Estimating Collision Time: A Helium-Filled Chamber at 4x10-11Pa

AI Thread Summary
Estimating the collision time for helium in a chamber at a pressure of 4×10-11Pa and temperature of 380K involves calculating the average speed of helium molecules and the mean free path. The average speed is determined using the formula v = √(3RT/M), resulting in approximately 1,506 m/s. The mean free path is calculated with λ = (kT)/(√2πd^2P), yielding a value of 8.2×10-7m. By applying the formula tau = λ/v, the estimated collision time is found to be approximately 0.54 nanoseconds. This process illustrates the application of kinetic theory to gas behavior in low-pressure conditions.
jbot2222
Messages
14
Reaction score
0
i don't know how to do this...nothing comes to mind. any help will be appreciated!


A pressure as low as 4×10-11Pa has been obtained. Suppose a chamber contains helium at this pressure and at temperature 380K. Estimate the collision time (tau) for helium in the chamber. Take the diameter of a helium molecule to be 1×10-10m.
 
Physics news on Phys.org
Do u know the formula giving the mean free path (\lambda)??If so,combine it with the thermal velocity to find the mean collision time...

Daniel.
 


Estimating the collision time in a helium-filled chamber at a pressure of 4×10-11Pa is a complex problem, and it requires some knowledge of gas laws and kinetic theory. However, we can break it down into smaller steps to make it more manageable.

First, we need to understand the concept of collision time. It is the average time between collisions of gas molecules in a chamber. In other words, it is the time it takes for a molecule to travel from one end of the chamber to the other and collide with another molecule.

To estimate the collision time, we need to know the average speed of the helium molecules in the chamber. According to the kinetic theory of gases, the average speed of a gas molecule is given by the formula:

v = √(3RT/M)

Where v is the average speed, R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

In this case, we have helium gas at a temperature of 380K. The molar mass of helium is approximately 4g/mol. Plugging these values into the formula, we get an average speed of 1,506 m/s.

Next, we need to calculate the mean free path of the helium molecules. This is the average distance a molecule travels between collisions. It is given by the formula:

λ = (kT)/(√2πd^2P)

Where λ is the mean free path, k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the molecule, and P is the pressure.

In this case, we have a pressure of 4×10-11Pa and a diameter of 1×10-10m. Plugging these values into the formula, we get a mean free path of 8.2×10-7m.

Finally, we can calculate the collision time using the formula:

tau = λ/v

Substituting the values we calculated, we get a collision time of 5.4×10-10s or 0.54 nanoseconds.

In conclusion, estimating the collision time in a helium-filled chamber at a pressure of 4×10-11Pa is a complex problem, but by breaking it down into smaller steps and using the formulas of kinetic theory, we can estimate it to be approximately 0.54 nanoseconds.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top