, $F_{net}=30.0\text{N}$

Formula Used:

Ideal gas law:

  $PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

  $F_b=\rho Vg$

here, $\rho$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

  $\begin{array}{l}{F}_{net}=F_b-F-w-w_{He}\\ F_b={\rho }_{a}Vg\\ w_{He}={\rho }_{He}Vg\\ \Rightarrow F_{net}={\rho }_{a}Vg-F-w-{\rho }_{He}Vg\\ \Rightarrow V=\frac{F_{net}+F+w}{\left({\rho }_a-{\rho }_{He}\right)g}\\ n=\frac{PV}{RT}\\ \Rightarrow n=\frac{P}{RT}\frac{F_{net}+F+w}{\left({\rho }_a-{\rho }_{He}\right)g}\end{array}$

Density of air, ${\rho }_a=1.293{\text{kg/m}}^{\text{3}}$

Density of helium, ${\rho }_{He}=0.179{\text{kg/m}}^{\text{3}}$

Substitute the values and solve:

  $\begin{array}{l}n=\frac{P}{RT}\frac{F_{net}+F+w}{\left({\rho }_a-{\rho }_{He}\right)g}\\ n=\frac{\left(\text{1atm}\right)\left(30.0\text{N}+110\text{N}+50.0\text{N}\right)}{\left(8.21\times {10}^{-2}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{273K}\right)\left(1.293{\text{kg/m}}^{\text{3}}-0.179{\text{kg/m}}^3\right)\left(\text{9}{\text{.81m/s}}^{\text{2}}\right)\left(\text{1L}/{10}^{-3}{\text{m}}^{\text{-3}}\right)}\\ n\approx 776\text{mol}\end{array}$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776\text{mol}$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

Explanation of Solution

Given:

Load, $F=110\text{N}$

weight of the balloon’s envelope, $w=50.0\text{}\text{N}$

Volume of the balloon when it is fully inflated, $V=32.0{\text{m}}^{\text{3}}$

Temperature of air, $T=0°\text{C = 273 K}$

Atmospheric pressure, $P=1\text{atm}\text{=1}{\text{.01×10}}^{\text{5}}\text{Pa}$

Net upward force on balloon, $F_{net}=30.0\text{N}$

Formula Used:

Variation of pressure with altitude ( h ):

  $\begin{array}{l}{P}_h=P_o{e}^{-Ch}\\ C=0.1\text{3}{\text{km}}^{\text{-1}}\end{array}$

From ideal gas law:

  $PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

  $\begin{array}{l}{P}_h=P_o{e}^{-Ch}\\ \Rightarrow h=\frac{1}{C}\ln \left(\frac{P_o}{P_h}\right)\\ P_h=\frac{nRT}{V}\\ h=\frac{1}{C}\ln \left(\frac{P_o}{\frac{nRT}{V}}\right)\\ \Rightarrow h=\frac{1}{C}\ln \left(\frac{P_{o}V}{nRT}\right)\end{array}$

Substitute the values and solve:

  $\begin{array}{c}h=\frac{1}{C}\ln \left(\frac{P_{o}V}{nRT}\right)\\ =\frac{1}{0.13{\text{km}}^{\text{-1}}}\ln \left(\frac{1.00\text{atm}\times 32\times \frac{1\text{L}}{{10}^{-3}{\text{m}}^{\text{3}}}}{\left(\text{776}\text{mol}\right)\left(8.206\times {10}^{-2}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(273\text{K}\right)}\right)\\ =4.7\text{km}\end{array}$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

Explanation of Solution

Given:

Load, $F=110\text{N}$

Weight of the balloon’s envelope, $w=50.0\text{}\text{N}$

Volume of the balloon when it is fully inflated, $V=32.0{\text{m}}^{\text{3}}$

Temperature of air, $T=0°\text{C = 273 K}$

Atmospheric pressure, $P=1\text{atm}\text{=1}{\text{.01×10}}^{\text{5}}\text{Pa}$

Net upward force on balloon, $F_{net}=30.0\text{N}$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $\left(F_b\right)$ must be at least equal to the total weight of the balloon $w_t$ or should be greater than it.

  $F_{net}=F_b-w_t\ge 0$

  $\begin{array}{l}{w}_t=F+w+w_{He}\\ w_t=F+w+{\rho }_{He}Vg\\ F_b={\rho }_{a,h}Vg\\ \frac{P_h}{P_o}=\frac{{\rho }_{a,h}}{{\rho }_a}\\ \Rightarrow {\rho }_{a,h}=\frac{P_h}{P_o}{\rho }_a\\ F_b={\rho }_{a,h}Vg\\ \Rightarrow F_b=\frac{P_h}{P_o}{\rho }_{a}Vg\\ P_h=P_o{e}^{-Ch}\\ F_{net}=e^{-Ch}Vg\left({\rho }_a-{\rho }_{He}\right)-F-w\end{array}$

Calculations:

Substitute the values and solve:

  $\begin{array}{c}{F}_{net}=e^{-Ch}Vg\left({\rho }_a-{\rho }_{He}\right)-F-w\\ =e^{-\left(0.13{\text{km}}^{\text{-1}}\right)\left(4.7\text{km}\right)}\left(32.0{\text{m}}^{\text{3}}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(1.293{\text{kg/m}}^3-0.179{\text{kg/m}}^{\text{3}}\right)-110\text{N}-5\text{0}\text{N}\\ =30\text{N}\end{array}$

Because $F_{net}>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

Explanation of Solution

Given:

Load, $F=110\text{N}$

Weight of the balloon’s envelope, $w=50.0\text{}\text{N}$

Volume of the balloon when it is fully inflated, $V=32.0{\text{m}}^{\text{3}}$

Temperature of air, $T=0°\text{C = 273 K}$

Atmospheric pressure, $P=1\text{atm}\text{=1}{\text{.01×10}}^{\text{5}}\text{Pa}$

Net upward force on balloon, $F_{net}=30.0\text{N}$

Formula Used:

The balloon will rise till the net force on it is zero.

  $F_{net}=F_b-w_t=0$

  $\begin{array}{l}h=\frac{1}{C}\ln \frac{{\rho }_a}{{\rho }_{a,h}}\\ {\rho }_{a,h}=\frac{F_b}{Vg}\\ h=\frac{1}{C}\ln \left[\frac{{\rho }_a}{\frac{F_b}{Vg}}\right]\\ h=\frac{1}{C}\ln \left[\frac{Vg{\rho }_a}{F_b}\right]\end{array}$

Calculations:

Substitute the values and solve:

  $\begin{array}{l}h=\frac{1}{0.13{\text{km}}^{\text{-1}}}\ln \left[\frac{\left(\text{32}\text{.0}{\text{m}}^{\text{3}}\right)\left(\text{9}{\text{.81m/s}}^{\text{2}}\right)\left(1.293{\text{kg/m}}^{\text{3}}\right)}{190.5\text{N}}\right]\\ h=5.8\text{km}\end{array}$

Conclusion:

The maximum altitude attained by the balloon is $5.8\text{km}$ .