Chapter 13, Problem 37P

To determine

To find: The value to the temperature of the air balloon.

To explain: The factors that limit the attainability to reach maximum altitude.

Explanation of Solution

Given:

The volume of the balloon $\mathrm{V}\text{=18}00{\text{m}}^{\text{3}}$

Weight of the balloon + Passenger $\mathrm{w}\text{=27}00\text{N}$

The temperature of the balloon outside = $\mathrm{T}=0°\text{C}$

Value of $\mathrm{g}\text{=9}.\text{8 m}/{\text{s}}^{\text{2}}$

Formula used:

  $\begin{array}{c}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\end{array}$

  $\begin{array}{c}=\frac{\mathrm{m}}{\mathrm{M}}\mathrm{R}\mathrm{T}\end{array}$

  $\begin{array}{c}\frac{\mathrm{P}\mathrm{M}}{\mathrm{R}}=\frac{\mathrm{m}}{\mathrm{V}}\mathrm{T}=\mathrm{\rho }\mathrm{T}\end{array}$

  $\rho T=\text{constant}={(\rho T)}_{\text{cold}}={(\rho T)}_{\text{hot}}$

Calculation:

In this, the three forces appear the buoyant force because of the cold air displaced by the balloon volume, the downward weight (force) of the air balloon+ the downward weight of the passenger in the balloon and apparatus.

When the balloon is going upper with the constant speed at that time the weight is equal to the buoyant force.

  $\begin{array}{l}{\mathrm{F}}_{\text{buoyant}}={\mathrm{m}}_{\text{hot}}+2700\text{N}\\ \mathrm{V}{\mathrm{\rho }}_{\text{cold}}\mathrm{g}=\mathrm{V}{\mathrm{\rho }}_{\text{hot}}\mathrm{g}+2700\text{N}\end{array}$

Now using the ideal gas equation in terms of the density of the gas $\mathrm{\rho }$ and M molecular mass

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ =\frac{\mathrm{m}}{\mathrm{M}}\mathrm{R}\mathrm{T}\\ \frac{\mathrm{P}\mathrm{M}}{\mathrm{R}}=\frac{\mathrm{m}}{\mathrm{V}}\mathrm{T}=\mathrm{\rho }\mathrm{T}\end{array}$

The air balloon inside and outside containing air. So molecular is the same outside as well as inside the air balloon. Hence pressure is also the same if we open the air balloon in the air.

In terms of density, the ideal gas equation is

  $\rho T=\text{constant}={(\rho T)}_{\text{cold}}={(\rho T)}_{\text{hot}}$

  $\begin{array}{c}\mathrm{V}{\mathrm{\rho }}_{\text{Cold}}\mathrm{g}=\mathrm{V}{\mathrm{\rho }}_{\text{hot}}\mathrm{g}+2700\text{N}\end{array}$

  $\begin{array}{c}=\mathrm{V}{\mathrm{\rho }}_{\text{cold}}\frac{{\mathrm{T}}_{\text{cold}}}{{\mathrm{T}}_{\text{hot}}}\mathrm{g}+2700\text{N}\end{array}$

  $\begin{array}{c}{\mathrm{T}}_{\text{hot}}=\frac{\mathrm{V}{\mathrm{\rho }}_{\text{cold}}{\mathrm{T}}_{\text{cold}}\mathrm{g}}{(\mathrm{V}{\mathrm{\rho }}_{\text{cold}}\mathrm{g}-2700\mathrm{N})}\end{array}$

  $\begin{array}{c}=\frac{(1800{\text{m}}^{\text{3}})(1.29{\text{kg/m}}^{\text{3}})(273\text{K})(9.8{\text{m/s}}^{\text{2}})}{[(1800{\text{m}}^{\text{3}})(1.29{\text{kg/m}}^{\text{3}})(9.8{\text{m/s}}^{\text{2}})-2700\text{N}]}\end{array}$

  $\begin{array}{c}=309.8\mathrm{K}\approx {37}^0\mathrm{C}\end{array}$

If the maximum height is limited for the balloon to go highthen density of the air becomes low and at the same time temperature is high. In that case, the air becomes hot and the balloon may be getting damaged.