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

[kuruman]

That's not what mean. To get a density, you need a mass over volume. You don't have a mass in your equation.

 

[morechem28]

Originally, I had: m(Kr)/V(Kr) = (ρ (water) * h * g * m (Kr))/(R * T), so I thought the masses cancel out each other.

 

[kuruman]

Originally, I had: m(Kr)/V(Kr) = (ρ (water) * h * g * m (Kr))/(R * T), so I thought the masses cancel out each other.

What you had originally was incorrect. If the masses cancel out, you have nothing relating the mass to the volume in the equation. There is no sense going back to what's incorrect especially after I gave you a step-by-step outline in post #23 for doing correctly. Follow it.

 

[morechem28]

So, should I get: m (Kr)/V (Kr) = (n * R * T)/(p * m (Kr)? Thanks.

 

[kuruman]

Your suggested equation is $$\frac{m_{\text{ Kr}}}{V_{\text{ Kr}}}=\frac{n~R~T}{p~m_{\text{ Kr}}}.$$Does this look right to you? If you continue not to follow my suggestions, I will stop offering them. In post #23 I wrote

1. Start with the ideal gas law either $pV=nRT$ or $pV =Nk_BT$.

You chose $pV=nRT$. That's fine. Next step says

2. Write the total mass of the gas m_{\text{ Kr}} in terms of the number of moles $n$ and the appropriate constants.

Clearly, the number of moles is proportional to the mass of the gas: twice the number of moles has twice the mass of Kr. What is the "appropriate constant" of proportionality? Give it a symbolic name. You will put in numbers later.

 

[morechem28]

Your suggested equation is $$\frac{m_{\text{ Kr}}}{V_{\text{ Kr}}}=\frac{n~R~T}{p~m_{\text{ Kr}}}.$$Does this look right to you? If you continue not to follow my suggestions, I will stop offering them. In post #23 I wrote

You chose $pV=nRT$. That's fine. Next step says

Clearly, the number of moles is proportional to the mass of the gas: twice the number of moles has twice the mass of Kr. What is the "appropriate constant" of proportionality? Give it a symbolic name. You will put in numbers later.

I know that: n =m/M or N × N(A).

 

[morechem28]

I know that: n =m/M or N × N(A).

It seems I quite don't understand what do you mean by naming this 'proportionality'. Does n change?

 

[kuruman]

O.K. To avoid confusion, use MW for the molecular weight of the gas. What do you get when you replace $n$ with $\dfrac{m_{\text{ Kr}}}{\text{MW}}$ in the ideal gas equation? Write it out then

3. Arrange your expression to have on the left side and the rest of the stuff on the right side of the equation.

 

[kuruman]

It seems I quite don't understand what do you mean by naming this 'proportionality'. Does n change?

It does if $m$ changes. The quantities $n$, $m$ and $N$ are different ways of writing the amount of gas you have. If you double one, you double the others. Therefore, you can write $n=\alpha N = \beta m$ where $\alpha$ and $\beta$ are constants. So if you want to take $n$ out of the equation and replace it with $m$, you cannot omit the proportionality constant.

 

[morechem28]

O.K. To avoid confusion, use MW for the molecular weight of the gas. What do you get when you replace $n$ with $\dfrac{m_{\text{ Kr}}}{\text{MW}}$ in the ideal gas equation? Write it out then

Thank you. So, now I've got: m(Kr)/V(Kr) = (RT)/(pM). Did I get it right?

 

[kuruman]

Thank you. So, now I've got: m(Kr)/V(Kr) = (RT)/(pM). Did I get it right?

You did not get it right. Because you did not show your work, i.e. how you got the result you wrote, I cannot tell where you went wrong. What I wanted you to do and to show was
1. Write the ideal gas law.
Work: $p~V=n~R~T.$

2. Replace $n$ with $\dfrac{m_{\text{ Kr}}}{M}.$
Work : $p~V=\dfrac{m_{\text{ Kr}}}{M}~R~T.$

Can you find an expression for the density of the gas using the result in 2?
Show your work in detail. It will help you get it right.

 

[morechem28]

I see. So, as you suggested I need m/V on the left to get the density. Therefore, p*V*M = m(Kr)*R*T, then: m(Kr)/V = (p*M)/(R*T). This would be it, I hope.

 

[kuruman]

I see. So, as you suggested I need m/V on the left to get the density. Therefore, p*V*M = m(Kr)*R*T, then: m(Kr)/V = (p*M)/(R*T). This would be it, I hope.

That is correct. Now proceed with the outline and make sure you post all details if you get stuck.

 

[morechem28]

Thank you. So, now I have to get the pressure: that would be: ρ (Kr) = (p * M)/(R * T), and then: ρ (Kr) * R * T = p * M; then I get: p = (ρ (Kr) * R * T)/M. But now, if that's correct, I can't proceed with the solution because I don't know the value of the density. Am I right?

 

[Chestermiller]

Thank you. So, now I have to get the pressure: that would be: ρ (Kr) = (p * M)/(R * T), and then: ρ (Kr) * R * T = p * M; then I get: p = (ρ (Kr) * R * T)/M. But now, if that's correct, I can't proceed with the solution because I don't know the value of the density. Am I right?

For the K to be neutrally buoyant, its density must be the same as that of the surrounding water.

 

[kuruman]

Am I right?

You are not right because, despite all my encouragements to the contrary, you still refuse to follow my outline in post #23. Here is what I had laid out for you.

3. Arrange your expression to have on the left side and the rest of the stuff on the right side of the equation.
4. Note that the left side is (that is why you don't need a value for ) and recognize that this equation gives you the density of the gas at any pressure and temperature. Here the temperature is given.
5. Use that equation to find the pressure (and hence the depth) at which the gas density matches the density of water.

You have to do the work yourself with our guidance but you seem to refuse to think about and put to use what we suggest. Instead, you keep trying to tease more and more out hoping that eventually the equation will be revealed to you. Am I right?

That's it for me. I have said all I needed to say on this thread. Maybe someone else will pick up the baton. Good luck with the completion of this problem.

 

[morechem28]

You are not right because, despite all my encouragements to the contrary, you still refuse to follow my outline in post #23. Here is what I had laid out for you.

You have to do the work yourself with our guidance but you seem to refuse to think about and put to use what we suggest. Instead, you keep trying to tease more and more out hoping that eventually the equation will be revealed to you. Am I right?

That's it for me. I have said all I needed to say on this thread. Maybe someone else will pick up the baton. Good luck with the completion of this problem.

What you call teasing, I call language barrier. I don't understand everything you suggest. But thank you for your effort, I appreciate it.

 

[kuruman]

What you call teasing, I call language barrier. I don't understand everything you suggest. But thank you for your effort, I appreciate it.

If you don't understand what I am asking you to do, then tell me (tell us) and it will be explained to you. We are here to help, not to mystify. You cannot simply ignore our suggestions because you do not understand them.

 

[morechem28]

If you don't understand what I am asking you to do, then tell me (tell us) and it will be explained to you. We are here to help, not to mystify. You cannot simply ignore our suggestions because you do not understand them.

Thanks to you, I managed to build up this equation: m(Kr)/V = (p*M)/(R*T). That was the last thing you approved of, and said to be correct. Now, you say:

“You don't need to know its numerical value, just use m.“ - Here I don't get the reason why I don't have to know the value of m. Should I just use the density of water (1040 kg/m³)? Because, as it was suggested, the density of Kr, in the end, must reach at least this density and above. Therefore, the p (Kr) would be 1040 kg/m³ and more.

“3. Arrange your expression to have m/V on the left side and the rest of the stuff on the right side of the equation.“ According to this piece of advice, here's the equation: m(Kr)/V = (p*M)/(R*T).

“4. Note that the left side is ρ (Kr) (that is why you don't need a value for m) and recognize that this equation gives you the density of the gas at any pressure and temperature. Here the temperature is given.“ I understand that we have a certain temperature that won't change (referring to the problem, the whole process must be isothermic). But how does that help me to get the pressure at which Kr would reach at least the density of seawater (1040 kg/m³)? I'd isolate p from the equation listed above as p = (ρ (Kr) * R * T)/M (Kr) and get 30 763 387,4 Pa. I presume that's incorrect, again, but I don't know why I can't do this.

“5. Use that equation to find the pressure (and hence the depth) at which the gas density matches the density of water.“ And then, how could I get the depth to which the balloon must be pressed to reach (at least) that particular density of water? I only know that p (hydrostatic) = ρ (water) * h * g. But I don't seem to get the way I could use this formula.

 

[kuruman]

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"?

 

[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.