What fraction of atoms in a sample of argon gas at 400 K have an energy of 12.5 kJ or greater

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 12.5 kJ or greater.

Im not completely sure how to work this problem out. I know that i have to use the following equation but everytime i use it i come up with different answers. I tried doing a problem from my book and i end up with a total different answer. Can someone help me an explain how to do this correctly, or if im using the wrong equation.

Choose one of the answers:
A.  None of the answers is correct. 
B.  0.023% 
C.  There is not enough information provided about Argon to answer the question. 
D.  10.7% 
E.  2.3% 
F.  99.6%

What equation are you supposed to be using?

Ideally you would need the Maxwell-Boltzmann distribution curve for argon in order to be able to answer this question.  However this involves calculating the integral for the energy distribution.  There may be a more simple equation, but I am not aware of one.

Nonetheless you can make an educated guess.  The curves show the relationship between the number of molecules and their speed or energy.The curves follow a typical Gaussian curve.  In a Gaussian curve every standard deviation gives a fraction of the occurrences that are within a deviation of the average. 

For this problem the average kinetic energy of gases is independent of mass and it only depends on the Kelvin temperature thus we have

KE = 3/2 RT          where R is the ideal gas constant in energy units (8.314 J/K*mol)

plugging in 400K yields just below 5 KJ of energy.  This would represent the top point on the Gaussian curve at this temperature.  Since they are asking for 12.5 KJ, two and a half times the average kinetic energy we can make a rough estimate by calculating 2 standard deviations.  At two starndard deviations you have less than 4.56 % of the molecules at three deviations you would have less than .28% of the molecules.

Looking at a typical curve I would say that 2.5 times the energy would put you beyond three deviations.

Chat with an Online Tutor
Get help immediately by chatting with an live tutor right now

If you find this answer useful please share it with other students.