Finding pressure, mean speed and mean free path

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Homework Statement



A right circular cylinder of ##4## cm (inner) radius and ##19.9## cm (inner) length contains ##4 \times 10^{14}## Argon atoms (atomic number ##18##). The cylinder is maintained at room temperature, nominally ##300## K.

(a) Estimate the momentum transfer per square meter per second to the curved wall due to gas collisions.

(b) Estimate the mean speed of an Argon atom in the container.

(c) Estimate the mean free path of an Argon atom in the container.

Show the formulas used, reasoning, and work for full credit. You may wish to recall that ##k_{B}= 1.38 \times 10^{-23}## J/K, ##1## amu ##= 1.66 \times 10^{-27}## kg. You may take the atomic weight of Argon to be ##40## amu.

Homework Equations



The Attempt at a Solution



(a) Assuming that the Argon gas in the right circular cylinder is an ideal gas, we can use the formula ##pV=NkT##,

where ##p## = pressure = force per square meter = momentum transfer per square meter per second.

So, ##p(\pi(0.04)^{2}(0.199))=(4\times 10^{14})(1.38 \times 10^{-23})(300) \implies p = 1.66## mPa.

Am I correct?

(b) Argon atom is a monoatomic gas, so using the equipartition theorem,

##\frac{3}{2}k_{B}T=\frac{1}{2}mv^{2} \implies v = 432## m/s.

Am I correct?

(c) Mean free path ##\lambda = \frac{1}{\sqrt{2}\pi d^{2}n_{V}} = \frac{k_{B}T}{\sqrt{2}\pi d^{2}p}##,

where ##d = ## diameter of the atoms, and ##n_{V} = ## number density of the atoms.

The diameter of the Argon atoms can be estimated to be on the order of ##10^{-10}## m.

So, mean free path ##\lambda = 56## m.

Am I correct?
 

Answers and Replies

  • #2
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I have not checked your calculations, but all your steps seem to be faultless.
 

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