# How Do You Calculate the New Density of Kr in a Balloon?

• morechem28
In summary: B?b) a pressure of 1 atm and a density of 1.04g/cm³ in B?c) a pressure of 1 atm and a density of 1.04g/cm³ and a volume of 22,41 L in B?In summary, B has to have a pressure of 1 atm and a density of 1.04g/cm³ in order to have a volume of 22,41 L.
As the balloon goes down, it's compressed more and more. If you hold it at any depth, the pressure of the gas must match the pressure outside. If it didn't the balloon would expand or be compressed until it does. Thus, as the balloon goes deeper, its density increases because the mass of the gas is constant while the volume decreases. There is a specific depth at which the pressure is such that the density of the gas matches the density of seawater. Say that with an equation and remember to use subscripts.

Thank you. And as doing so, will the sum: p (atm) + p (hydr) be the right substituent for p?

And what is p(hydr) equal to?

morechem28
The density of the fluid * the depth to which the balloon would be pressed * gravitational acceleration. Is that right?

Let's call the critical pressure at which the density of the gas matches the density of water ##p_{cr.}##. Write an equation for it.

I have to quit now and tend to other business. See how far you can go.

Thank you, I really appreciate your help!

So, it would be, as I've indicated before: ρ (Kr) = (p*M)/(R*T), and then: p (cr) = (ρ (Kr) * R * T)/M (Kr). Then the p (cr) = 307 633,87 Pa. Could this be right?

Then, p (hydr) = p (cr) - p (atm) = 307 633,87 - 101 325 = 206 308,87 Pa. Now, I would calculate the depth: p (hydr) = ρ (Kr) * h * g; h = p (hydr)/(ρ (Kr) * g). It seems way too easy to be right. :D But at least I've tried.

morechem28 said:
Then, p (hydr) = p (cr) - p (atm) = 307 633,87 - 101 325 = 206 308,87 Pa. Now, I would calculate the depth: p (hydr) = ρ (Kr) * h * g; h = p (hydr)/(ρ (Kr) * g). It seems way too easy to be right. :D But at least I've tried.
That's not what I get. I get about 30 MPa = 300 Bars. Let's see your arithmetic and your management of units.

morechem28 and hutchphd
Chestermiller said:
That's not what I get. I get about 30 MPa = 300 Bars. Let's see your arithmetic and your management of units.

morechem28, hutchphd and Chestermiller
kuruman said:
Thank you, yes. I should divide (density*R*T) by 0,0838 kg/mol, not 83,8 g/mol. Now, I get 30.76 (p (cr)) MPa. And now, I'd do this: p(cr) - p(atm) = p(hydr), p(hydr) = 30 763 387.4 - 101 325 = 30 662 062.4 Pa. And then, given that: p (hydr) = density*h*g; h = p (hydr)/(density*g) = 3005.38 m. Is that right, please?

morechem28 said:
Thank you, yes. I should divide (density*R*T) by 0,0838 kg/mol, not 83,8 g/mol. Now, I get 30.76 (p (cr)) MPa. And now, I'd do this: p(cr) - p(atm) = p(hydr), p(hydr) = 30 763 387.4 - 101 325 = 30 662 062.4 Pa. And then, given that: p (hydr) = density*h*g; h = p (hydr)/(density*g) = 3005.38 m. Is that right, please?
Do you really feel that 6 significant figures is justified for this calculation? What would your answer be for a realistic number of significant figures?

What is the critical pressure of Krypton? Based on this, do you feel that use of the ideal gas law was accurate enough for this calculation? How would you modify the calculation to take into account the deviation from ideal gas behavior?

morechem28
Chestermiller said:
Do you really feel that 6 significant figures is justified for this calculation? What would your answer be for a realistic number of significant figures?

What is the critical pressure of Krypton? Based on this, do you feel that use of the ideal gas law was accurate enough for this calculation? How would you modify the calculation to take into account the deviation from ideal gas behavior?
Yes, I think I get it. But in accordance with the original assignment, I should consider Kr an ideal gas. I know that is not real, but the whole situation is meant to be hypothetical.

morechem28 said:
Yes, I think I get it. But in accordance with the original assignment, I should consider Kr an ideal gas. I know that is not real, but the whole situation is meant to be hypothetical.
@morechem28, note that @Chestermiller asked you some specific questions in Post #82. But you didn’t answer any of them!

If I were replying to him, here’s how I would have answered the first two questions:

Chestermiller said:
Do you really feel that 6 significant figures is justified for this calculation?
No. 3005.38 m has 6 significant which is too many. The depth can’t be determined to the nearest centimetre because the data used in the calculation are not precise enough.

Chestermiller said:
What would your answer be for a realistic number of significant figures?
Two or three significant figures is realistic, since the values used (see Post #1) are given to only three significant figures (and only multiplications and divisions have been used).

So I should have given my answer as 3.0x10⁴m or 3.01x10⁴m.

kuruman
Okay, thank you for your help. Now I get it.

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