The Rising Balloon's

We know that when $\text{temperature increases the space between the molecules (inter molecular space) increases}$ . Also, when temperature increases the volume increases. So, the balloon filled with hot helium will have more volume than the balloon filled with cold helium. Further it is given that same mass of helium is filled so mass is constant .

Now, Density = $\dfrac{Mass}{Volume}$

Now it is clear that when mass is constant, density decreases with increase in volume and vice-versa .

Therefore, the density of the balloon filled with hot helium is less than that of the balloon filled with cold helium. Hence the hot helium balloon will rise higher and faster.

Note : This is the theory behind hot air balloons.

Ideal gas law: PV=nRT

If T increases, what happens? n and R obviously don't change, so it must be P, V, or both. Regarding buoyant force, of P and V, it is volume that matters, because the buoyant force is equivalent to the weight of an equivalent amount of normal atmospheric air that would fill the volume. More volume = more air = more weight = more buoyant force. Balloon skin flexibility tends to allow stable increases in volume more easily than stable increases in pressure (this is why they are inflatable). A rigid container on the other hand, unlike a balloon, might tolerate increases in pressure but would better maintain volume. So higher temperature for a balloon should increase volume, therefore increase buoyant force, therefore increase upward acceleration all else being equal.

Hmmm from experience I said hot air rises, hence the hot balloon rises faster then the cold balloon.

Let's begin by characterising the property which makes the balloons rise.

This phenomenon is called buoyancy , which means fluid (air) exerts forces on objects (balloons) immersed in it.
The strength of this force is dependent on the volume of the fluid (air) the object (balloon) displaces.

Since both of the balloons have the same mass, effect of gravity onto them is the same,
meaning that the one which experiences higher buoyant force will rise faster.
Higher the volume, higher the buoyant force .

Charles's law states that for gases $V \propto T$ when pressure is held constant,
where $V$ and $T$ denote the volume and (kelvin) temperature of the gas.

We can assume that pressure inside the balloons is pretty much constant,
because balloons are made of very flexible materials which would equalize the pressure by just expanding or contracting,
meaning that Charles's law holds for the gas inside the balloons.

This means that the balloon which has higher temperature will have higher volume
which will displace more air creating more buoyant force making the Hot balloon rise faster.

Like ships in water, the balloons in air experience an up thrust equal to the mass of fluid (air in this case) displaced. The warmer balloon has greater volume (air expands with increasing temperature and the balloon stretches) and therefore experiences a greater upthrust and rises faster.

We know that ${\color{#D61F06}hot}$ air is lighter then ${\color{#3D99F6}cold}$ air so the balloon that is heated will rise faster.

The density of hot fluid is always lesser than that of cold fluid,which is why baloon filled with hot air will soar higher compared to the other one filled with cool air.

the heated air moliculs will vibrate and air makes stuff simple

We know that change in temperature occurs due to enrgy change. If we put balloon in cold environment then it will lose some of it's energy and of we put it in hot environment then it will get some more energy which will accelerate the balloon more faster.

If you have ever seen a real balloon that carries people...you have seen they heat it up to ascend! the same principle applies here.

The heated balloon expands whereas the cooled balloon contracts. The gas in the heated (expanded) balloon is less dense than the cooled (contracted) balloon. Hence, the heated balloon will have greater eternal atmospheric pressure exerted on it, causing it to rise at a quicker rate towards the less dense atmosphere located at a higher altitude.

When you heat the helium in the balloon, it becomes less dense as compared to surrounding air. So, the balloon with hot helium tends to float on comparatively cold air because of its less density.

One variation of this question would be which balloon has a faster acceleration.

On hot ballon, volume becomes more than the cold due to thermal expansion of gas so more buoyant force act on it while weight s are same so net upward force on hot ballon is more than the cold ballon.

When molecules cool, they become denser, so the cold balloon will rise slower.

I answered based on my science teacher, when in a pot on the stove the warm water rises and the cold water sinks. This is a similar problem only with different objects.

helium will expand when heated and contracts when cooled, therefore the hot balloon will rise faster.

Conocemos de antemano que los globos con helio flotan porque su densidad es menor que la del aire, así que subirá más rápido el que tenga menor densidad, tenemos que densidad=masa/volumen, así que como las masas son iguales subirá más rápido el que tenga mayor volumen, el volumen se determina con las leyes de los gases que dice que a mayor temperatura mayor volumen, así que subirá más rápido el globo caliente.

Vro y'alls discussion complicated asf. Heat rises so the hot balloon finna rise faster

The kinetic theory of gases: 1/2mv^2=3/2kT

Now, in the case of the same gasses, m = const. Needless to say, k = const. If T increases, v increases. So higher temperature for a balloon should increase speed.

I have been struggling whether the drive force or the air resistance will increase more... which is beyond my knowledge, however , I find the air resistance is proportional to the πr^2 or less( because of its shape), while the motivate force is exactly proportional to (4/3) πr^3 ,for F(up)=V ρ(air) g-V ρ(in)*g=Vg[ρ(air)-ρ(in)] so, as the radius increases, the increase of air resistance is smaller than the increase of drive force by an order of magnitude.

we know that higher temperature increases the volume of the hotter balloon since the volume of gases is directly proportional to the temperature. Since the hotter balloon higher volume as compared to the colder balloon, it will experience more buoyant force by the air surrounding it.

FB = ρf *Vf *g where FB is the buoyant force, ρf is the density of the displaced fluid, Vf is the volume of the displaced fluid, and g is the acceleration due to gravity, 9.8 m/s2.

Also since they have the same mass they will experience the same downward force from the gravity. So the net upward force experienced by the hotter balloon is higher as compared to the colder balloon. Therefore the hotter ballon will rise higher.