Solving for Two Speeds of an Ideal Gas He at a Given Temperature

• j2dabizo
In summary: The value of f(v*), which is v*=\sqrt{2kT/m}, is 847.06 m/s. The half value of that, or 473.14 m/s, is the speed at which v* is most probable.
j2dabizo

Homework Statement

For an ideal gas He at T = 328 K find the two speeds v that satisfy the equation 2F(v) = F(v*).

m/s (lower speed)
m/s (higher speed)

Which of the two speeds you found is farther from v*?
the lower speed or the upper speed

Explain this result.

The Attempt at a Solution

Not even sure here!

What is v* supposed to represent? What kind of function is F?

AM

Again not really sure where to start here.

All the teacher told me was to use the Maxwell Speed Distrubution.

Any help will be greatly appreciated...I'm stuck.

Andrew Mason said:
What is v* supposed to represent? What kind of function is F?

AM

v* = most probable speed

http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

I assume you use the maxwell speed distrtibution equation but for f(v) you but in f(v*) and solve the equation using v*. Then after you find the answer form f(v*) i guess you divide it by two to get the left side of the 2f(v)=f(v*) equation?

Last edited:

j2dabizo said:
v* = most probable speed

http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

I assume you use the maxwell speed distrtibution equation but for f(v) you but in f(v*) and solve the equation using v*. Then after you find the answer form f(v*) i guess you divide it by two to get the left side of the 2f(v)=f(v*) equation?

It looks like you have the right idea. But we can't say for sure, and can't see what went wrong, if you don't show your work:
• Show explicitly what f(v) is -- show the equation.
• What is the equation and the value of v* that you used?
• What value did you get for f(v*)?
• Show how you got the answer you got, even though it is wrong.
If we don't see what you did, we can't help.

Ok..I went back to this problem and still pretty stumped!

the given is ideal gas He at temp= 328K

This is a maxwell speed distribution problem.

we need to give 2 values of v that are in m/s for the equation 2F(v)=F(v*)

The maxwell speed distribution equation is given as

F(v) dv = 4$\Pi$Ce-1/2Bmv2dv

m=mass of He
v=velocity

C= (Bm/2$\Pi$)3/2

B(beta)= (kT)-1; with k(constant) = 1.38E-23J/K

we know v*= $\sqrt{}2kT/m$; k is the constant from above; T is tempreture in K; m is mass of He

For He @ 328K, the v*(most probable speed) I calculated was 1167.34 m/s.

I have all this information and don't know how to solve...I am not sure where to go with this as I am not great with intergrals. If someone can get me to an equation that I can solve and a brief explanation of where I am going with this problem that would be a great help.

Thank you all for your time once again..physics forum has been a wonderful help.

Last edited:

j2dabizo said:
The maxwell speed distribution equation is given as

F(v) dv = 4$\pi$Ce-1/2Bmv2dv

m=mass of He
v=velocity

C= (Bm/2$\pi$)3/2

B(beta)= (kT)-1; with k(constant) = 1.38E-23J/K

we know v*= $\sqrt{}2kT/m$; k is the constant from above; T is tempreture in K; m is mass of He

For He @ 328K, the v*(most probable speed) I calculated was 1167.34 m/s.
Something is wrong, the f(v) expression you wrote should have another factor of v2 in it.

So, what is the value of f(v*)? And, what is one half of that value?

1. What is an ideal gas?

An ideal gas is a theoretical gas that is made up of particles that have no volume and do not interact with each other. It follows the ideal gas law, which describes the relationship between pressure, volume, temperature, and number of moles of the gas.

2. How do you solve for two speeds of an ideal gas?

To solve for two speeds of an ideal gas, you will need to use the Maxwell-Boltzmann distribution, which is a probability distribution that describes the speeds of particles in a gas at a given temperature. This distribution can be used to find the most probable speed and the average speed of the gas particles.

3. Why is temperature important in solving for two speeds of an ideal gas?

Temperature is a crucial factor in solving for two speeds of an ideal gas because it directly affects the kinetic energy of the gas particles. The Maxwell-Boltzmann distribution is dependent on temperature, and a change in temperature will result in a change in the speeds of the gas particles.

4. Can two speeds of an ideal gas be the same?

Yes, two speeds of an ideal gas can be the same. The Maxwell-Boltzmann distribution shows a continuous range of speeds, so it is possible for two particles to have the same speed at a given temperature. However, this is less likely to occur as the number of particles in the gas increases.

5. What are the units of measurement for the speeds of an ideal gas?

The units of measurement for the speeds of an ideal gas are typically meters per second (m/s) or centimeters per second (cm/s). However, the units may vary depending on the system of units used for other variables in the problem.

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