Getting moles number of Oxygen in Balloon - 65 characters

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In summary: I was able to solve it. In summary, the oxygen pressure is always higher than the helium pressure due to the elasticity of the rubber.
  • #1
cathode-ray
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Homework Statement


A box with adiabatic and rigid walls, has a volume of 5L. The box is filled with helium and has a balloon inside filled with oxygen. Initially the helium occupies a volume of 4L, is under a pressure of 1atm and is at a temperature of 300 K. The oxygen inside the balloon (with a volume of 1L) is in thermal balance with the helium. Due to the elasticity of the rubber, the oxygen pressure is always higher than the helium pressure.

Suppose that the pressure done by the rubber is given by P_rubber=K/V_balloon, K=0.1 [atm L]. Furthermore in the experience conditions the molar heat capacity(at a constant volume) of the oxygen is equal to 5R/2, R=8.314 [J K^-1 mol^-1]

Homework Equations



Calculate the number of moles of the oxygen inside the balloon.

The Attempt at a Solution



I tried to think, considering the oxygen a ideal gas, but it gives me a different result from the solution( where n=0.045 mol). I also tried to think in some other approach but I wasn't able to get another.
 
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  • #2
Can you show your work? Specifically, did you calculate the pressure of the oxygen? What answer did you get?
 
  • #3
I did it this way:

[tex]n_{O_{2}}=\frac{P_{O_{2}}V_{O_{2}}}{RT_{O_{2}}}=\frac{(P_{rubber}-P_{He})V_{O_{2}}}{RT_{O_{2}}}=(\frac{K}{V_{O_{2}}}-P_{He})\frac{V_{O_{2}}}{RT_{O_{2}}}=\frac{K}{RT_{O_{2}}}-\frac{P_{He}V_{O_{2}}}{RT_{O_{2}}} [/tex]

where:

[tex] V_{O_{2}}=1 L = 1 \times 10^{3} m^{3}[/tex]
[tex] P_{He}=1 atm = 1.013\times10^{5} Pa[/tex]
[tex] R = 8.314 J K^{-1} mol^{-1} [/tex]
[tex] T_{O_{2}}=300 K [/tex]
[tex] K=0.1 atm L [/tex]

The final result is n=4.02 moles. Even if I hade made a mistake and was missing a factor of [tex] 10^{-2} [/tex] the result would still be wrong.
 
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  • #4
cathode-ray said:
I did it this way:

[tex]n_{O_{2}}=\frac{P_{O_{2}}V_{O_{2}}}{RT_{O_{2}}}=\frac{(P_{rubber}-P_{He})V_{O_{2}}}{RT_{O_{2}}}=(\frac{K}{V_{O_{2}}}-P_{He})\frac{V_{O_{2}}}{RT_{O_{2}}}=\frac{K}{RT_{O_{2}}}-\frac{P_{He}V_{O_{2}}}{RT_{O_{2}}} [/tex]

Lets' think about whether it makes sense to equate PO2 with Prubber-PHe:

If Prubber is smaller than PHe, we get a negative pressure for PO2. This is an impossible, unphysical result.

If Prubber equals PHe, then we get PO2=0. But that contradicts the statement that "the oxygen pressure is always higher than the helium pressure".

So, can you rethink how the rubber baloon affects the oxygen pressure, to make it (the oxygen) at a higher pressure than the helium?
 
  • #5
You're right. I didn't thought very well :(

I tried to analyze the situation again.

My thoughts were based in three ideas: the helium makes a pressure on the rubber, equals to the atmospheric pressure; the rubber makes a pressure on the oxygen inside it. Besides there are the pressures that appear as a consequence of the forces originated by the Newton's third law.

I suppose that these ideas were correct to develop my logic, and that they are enough, but I still wasn't able to get the solutions result.

Sorry but can you give me one more hint? I still didn't find the way :(
 
  • #6
If the balloon had no effect, the oxygen pressure would simply be equal to the helium pressure. However, the balloon does have an effect; it exerts additional pressure on the oxygen, in the amount K/Vballoon.
 
  • #7
That was the first thing that comes to my mind after your previous post :smile:. But it still gave me a wrong result.

Now i finally saw where was my mistake: unit conversion in the K constant. After all the problem was just numerics and not logic :smile:.

Thanks for your help!
 

FAQ: Getting moles number of Oxygen in Balloon - 65 characters

What is the process for getting the number of moles of oxygen in a balloon?

To determine the number of moles of oxygen in a balloon, first measure the volume of the balloon using a ruler or measuring cup. Then, calculate the mass of the balloon by weighing it on a scale. Next, use the ideal gas law (PV = nRT) to calculate the number of moles of oxygen present in the balloon.

What is the ideal gas law and how is it used to calculate the number of moles?

The ideal gas law is a mathematical equation that relates the pressure, volume, temperature, and number of moles of a gas. It is written as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. This equation can be rearranged to solve for the number of moles (n = PV/RT).

Why is it important to know the number of moles of oxygen in a balloon?

Knowing the number of moles of oxygen in a balloon is important for several reasons. It can help determine the amount of gas present in the balloon, which is useful for safety purposes. It can also be used to calculate the density of the gas and compare it to other gases. Additionally, it is a fundamental concept in chemistry and can be applied to various real-world scenarios.

What are the units of measurement for the number of moles?

The number of moles is typically measured in moles (mol) or in some cases, in millimoles (mmol). Both units represent the amount of substance, with 1 mol equaling 6.022 x 10^23 particles and 1 mmol equaling 6.022 x 10^20 particles.

Are there any assumptions or limitations when using the ideal gas law to calculate the number of moles?

Yes, there are a few assumptions and limitations when using the ideal gas law. First, it assumes that the gas behaves ideally, which means there are no intermolecular forces or volume occupied by the gas particles. Additionally, it assumes that the gas is at a constant temperature and pressure, and that the gas particles are not interacting with each other. In reality, these conditions may not always be met, so the calculated value for the number of moles may not be completely accurate.

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