# The question: Can nitrogen molecules escape from Europa and into space?

• Sudikshya Pant
In summary: I see. So, I have to integrate the probability density from 2025 m/s to infinity. But how do I do that?In summary, an astronaut visiting Jupiter’s satellite Europa leaves a canister of 1.20 mol of nitrogen gas 28.0 g/mol^2 at 25.0°C on the satellite’s surface. The maximum height reached by a nitrogen molecule with a speed equal to the root mean square velocity is 102 km. To determine if any nitrogen molecules can escape from Europa into space, we must calculate the cumulative probability of molecules with a speed greater than or equal to the escape speed of 2025 m/s. This involves integrating the probability density from 2025 m/s to infinity.
Sudikshya Pant

## Homework Statement

:[/B]
An astronaut visiting Jupiter’s satellite Europa leaves a canister of 1.20 mol of nitrogen gas 28.0 g/mol^2 at 25.0°C on the satellite’s surface. Europa has no significant atmosphere, and the acceleration due to gravity at its surface is 1.30 m/s2. The canister springs a leak, allowing molecules to escape from a small hole. (a) What is the maximum height (in km) above Europa’s surface that is reached by a nitrogen molecule whose speed equals the rms speed? Assume that the molecule is shot straight up out of the hole in the canister, and ignore the variation in g with altitude. (b) The escape speed from Europa is 2025 m/s. Can any of the nitrogen molecules escape from Europa and into space?

## Homework Equations

:
[/B]
v_rms =sqrt(3kT/m)
where, m is the mass of a molecule in kg

OR

v_rms =sqrt(3RT/M)
where, M is the molar mass in kg

## The Attempt at a Solution

;[/B]

Basically, I used the above formula and got the value for root mean square velocity as 515 m/s. Then, I tried to use Newton's law to get the height which is 102 km. For part (b) we might have to find the maximum velocity that a nitrogen molecule can have and when that molecule escapes the chances of nitrogen might be high. But how to get that velocity?I need some help for part (b). The velocity obtained must greater than the escape velocity. But how do we obtain the maximum velocity?

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Sudikshya Pant said:
But how do we obtain the maximum velocity?
The mathematical speed distribution doesn't have a maximal speed, but you can compare the required speed to the rms speed. Will some molecules have ~4 times the rms speed?

mfb said:
The mathematical speed distribution doesn't have a maximal speed, but you can compare the required speed to the rms speed. Will some molecules have ~4 times the rms speed?
It's very likely. But sorry, I can't come up with a mathematical way to ascertain this. Is there a mathematical way to get through part 'b' of the question? Or could we just give such logical answers to it?

Check the Maxwell-Boltzmann distribution.

mfb said:
Check the Maxwell-Boltzmann distribution.
mfb said:
Check the Maxwell-Boltzmann distribution.
I got the probability of 1.075*10^(-11). Can such a small probability be enough to say yes? It's almost zero.

There are more than 1/(1.075*10^(-11)) particles in the bottle. The number should be larger, however.

mfb said:
There are more than 1/(1.075*10^(-11)) particles in the bottle. The number should be larger, however.
mfb said:
There are more than 1/(1.075*10^(-11)) particles in the bottle. The number should be larger, however.
Oh! I forgot that this value is just the fraction of number of molecules that can have the velocity 2025m/s. That means I need to multiply this value by (1.20*6.022*10^23) to get the actual number of particles having the required velocity. Which will be almost in order of 10^13. Hence, yes there are these many particles (10^3) that could escape the planet.Am I right?

That is the fraction at 2025 m/s. What you need is the cumulative fraction for all speeds >= 2025 m/s.

mjc123 said:
That is the fraction at 2025 m/s.
Maybe not even that. It is the probability density at 2025 m/s, normalized to something.
The fraction at "exactly" 2025 m/s is zero.

mfb said:
Maybe not even that. It is the probability density at 2025 m/s, normalized to something.
The fraction at "exactly" 2025 m/s is zero.
If so there will any nitrogen atom be able to escape Jupiter?

Did you calculate it? What was the result?

mfb said:
Did you calculate it? What was the result?
Please have a look on my last post where I've written on what value I got and have asked if it's right.

I thought that was a calculation for Europa, not Jupiter.
And, as @mjc123 pointed out, the number is not right.

mfb said:
I thought that was a calculation for Europa, not Jupiter.
And, as @mjc123 pointed out, the number is not right.
Sorry that was my mistake to write Jupiter instead of Europa.

I applied the formula for probability per change in temperature given by maxwell Boltzmann distribution formula, where I kept the respective values for m, k, T, and v (2025m/s) and multiplied the result with (1.20 mol* Avogadro's number) to get the number of particles having the velocity 2025m/s. Where did I go wrong?

I told you. You need the total number of molecules with speed (not velocity) >= 2025 m/s. For that you need the cumulative form of the distribution - see e.g. Wikipedia article on Maxwell-Boltzmann distribution.

We must calculate: 1 - integral f(v)dv from 0 to vesc
This is the percentage of molecules with v >= vesc and is about 10^(-6)
Percentage can be calculated using this VB program

Dim gdPigreco As Double
Dim dNA As Double ' numero di Avogadro
Dim dK As Double ' costante di Boltzmann
Dim fMmaj As Single
Dim fM As Single
Dim fT As Single
Dim fVesc As Single
Dim fVmp As Single
Dim fVav As Single
Dim fVrms As Single
Dim fV As Single
Dim fDeltaV As Single
Dim fA As Single
Dim fDeltaA As Single
Dim fF1 As Single
Dim fF2 As Single

gdPigreco = 4 * Atn(1)
dNA = 6.022 * 10 ^ 23
dK = 1.381 * 10 ^ (-23)
fMmaj = 28 * 10 ^ (-3)
fM = fMmaj / dNA
fT = 273.15 + 25
fVesc = 2025
fVmp = Sqr(2 * dK * fT / fM)
fVav = Sqr(8 * dK * fT / (gdPigreco * fM))
fVrms = Sqr(3 * dK * fT / fM)
fV = 0
fDeltaV = 1 ' or = .1 or = .01 to improve accuracy
fA = 0
fDeltaA = 0
fF1 = 0
fF2 = 0
Do
fF1 = fF2
fV = fV + fDeltaV
fF2 = 4 * gdPigreco * (fM / (2 * gdPigreco * dK * fT)) ^ (3 / 2)
fF2 = fF2 * fV ^ 2 * Exp(1) ^ (-(fM * fV ^ 2) / (2 * dK * fT))
fDeltaA = (fF1 + fF2) * fDeltaV / 2
fA = fA + fDeltaA
Debug.Print fV, fV / fVrms, fF1, fF2, fDeltaA, fA, 1 - fA
' If fV > fVmp Then Stop
' If fV > fVav Then Stop
' If fV > fVrms Then Stop
' If fA > 0.5 Then Stop
If fA > 1 Then Stop
If fV >= fVesc Then Stop
Loop
End

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## 1. What is the composition of gas on Jupiter's Moon Europa?

The gas on Jupiter's Moon Europa primarily consists of oxygen, hydrogen, and carbon dioxide. However, there are also trace amounts of gases such as methane, nitrogen, and sulfur dioxide.

## 2. How does the gas on Europa compare to the gas on Jupiter?

The gas on Europa is significantly different from the gas on Jupiter. While Jupiter's atmosphere is mostly made up of hydrogen and helium, Europa has a much thinner atmosphere with a higher concentration of oxygen and other trace gases.

## 3. Is there a specific source for the gas on Europa?

The gas on Europa is believed to come from a combination of processes, including internal geological activity and the bombardment of particles from Jupiter's magnetic field. It is also possible that the gas is being released from subsurface oceans on the moon.

## 4. Can the gas on Europa sustain life?

While the gas on Europa does have some crucial components for sustaining life, such as oxygen and hydrogen, it is not enough to support complex life forms. However, some scientists believe that the subsurface oceans on Europa may have the potential to harbor microbial life.

## 5. How does the gas on Europa affect the moon's surface?

The gas on Europa does not have a significant impact on the moon's surface, as the atmosphere is too thin to cause erosion or weathering. However, the gas may play a role in the formation of surface features, such as the reddish-brown streaks known as "chaos terrain."

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