Getting moles number of Oxygen in Balloon - 65 characters

[cathode-ray]

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.

 

[Redbelly98]

Can you show your work? Specifically, did you calculate the pressure of the oxygen? What answer did you get?

 

[cathode-ray]

I did it this way:

$$n_{O_{2}}=\frac{P_{O_{2}}V_{O_{2}}}{RT_{O_{2}}}=\frac{(P_{rubber}-P_{He})V_{O_{2}}}{RT_{O_{2}}}=(\frac{K}{V_{O_{2}}}-P_{He})\frac{V_{O_{2}}}{RT_{O_{2}}}=\frac{K}{RT_{O_{2}}}-\frac{P_{He}V_{O_{2}}}{RT_{O_{2}}}$$

where:

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

The final result is n=4.02 moles. Even if I hade made a mistake and was missing a factor of $$10^{-2}$$ the result would still be wrong.

 

[Redbelly98]

I did it this way:

$$n_{O_{2}}=\frac{P_{O_{2}}V_{O_{2}}}{RT_{O_{2}}}=\frac{(P_{rubber}-P_{He})V_{O_{2}}}{RT_{O_{2}}}=(\frac{K}{V_{O_{2}}}-P_{He})\frac{V_{O_{2}}}{RT_{O_{2}}}=\frac{K}{RT_{O_{2}}}-\frac{P_{He}V_{O_{2}}}{RT_{O_{2}}}$$

Lets' think about whether it makes sense to equate P_O2 with P_rubber-P_He:

If P_rubber is smaller than P_He, we get a negative pressure for P_O2. This is an impossible, unphysical result.

If P_rubber equals P_He, then we get P_O2=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?

 

[cathode-ray]

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 :(

 

[Redbelly98]

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/V_balloon.

 

[cathode-ray]

That was the first thing that comes to my mind after your previous post . 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 .

Thanks for your help!