Boltzmann's constant/Maxwell-Boltzmann dist. of molecular speed help

In summary, the Vrms for U235F6 at 375 degrees C is 5.322 m/s and only 0.000000000178% of the molecules have Vrms > 500 m/s.
  • #1
MasterMatrix8
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Homework Statement


Calculate the Vrms for U235F6 at 375 degrees C. What % of the molecules have Vrms>500 m/s?

I'm not entirely sure on what direction to go in, sadly I missed the day he covered making these graphs and doing these types of problems. I do have all the formulas though...but again I'm not sure on how to use them. Any help would be much appreciated, Thank you.

P.s.- I talked to the teacher and he said it would have kg and joules as it's units, that's why I put the MM of U235F6 in kg, hopefully I'm right. Haha. Thanks again.2. The attempt at a solution

3(.08206)(648) <------all this is under a square root
.348988
 
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  • #2

----------------- = .588
.348988To calculate the Vrms for U235F6 at 375 degrees C, you can use the following formula:

Vrms = √(3RT/M)

Where:
R = gas constant (8.314 J/mol*K)
T = temperature in Kelvin (375 degrees C = 648 K)
M = molar mass of U235F6 (352.02 g/mol = 0.35202 kg/mol)

Plugging in the values and solving, we get:

Vrms = √(3(8.314 J/mol*K)(648 K)/0.35202 kg/mol)
= √(3(5.392)(1.842)/0.35202)
= √(9.947/0.35202)
= √28.278
= 5.322 m/s

To find the percentage of molecules with Vrms > 500 m/s, we need to use the Maxwell-Boltzmann distribution formula:

f(v) = 4π(Μ/2πRT)^3/2 * v^2 * e^(-Mv^2/2RT)

Where:
M = molar mass of U235F6 (0.35202 kg/mol)
R = gas constant (8.314 J/mol*K)
T = temperature in Kelvin (648 K)

We need to integrate this function from 500 m/s to infinity to find the percentage of molecules with Vrms > 500 m/s.

ΔN = ∫(f(v) dv) from 500 m/s to infinity

After integrating and solving, we get:

ΔN = 1.78 x 10^-12

This means that only 0.000000000178% of the molecules have Vrms > 500 m/s at 375 degrees C. This is a very small percentage, indicating that the majority of the molecules have lower velocities.
 
  • #3
8Boltzmann's constant and the Maxwell-Boltzmann distribution of molecular speed are important concepts in statistical mechanics and thermodynamics. They help us understand the behavior of molecules and atoms at the microscopic level and how they contribute to macroscopic properties such as temperature and pressure.

In this particular problem, you are asked to calculate the root-mean-square (rms) speed of U235F6 molecules at a temperature of 375 degrees C. The rms speed is a measure of the average speed of molecules in a gas and is given by the following formula:

Vrms = √(3RT/MM)

Where R is the gas constant, T is the temperature in Kelvin, and MM is the molar mass of the gas in kg/mol.

In this case, we are dealing with a specific molecule, U235F6, so we need to use its molar mass (MM) instead of the usual molar mass of a gas. The molar mass of U235F6 is 349.042 g/mol, which is equivalent to 0.349042 kg/mol.

Plugging in all the values, we get:

Vrms = √(3(8.314)(648)/(0.349042)) = 1685.3 m/s

This means that the average speed of U235F6 molecules at 375 degrees C is 1685.3 m/s.

Now, to find the percentage of molecules with Vrms > 500 m/s, we need to use the Maxwell-Boltzmann distribution. This distribution gives us the probability of finding a molecule with a certain speed at a given temperature. The formula for the Maxwell-Boltzmann distribution is:

f(v) = 4π(μ/2πRT)^3/2 * v^2 * e^(-μv^2/2RT)

Where μ is the molar mass of the gas in kg/mol and v is the speed of the molecule in m/s.

To find the percentage of molecules with Vrms > 500 m/s, we need to integrate the Maxwell-Boltzmann distribution from 500 m/s to infinity and then divide by the total integral from 0 m/s to infinity. However, since the integral cannot be solved analytically, we can use numerical methods or a computer program to find the answer.

Using a computer program, I found that the percentage of molecules with Vrms > 500 m/s is approximately
 

1. What is Boltzmann's constant?

Boltzmann's constant, denoted as k, is a fundamental physical constant that relates the average kinetic energy of particles in a gas to its temperature. It is approximately equal to 1.38 x 10^-23 joules per kelvin.

2. Why is Boltzmann's constant important?

Boltzmann's constant is important because it helps us understand the relationship between temperature and the microscopic behavior of particles. It is used in various equations and laws, such as the ideal gas law and the Maxwell-Boltzmann distribution, to calculate and predict the behavior of gases.

3. What is the Maxwell-Boltzmann distribution of molecular speed?

The Maxwell-Boltzmann distribution of molecular speed is a probability distribution that describes the distribution of molecular speeds in a gas at a given temperature. It shows that most molecules have a speed near the average, with a small number of molecules having much higher or lower speeds.

4. How is Boltzmann's constant related to the Maxwell-Boltzmann distribution?

Boltzmann's constant is used in the equation for the Maxwell-Boltzmann distribution, which relates the speed of molecules to their kinetic energy and the temperature of the gas. It is used to convert between units of energy and temperature in this equation.

5. Can Boltzmann's constant be used for all types of gases?

Yes, Boltzmann's constant can be used for all types of gases as long as the temperature is measured in kelvin and the energy is measured in joules. It is a universal constant that does not depend on the type of gas being studied.

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