# Do Different Volumes of Helium Affect Molecular Speed in Balloons?

• Vibhor
In summary, the balloon with 20 liters of Helium gas will have molecules moving at a slower average speed compared to the balloon with 10 liters of Helium gas, assuming the same pressure and ambient temperature. This is because the number of moles and volume are directly proportional, but the temperature is constant. In a different scenario with two party balloons filled with air, the balloon with 2 liters of air will have a higher number of moles and the same temperature as the balloon with 1 liter of air, resulting in molecules moving at the same average speed.
Vibhor

## Homework Statement

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Two balloons with Helium gas are filled, first with 10 liters of He and second with 20 liters. Molecules of which balloon will be moving faster as compared to the other?

## The Attempt at a Solution

Speed of molecules is directly proportional to temperature . Consider the fact that pressure inside balloon is same as ambient pressure i.e pressure is constant ( Atm. pressure) . Applying Ideal gas law PV = nRT ,temperature depends on 'n' and V . Volume is given but , 'n' is not given .

Is the question incomplete or am I missing something ?

Many Thanks

Vibhor said:
10 liters
Vibhor said:
20 liters
Vibhor said:
'n' is not given
Vibhor said:
or am I missing something ?
You "am" missing something.

Are you suggesting that 'n' i.e number of moles can be calculated from the volume given ? Should n1 = 10/22.4 and n2 = 20/22.4 ?

Preference would be some unspecified constant, but otherwise, yes.

But what I did (division by 22.4) was under the assumption that 1 mole of an ideal gas occupies 22.4 liters at "STP" i.e at 273K . Why should we assume that gas is filled at a constant temperature of 273K ?

Vibhor said:
Why should we assume that gas is filled at 273K ?
Hence, the UN-specified constant.

Sorry . I do not understand . Could you please explain ?

Vibhor said:
You multiply by a constant; you then divide by that same constant; what do you get? Rid of the constant.

There is not enough information in the problem statement. I assume the balloons are initially identical, but are we considering them immediately after filling or on return to ambient temperature?
In the attempted solution it says the balloon contents are at atmospheric pressure, which seems unlikely. Is this a given or just part of the attempted solution?

haruspex said:
There is not enough information in the problem statement. I assume the balloons are initially identical, but are we considering them immediately after filling or on return to ambient temperature?
I interpret this as immediately after filling .

haruspex said:
In the attempted solution it says the balloon contents are at atmospheric pressure, which seems unlikely. Is this a given or just part of the attempted solution?

I think the gas inside the rubber party balloons are always at a pressure just above the ambient pressure (approx. equal to ambient pressure ) . It is part of the attempted solution.

Vibhor said:
I interpret this as immediately after filling .
Well, that is rather a critical question, no? But if so, what would that say about the temperatures?

Vibhor said:
I think the gas inside the rubber party balloons are always at a pressure just above the ambient pressure (approx. equal to ambient pressure ) . It is part of the attempted solution.
Either the volumes are quite different or the pressures are. Or both. Without knowing the relaxed volumes of the balloons it is impossible to say how the two volumes and pressures will change. But as I hinted above, I don't think you need to investigate that beyond some basic assumptions.

haruspex said:
Well, that is rather a critical question, no? But if so, what would that say about the temperatures?

This is what we need to find . Translational speed of molecules is directly proportional to temperature . Higher the temperature higher the speed.

haruspex said:
Either the volumes are quite different or the pressures are.

Volume of one is double that of the other . Pressures are same . This is my thinking .

Vibhor said:
This is what we need to find . Translational speed of molecules is directly proportional to temperature . Higher the temperature higher the speed.
Sure, but you are only assuming the question relates to just after filling the balloons. That assumption affects the answer.
How far do you think 'adiabatic' would apply here?
Vibhor said:
Volume of one is double that of the other . Pressures are same . This is my thinking .
It would be somewhat strange for a balloon to be filled to double the size without some increase in pressure. It need not be much to affect the answer.

They are not party balloons I am sure. Can be meteorologic ones - of the same volume when filled, which means when they are taut. That also means about the same pressure, a bit higher than atmospheric. But then the temperature must be different

Vibhor
ehild said:
They are not party balloons I am sure. Can be meteorologic ones - of the same volume when filled, which means when they are taut. That also means about the same pressure, a bit higher than atmospheric. But then the temperature must be different

In that case the 20L would have double the number of moles and half the temperature as compared to 10L one ??

Vibhor said:
In that case the 20L would have double the number of moles and half the temperature as compared to 10L one ??
I think so.

Vibhor
ehild said:
I think so.

Fine . And molecules in 20L balloon would be moving at slower(half) average speed than the 10L balloon ??

What if we change the question a bit and take two party balloons and we put in 1L and 2L of air ( i.e volume of air inside =volume of balloon) . Do you think we would consider temperatures to be same (ambient temperatures) ??

Since Pressure and temperatures would be same in this case , the Volume would be directly proportional to the no of moles of air put in . The number of moles in 2L balloon would be double that of 1L balloon .But since temperatures are equal , in this case molecules in 2L balloon would be moving at same average speeds as that in 1L balloon

Does that seem correct ??

Vibhor said:
Since Pressure and temperatures would be same in this case , the Volume would be directly proportional to the no of moles of air put in . The number of moles in 2L balloon would be double that of 1L balloon .But since temperatures are equal , in this case molecules in 2L balloon would be moving at same average speeds as that in 1L balloon

Does that seem correct ??
Yes, in case of ideal gas.
But how can you fill one party balloon with twice as much gas so as the pressure is the same in both? The pressure should balance the tension of the rubber wall, which must be higher if the volume is greater.

Vibhor
Ok. How would two meteorological balloons as we interpret in the OP have equal gas pressures i.e same as ambient pressure ? Are they closed or do they have an opening ?

Vibhor said:
In that case the 20L would have double the number of moles and half the temperature as compared to 10L one ??
How can it have half the temperature?!
You put 10L in each balloon, so far they're the same. Now you put another 10L in one of them. Why should the temperature go down dramatically? Half the temperature (K) would be extremely cold.

Pretend you blow the content of your lung in one party balloon. It is not flat any more, but It is small and easy to press. .
You blow air twice in the other balloon. It becomes round and hard to press. What you need to do with the first one to make it as big and taut as the second balloon?

ehild said:
Pretend you blow the content of your lung in one party balloon. It is not flat any more, but It is small and easy to press. .
You blow air twice in the other balloon. It becomes round and hard to press. What you need to do with the first one to make it as big and taut as the second balloon?
I started off down that line, but unfortunately we do not know whether that model applies. At 10L plus, they do sound more like meteorological balloons, which don't become taut. So now I'm thinking they could be, as originally suggested, both at pretty much atmospheric pressure. Several ways in which the information is inadequate.

I agree that the information is inadequate.

Vibhor

Vibhor said:
Ok. How would two meteorological balloons as we interpret in the OP have equal gas pressures i.e same as ambient pressure ? Are they closed or do they have an opening ?
The weather balloons are closed. The hot-air balloons are open.
The pressure in the balloon should balance the ambient pressure + the tension of the wall. The weather balloons are blown up to a certain size.
I do not really know what was in the mind of the problem writer.

Vibhor

## 1. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is represented as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature.

## 2. How does the Ideal Gas Law apply to balloons?

The Ideal Gas Law applies to balloons because balloons are filled with gas, which behaves like an ideal gas. As the gas inside the balloon is heated, it expands, causing the volume of the balloon to increase. Similarly, if the gas inside the balloon is cooled, it contracts, causing the volume of the balloon to decrease. This relationship between pressure, volume, and temperature is described by the Ideal Gas Law.

## 3. Can the Ideal Gas Law be applied to all gases?

The Ideal Gas Law can only be applied to ideal gases, which are gases that follow all the assumptions of the kinetic molecular theory. These assumptions include that the gas particles are point masses with no volume, there are no intermolecular forces, and the collisions between particles are perfectly elastic. Real gases deviate from these assumptions and may require different equations to accurately describe their behavior.

## 4. How do changes in temperature affect the volume of a balloon?

According to the Ideal Gas Law, as the temperature of a gas increases, its volume also increases. This is because the increased temperature causes the gas particles to move faster and collide with the walls of the balloon more frequently and with more force, leading to an increase in volume. Conversely, as the temperature decreases, the volume of the balloon decreases as well.

## 5. Can the Ideal Gas Law be used to predict the behavior of a balloon at all altitudes?

The Ideal Gas Law can be used to predict the behavior of a balloon at different altitudes, as long as the gas inside the balloon behaves like an ideal gas. However, at higher altitudes, the air pressure decreases, which can cause the balloon to expand and potentially burst if not designed properly. This is due to the decrease in external pressure, allowing the gas inside the balloon to expand more freely. Other factors such as temperature and humidity can also affect the behavior of a balloon at different altitudes.

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