What is the mean rate of flow through the hole?

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The discussion revolves around calculating the mean rate of flow of mercury atoms through a small hole in a vessel containing liquid mercury at 0°C. After losing 24mg of mercury over 30 days, the average speed of mercury atoms in vapor was determined to be 169 m/s. The mean rate of flow was calculated to be approximately 2.76 x 10^13 atoms per second. To find the vapor pressure at 0°C, participants suggested using the flux and the ideal gas law, leading to a result of 2.5 x 10^(-2) N/m^2. The conversation highlights the importance of understanding molecular flow and gas laws in solving such problems.
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Hello everyone! I have trouble finding the answer to the following problem - this really bugs me. Can anyone give me a hint?

Here is the problem:

A vessel contains a hole of 0.1mm^2 above the liquid level (contains liquid mercury) at T=0°C. The vessel is contained within a very much larger vessel in which a very high vacuum is maintained. After 30 days it is found that 24mg of mercury have been lost from the inner vessel. The relative atomic mass of mercury is 201.


I have already calculated the average speed of the mercury atoms in the vapor to be

169 m/s

now the questions:

1) What is the mean rate of flow of mercury atoms through the hole?

2) hence calculate the vapour pressure at T=0°C


note:
the numerical answers are:

1) 2.76 * 10^13 atoms/sec

2) 2.5 * 10^(-2) N/m^2
 
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You only need to divide the number of atoms of mercury that have been transported by the total time it took for them to be transported to find the average rate of flow. The number you showed is correct.

To find the number of atoms, divide the gram molecular weight of Hg by the number of grams transported and multiply by Avogadro's number.
 
Thank you Tide!

That was indeed easier than expected!
However, I can still not calculate the pressure. I was trying to do it with the relationship p=nkT but that does not work. Would you mint giving me another hint?
 
thanks for this! Using the flux and the ideal gas law one can now find the pressure - the result it gives is correct!
 
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