Hot air balloon question with bouyancy

[Fadel_K]

Homework Statement

A hot air balloon of mass 350kg is carrying 5 people each of mass 70kg. The total volume of the balloon is 2800m[SUP]3[/SUP]. The balloon flies horizontally in dry air 1km above sea level. The atmospheric pressure at this altitude is 89.9kPa and the surrounding temperature is 9∘C. Given that the molar mass of dry air is 28.97g/mol, work out the temperature of the heated air inside the balloon. (You can take gas constant R=8.31J/mol K and you may assume that air behaves as an ideal gas).

Relevant Equations

mg=ρVg
PV=nRT

In the solution, it is given that the mass of the air inside the balloon at 1 km is 700 kg less than the mass of the air on the ground. I don't understand this intuitively, could someone help me?

 

[BvU]

In the solution, it is given that the mass of the air inside the balloon at 1 km is 700 kg less than the mass of the air on the ground.

Not that mass does NOT depend on height. The book answer calculates the mass of the given volume of air under the conditions that are valid at the position of the balloon (but at the outside temperature 9 $^\circ$C). If the balloon doesn't accelerate upwards or downwards, the 700 kg buoyancy must be the difference between that mass and the mass of the same volume and pressure at the higher temperature inside the balloon .

The 1 km height doesn't appear in the calculation (only indirectly, in the pressure).

And I don't see the word 'ground' appear either

$\$

 

[Steve4Physics]

In the solution, it is given that the mass of the air inside the balloon at 1 km is 700 kg less than the mass of the air on the ground.
View attachment 354404

Hi @Fadel_K. It looks like you have misinterpreted the meaning of $m_{balloon}$ (and possibly the meaning of $m_{atm}$) in the supplied solution.

For the balloon at 1km altitude:
$m_{atm}$ is the mass of air being displaced by the balloon;
$m_{balloon}$ is the mass of hot air inside the balloon.

The total mass of the balloon at 1km altitude is 700kg + $m_{balloon}$ and its total weight is balanced by the upthrust (which equals the weight of displaced air). In terms of masses this gives:
$m_{atm}$ =700kg + $m_{balloon}$

Ideally, the supplied solution should define $m_{atm}$ and $m_{balloon}$ - but doesn't. Also, IMO, the variable names are poor, e.g. they could have better been called $m_{displaced}$ and $m_{hot~air}$.

 

[Lnewqban]

In the solution, it is given that the mass of the air inside the balloon at 1 km is 700 kg less than the mass of the air on the ground. I don't understand this intuitively, could someone help me?

The interior cavity of the more or less rigid balloon is connected to the atmosphere outside it.
As the density of the latter decreases with altitude, so does the former in order to keep balance of the pressure differential, which means less mass is contained in the same interior volume of the balloon.

 

[kuruman]

In the solution, it is given that the mass of the air inside the balloon at 1 km is 700 kg less than the mass of the air on the ground.

That is not given. It is the equation that you are supposed to come up with. Perhaps thinking in terms of forces will help.

According to Archimedes's principle, the buoyant force BF is directed up and is equal to the weight of the displaced air: $$BF=M_{\text{displaced}}~g.$$ This force has to support the weight of the hot air inside the balloon and the weight of the load which consists of the weight of the people and weight of the empty balloon:$$BF=M_{\text{hot air}}~g+M_{\text{people}}~g+M_{\text{balloon}}~g.\tag{1}$$
Now using $\mu$ for the molar mass (28.97 g/mol) to distinguish it from all the other mass symbols,
$$ M_{\text{displaced}}=\frac{\mu P_A V}{RT_A}~;~~M_{\text{hot air}}=\frac{\mu P_A V}{RT}~;~~M_{\text{people}}=350~\text{kg};~~M_{\text{balloon}}=350~\text{kg}.$$Putting all thi in equation (1) gives, after cancelling the common factor $g$, $$\begin{align} & \frac{\mu P_A V}{RT_A}=\frac{\mu P_A V}{RT}+350~\text{kg}+350~\text{kg} \nonumber \\
& \implies \frac{\mu P_A V}{R}\left(\frac{1}{T_A}-\frac{1}{T}\right)=700~\text{kg}. \nonumber
\end{align}$$

 

[Fadel_K]

That is not given. It is the equation that you are supposed to come up with. Perhaps thinking in terms of forces will help.

According to Archimedes's principle, the buoyant force BF is directed up and is equal to the weight of the displaced air: $$BF=M_{\text{displaced}}~g.$$ This force has to support the weight of the hot air inside the balloon and the weight of the load which consists of the weight of the people and weight of the empty balloon:$$BF=M_{\text{hot air}}~g+M_{\text{people}}~g+M_{\text{balloon}}~g.\tag{1}$$
Now using $\mu$ for the molar mass (28.97 g/mol) to distinguish it from all the other mass symbols,
$$ M_{\text{displaced}}=\frac{\mu P_A V}{RT_A}~;~~M_{\text{hot air}}=\frac{\mu P_A V}{RT}~;~~M_{\text{people}}=350~\text{kg};~~M_{\text{balloon}}=350~\text{kg}.$$Putting all thi in equation (1) gives, after cancelling the common factor $g$, $$\begin{align} & \frac{\mu P_A V}{RT_A}=\frac{\mu P_A V}{RT}+350~\text{kg}+350~\text{kg} \nonumber \\
& \implies \frac{\mu P_A V}{R}\left(\frac{1}{T_A}-\frac{1}{T}\right)=700~\text{kg}. \nonumber
\end{align}$$

Thank you so much, your explanation helped me understand it perfectly!