# Why do hot air balloons float

## General gas law

"For a body to swim in water, it has to be lighter than water." As a physically trained person, you must immediately protest and correct: "For a body to swim in water, its density must be less than the density of water."

The observation made with liquids can now be transferred to the "sea of ​​air" in which we live: "In order for a body to rise in the air, its density must be less than the density of the air." Colloquially (not physically correct) one often hears: "The body has to be lighter than air."

For a more precise understanding of balloon physics, one has to use two physical laws: The law of buoyancy of ARCHIMEDES and the general gas law.

Archimedes' Law - or - Why does a balloon have to be so big?

We want to do that again first ARCHIMEDES'S LAW Remember: The buoyancy \ ({F _ {\ rm {A}}} \) of a balloon is equal to the weight of the displaced (outer) air. In formulas
\ [{F _ {\ rm {A}}} = {F _ {{\ rm {G, displaced \; air}}}} = {m _ {{\ rm {displaced \; air}}}} \ cdot g \ ]
If the balloon has the volume \ (V \) and the displaced (outer) air has the density \ ({\ rho _ {\ rm {a}}} \) then also applies
\ [{F _ {\ rm {A}}} = {\ rho _ {\ rm {a}}} \ cdot V \ cdot g \]
The density of the air under normal conditions (\ (\ vartheta = 0 ^ \ circ C \) and \ (p = 1013 {\ rm {hPa}} \)) is \ ({\ rho _ {\ rm {0}}} = 1.3 \ frac {{{\ rm {kg}}}} {{{{\ rm {m}} ^ {\ rm {3}}}}} \). If, for example, \ (1 \ rm {m ^ 3} \) air is displaced under normal conditions, a buoyancy force of
\ [{F _ {\ rm {A}}} = 1.3 \ frac {{{\ rm {kg}}}} {{{{{\ rm {m}} ^ {\ rm {3}}}}} \ cdot 1 {{\ rm {m}} ^ {\ rm {3}}} \ cdot 10 \ frac {{\ rm {N}}} {{{\ rm {kg}}}} = 13 {\ rm {N}} \]

Calculate the volume that a body of mass \ (1 \ rm {kg} \) would have to have so that it has the same density as air, i.e. it would float in air.

\ [m = \ rho \ cdot V \ Leftrightarrow V = \ frac {m} {\ rho} \ Rightarrow V = \ frac {{1 {\ rm {kg}}}}} {{1,3 \ frac {{{ \ rm {kg}}}} {{{{{\ rm {m}} ^ {\ rm {3}}}}}}} = 0.8 {{\ rm {m}} ^ {\ rm {3} }} \]

Why does a balloon need a gas filling?

From the result of the task it is clear that for ballooning only a lot voluminous body come into question. Very early on (1670) the idea was to take a large evacuated ball into the air, However, such a ball would only withstand the external air pressure if its shell were made of very stiff (and therefore very heavy) material.

A way out of the dilemma is to fill the balloon with gas. So it is possible to use a slack (and therefore very light) cover. The gas filling has the task of generating the necessary internal pressure so that the balloon does not collapse. In addition, the filling must be lighter than the displaced air.

Condition for the balloon to take off

In order for the balloon to take off, the buoyancy force must be greater than the weight. With the specified counting direction, the resulting force \ (F _ {\ rm {res}} \) must be greater than zero
\ [F _ {\ rm {res}} = {F _ {\ rm {A}}} - {F _ {\ rm {G}}}> 0 \]
The weight of the balloon is made up of several components:
• the weight of the balloon envelope
• the weight of the gas filling (\ (\ rho = {\ rho _i} \)) in the balloon
If one assumes for an initial calculation that the weight of the shell, basket and load is negligible, then the following applies
\ [{F _ {{\ rm {res}}}} = {\ rho _a} \ cdot V \ cdot g - {\ rho _i} \ cdot V \ cdot g = \ left ({{\ rho _a} - { \ rho _i}} \ right) \ cdot V \ cdot g> 0 \]
This relationship clearly shows that the resulting force is only greater than zero if the density of the outer air is greater than the density of the (inner) gas filling. Here are examples for some density values ​​of gases under normal conditions ((\ (\ vartheta = 0 ^ \ circ C \) and \ (p = 1013 {\ rm {hPa}} \)))

 gas air helium hydrogen density \ (1.3 \ frac {{{\ rm {kg}}}} {{{{{\ rm {m}} ^ {\ rm {3}}}}}} \) \ (0.18 \ frac {{{\ rm {kg}}}} {{{{\ rm {m}} ^ {\ rm {3}}}}} \) \ (0.090 \ frac {{{\ rm {kg}}}} {{{{\ rm {m}} ^ {\ rm {3}}}}}} \)

a)

Calculate the resulting force on a balloon with \ (100 {{\ rm {m}} ^ {\ rm {3}}} \) volume under normal conditions and (i) a hydrogen filling and (ii) a helium filling. For the sake of simplicity, it is assumed that the weight of the shell and the payload are negligible.

(i) \ ({F _ {{\ rm {res}} {\ rm {, H}}}} = \ left ({1,3 \ frac {{{{\ rm {kg}}}} {{{\ rm {1 }} {{\ rm {m}} ^ {\ rm {3}}}}} - 0.090 \ frac {{{\ rm {kg}}}} {{{\ rm {1}} {{\ rm { m}} ^ {\ rm {3}}}}}} \ right) \ cdot 100 {{\ rm {m}} ^ {\ rm {3}}} \ cdot 10 \ frac {{\ rm {N} }} {{{\ rm {kg}}}} = 1200 {\ rm {N}} \)

(ii) \ ({F _ {{\ rm {res}} {\ rm {, He}}}} = \ left ({1,3 \ frac {{{{\ rm {kg}}}} {{{\ rm {1 }} {{\ rm {m}} ^ {\ rm {3}}}}} - 0.18 \ frac {{{\ rm {kg}}}} {{{\ rm {1}} {{\ rm {m}} ^ {\ rm {3}}}}}} \ right) \ cdot 100 {{\ rm {m}} ^ {\ rm {3}}} \ cdot 10 \ frac {{\ rm { N}}} {{{\ rm {kg}}}} = 1100 {\ rm {N}} \)

b)

A \ (100 {{\ rm {m}} ^ {\ rm {3}}} \) balloon, which is filled with luminous gas, experiences part of the task under the simplifying conditions a) a resulting force of \ (490 {\ rm {N}} \). From this, calculate the density of luminous gas.

\ [{F _ {{\ rm {res}}}} = \ left ({{\ rho _ {\ rm {a}}} - {\ rho _ {\ rm {i}}}} \ right) \ cdot V \ cdot g \ Leftrightarrow {\ rho _ {\ rm {a}}} - {\ rho _ {\ rm {i}}} = \ frac {{{F _ {{\ rm {res}}}}}} {{V \ cdot g}} \ Leftrightarrow {\ rho _ {\ rm {i}}} = {\ rho _ {\ rm {a}}} - \ frac {{{F _ {{\ rm {res}} }}}} {{V \ cdot g}} \]
Substituting the given values ​​yields
\ [{\ rho _ {\ rm {i}}} = 1.3 \ frac {{{\ rm {kg}}}} {{{{\ rm {m}} ^ {\ rm {3}}} }} - \ frac {{490 {\ rm {N}}}} {{100 {{\ rm {m}} ^ {\ rm {3}}} \ cdot 10 \ frac {{\ rm {N}} } {{{\ rm {kg}}}}}} = 0.80 \ frac {{{\ rm {kg}}}} {{{{{\ rm {m}} ^ {\ rm {3}}} }} \]

The trick with the hot air balloons

In addition to filling a closed balloon with a gas of lower density (gas balloon), there is also the option of heating the air in the balloon (hot air balloon). As you know, at higher temperatures (and the same pressure) air takes up a larger volume and therefore has a lower density. In order to be able to quantitatively record the conditions in hot air balloons, we need the general gas law. This reads
\ [\ frac {{p \ cdot V}} {T} = {\ rm {const}} {\ rm {.}} \; \; \; {\ rm {or}} \; \; \; \ frac {{{p_1} \ cdot {V_1}}} {{{T_1}}} = \ frac {{{p_2} \ cdot {V_2}}} {{{T_2}}} \]
If the air in the balloon (\ ({{p_1}} \), \ ({{V_1}} \), \ ({{T_1}} \)) is warmed up, it changes to the new state (\ ({{{ p_2}} \), \ ({{V_2}} \), \ ({{T_2}} \)) over. Since the hot air balloon is open, \ ({p_1} = {p_2} = {p_a} \) applies. This simplifies the general gas equation and you can now calculate the new volume of the hot gas according to the so-called law of GAY and LUSSAC:
\ [{V_2} = {V_1} \ cdot \ frac {{{T_2}}} {{{T_1}}} \]
Hot air with the volume escapes from the balloon
\ [\ Delta V = {V_2} - {V_1} = {V_1} \ cdot \ frac {{{T_2}}} {{{T_1}}} - {V_1} = {V_1} \ cdot \ left ({\ frac {{{T_2}}} {{{T_1}}} - 1} \ right) \]
For example, if a third of the gas originally present escapes, the mass of the gas still in the balloon is two thirds of the initial mass. The following applies to the density of the gas in the heated balloon
\ [{\ rho _ {\ rm {i}}} = \ frac {{\ frac {2} {3} \ cdot m}} {V} = \ frac {2} {3} \ cdot \ frac {m } {V} = \ frac {2} {3} \ cdot {\ rho _ {\ rm {a}}} \]
Now you can use the formulas developed above to calculate the resulting force on the balloon.

If you feel like an expert and don't just want to be satisfied with the numerical example, but want to go a little deeper, you can display and understand the conversion of the general gas equation.

As already explained above, one needs the general gas equation for calculations on the hot air balloon and later the density of the air hot air \ ({\ rho _ {\ rm {i}}} \) inside the balloon and to calculate the resulting force on the balloon the density \ ({\ rho _ {\ rm {a}}} \) of the outside air. The following shows that the general gas equation can be rewritten in such a way that the density \ ({\ rho _ {\ rm {i}}} \) can be calculated directly with it.

First, the constant in the general gas law should be examined more closely. It can be shown that it is proportional to the mass of the enclosed gas. This is suggested by the following thought experiment:

Imagine two containers with the same volume, each of which contains gas of mass \ (m \) at pressure \ (p \) and temperature \ (T \). For every container \ (\ frac {{p \ cdot V}} {T} = {{\ rm {c}} _ 1} \) applies.

Now a connection is made between these two containers (the volume of the connection is negligible). How does this change the pressure and temperature in the containers? Correct! Nothing changes at all.

One can imagine the two containers fused into one, which has twice the volume and contains gas of twice the mass - with the same \ (p \) and \ (T \). Then \ (\ frac {{p \ cdot 2 \ cdot V}} {T} = {{\ rm {c}} _ 2} \) applies.

Conclusion: At the same pressure and temperature, a gas with \ (n \) times the volume also has \ (n \) times the mass. We have \ (V \ sim m \). This result can be taken into account in the general gas equation as follows:
\ [\ frac {{p \ cdot V}} {T} = k \ cdot m \ Leftrightarrow \ frac {m} {V} = \ frac {p} {{T \ cdot k}} \ quad (1) \ ]
Using \ ((1) \) the following then applies to the density of the gas
\ [m = \ rho \ cdot V \ Leftrightarrow \ rho = \ frac {m} {V} = \ frac {p} {{T \ cdot k}} \ quad (2) \]
If one writes the equation \ ((2) \) for normal conditions, then the following applies
\ [{\ rho _0} = \ frac {{{p_0}}} {{{T_0} \ cdot k}} \ Leftrightarrow \ frac {1} {k} = \ frac {{{\ rho _0} \ cdot { T_0}}} {{{p_0}}} \ quad (3) \]
If one replaces the factor \ (\ frac {1} {k} \) in equation \ ((2) \) according to equation \ ((3) \), it follows
\ [\ rho = \ frac {p} {{T \ cdot k}} = \ frac {p} {T} \ cdot \ frac {1} {k} = \ frac {p} {T} \ cdot \ frac {{{\ rho _0} \ cdot {T_0}}} {{{p_0}}} \ Leftrightarrow \ frac {\ rho} {{{\ rho _0}}} = \ frac {p} {{{p_0}} } \ cdot \ frac {{{T_0}}} {T} \ quad (4) \]
Equation \ ((4) \) represents the general gas law. In this form, one can calculate the density of the gas at the temperature \ (T \) and the pressure \ (p \) (both measurands) by using the density at Knows normal conditions.

If one assumes that the density of the outer air \ ({{\ rho _ {\ rm {0}}}} \), the outside temperature \ ({{T_0}} \) and the outside pressure \ ({{{ p_0}} \), one can write for the density of the gas inside due to the equality of \ (p \) and \ ({{p_0}} \) (hot air balloon is open)
\ [\ rho = {\ rho _ {\ rm {i}}} = \ frac {{{T_0}}} {{{T _ {\ rm {i}}}}} \ cdot {\ rho _0} \ quad (5) \]
We remember the formula for the resulting force on the balloon (if envelope and payload can be neglected):
\ [{F _ {{\ rm {res}}}} = \ left ({{\ rho _ {\ rm {a}}} - {\ rho _ {\ rm {i}}}} \ right) \ cdot V \ cdot g \ quad (6) \]
If one substitutes \ ((5) \) in \ ((6) \) and takes into account that the density of the outer air is equal to the normal density, then finally applies
\ [{F _ {{\ rm {res}}}} = \ left ({{\ rho _ {\ rm {0}}} - \ frac {{{T_0}}} {{{T _ {\ rm {i }}}}} \ cdot {\ rho _0}} \ right) \ cdot V \ cdot g = \ left ({1 - \ frac {{{T_0}}} {{{T _ {\ rm {i}}} }}} \ right) \ cdot {\ rho _ {\ rm {0}}} \ cdot V \ cdot g \]
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