Recent content by morechem28

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

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.

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...

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.

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?

Thank you, I really appreciate your help!

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

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

Could it look like that: ρ (Kr) = ((p (atm) + p (hydr.))*M)/(R*T)? But then I think about the fact that the final pressure would be the sum of p (atm) and the p caused by buoyancy (which would give zero, since the balloon won't move, just stay at one point... I must say, I'm completely confused now.

Yes, I see. Of course, we're looking for the depth at which the density of Kr would be the same as the density of water. But I still can't think of a formula that would help me out in expressing p in the equation.

Couldn't we say about the p on the right that it is the sum of p (atm) and the p (hydrostatic)?

Pressure at a specific depth, I think.

Firstly, I'd replace m/V on the left with that number. And now I'd be closer to isolating the pressure, but I still don't know what it does stand for. I can't just replace p with density * h * g, that wouldn't make sense, right?

The density of Kr must be the same (or higher than) as the density of seawater 1040 kg/m3. Is that right?

Yes, I see your point. It would give me the density at 101,325 kPa, which I understand. I'm looking for a certain pressure at which the density of Kr would be higher than the density of water (and the balloon with Kr would be submerged in water so that it would flow). So, I wonder whether it has...