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

[morechem28]

Let's look at step 3. You have the ratio of the mass of the gas to its volume. What is another name for "mass of Kr over volume of Kr"?

It's the density of Kr. :) So: ρ (Kr) = (p*M)/(R*T).

 

[morechem28]

Please, should I replace ρ (Kr) with 1040 (which is the density of seawater), and then just find the value of p? Thank you.

 

[kuruman]

Please, should I replace ρ (Kr) with 1040 (which is the density of seawater), and then just find the value of p? Thank you.

Don't replace anything yet. Before you go to step 4 it is very important to understand what you have already and your question shows to me that you do not understand why you are doing what you are doing. Please explain to me, in your own words, what this equation is giving you $$\rho_{\text{ Kr}}=\frac{p~M_{\text{ Kr}}}{R~T}.$$ Begin your answer by saying "It is the density of Kr when $\dots$"

 

[morechem28]

Don't replace anything yet. Before you go to step 4 it is very important to understand what you have already and your question shows to me that you do not understand why you are doing what you are doing. Please explain to me, in your own words, what this equation is giving you $$\rho_{\text{ Kr}}=\frac{p~M_{\text{ Kr}}}{R~T}.$$ Begin your answer by saying "It is the density of Kr when $\dots$"

It's the density of Kr when it's outside (not submerged in water) the temperature is 298,15 K, and there's a certain pressure inside the balloon.

 

[morechem28]

It's the density of Kr when it's outside (not submerged in water) the temperature is 298,15 K, and there's a certain pressure inside the balloon.

So, is it the density of Kr when the pressure in the balloon is the same as the atm. pressure, 101 325 Pa?

 

[kuruman]

So, is it the density of Kr when the pressure in the balloon is the same as the atm. pressure, 101 325 Pa?

I think I have diagnosed a problem you have with equations. In equations, symbols without subscripts are placeholders that stand for the word "any". Subscripts are used to narrow down the "any" to a specific case. To see what I mean consider the density equation without subscripts $$\rho=\frac{p~M}{R~T}.$$This equation gives you the density $\rho$ of any ideal gas of any molecular weight $M$ at any pressure $p$ and any temperature $T$. Of course, $R$ is the gas constant.

Now you start narrowing things down using subscripts. You don't have any gas, you have Kr. Also its temperature is not anything but it has a specific value, call it $T_0$ of 298 K. if you replace the placeholders with their specific values, you get $$\rho_{\text{ Kr}}=\frac{p~M_{\text{ Kr}}}{R~T_0}.$$ So far, so good. Now you need to replace "any pressure" with a specific pressure. What would that be? Answer it by starting with "It is the pressure at which $\dots~$" and give it a subscript of your choice.

 

[morechem28]

I think I have diagnosed a problem you have with equations. In equations, symbols without subscripts are placeholders that stand for the word "any". Subscripts are used to narrow down the "any" to a specific case. To see what I mean consider the density equation without subscripts $$\rho=\frac{p~M}{R~T}.$$This equation gives you the density $\rho$ of any ideal gas of any molecular weight $M$ at any pressure $p$ and any temperature $T$. Of course, $R$ is the gas constant.

Now you start narrowing things down using subscripts. You don't have any gas, you have Kr. Also its temperature is not anything but it has a specific value, call it $T_0$ of 298 K. if you replace the placeholders with their specific values, you get $$\rho_{\text{ Kr}}=\frac{p~M_{\text{ Kr}}}{R~T_0}.$$ So far, so good. Now you need to replace "any pressure" with a specific pressure. What would that be? Answer it by starting with "It is the pressure at which $\dots~$" and give it a subscript of your choice.

It's the pressure when Kr has a certain density (we are up to find the pressure when the density will be at least equal to the density of water). And I think we're referring to the state of Kr when it hasn't been submerged, so it's in the air. Therefore, I gather the value of p would same as the atmospheric pressure, 101 325 Pa. Isn't it?

 

[kuruman]

If you replace $p$ with the atmospheric pressure $p_A$ in that equation, you will get the density of Kr at atmospheric pressure. Why is that useful to you? What are you looking for?

 

[morechem28]

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 to do something with the hydrostatic pressure.

 

[kuruman]

So, I wonder whether it has to do something with the hydrostatic pressure.

It does have a lot to do with hydrostatic pressure. If the density of the gas is that at atmospheric pressure, the balloon will float. What must its density be so that it will barely float and be at the threshold of sinking? The answer has been mentioned on this thread many times.

 

[morechem28]

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

 

[kuruman]

That is right. Now what?

 

[morechem28]

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?

 

[kuruman]

Firstly, I'd replace m/V on the left with that number.

Why? Because you have it? If you choose a specific value $\rho_{\text{ H}_2\text{O}}$ for the density on one side of the equation, then symbol $p$ must also acquire a subscript. It needs to be narrowed down as well. If the left-hand side has no placeholders, the right-hand side must also have no placeholders. What is an appropriate subscript for $p$? What specific pressure is it when the left-hand side is the density of seawater?

 

[morechem28]

Why? Because you have it? If you choose a specific value $\rho_{\text{ H}_2\text{O}}$ for the density on one side of the equation, then symbol $p$ must also acquire a subscript. It needs to be narrowed down as well. If the left-hand side has no placeholders, the right-hand side must also have no placeholders. What is an appropriate subscript for $p$? What specific pressure is it when the left-hand side is the density of seawater?

Pressure at a specific depth, I think.

 

[morechem28]

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

 

[kuruman]

Which specific depth? What makes it so special of all depths that you could consider? "It is the depth at which $\dots~$" (see step 5 in #23.)

 

[morechem28]

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.

 

[kuruman]

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

That's exactly what we should say. You want all symbols to have subscripts. So let's see your equation.

 

[morechem28]

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.

 

[kuruman]

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.

 

[morechem28]

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

 

[kuruman]

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?

 

[kuruman]

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.

 

[morechem28]

Thank you, I really appreciate your help!

 

[morechem28]

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?

 

[morechem28]

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.

 

[Chestermiller]

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.

 

[kuruman]

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

My answer agrees with @Chestermiller's answer.

 

[morechem28]

My answer agrees with @Chestermiller's answer.

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?

 

[Chestermiller]

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]

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.

 

[Steve4Physics]

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:

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.

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.

 

[morechem28]

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