Respuesta :
Answer:
The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa
Explanation:
Step 1: Calculate the volume of a spherical container V
V = (4π*r³)/3
r = (3V/4π)^1/3
2r = d = 2*(3V/4π)^1/3
with r= radius
with d= diameter
The diameter is:
d= 2*(3V/4π)^1/3
d= 2*(3*100cm³/4π)^1/3
d= 5.76 cm
Step2 : Define the free path lambda λ of argon
with λ =k*T/ σp
with p = kT/σλ
with T= temperature = 20°C = 293.15 Kelvin
with k = Boltzmann's constant = 1.381 * 10^-23 J/K
with p = the atmospheric pressure
with σ = 0.36 nm²
p = kT/σλ
p = (1.38 * 10^-23 J*K^-1 * 1Pa *m³/1J)*(293,15K) /(0.36 nm²*(10^-9/ 1nm)² *(5.76cm* 10^-2m/1cm)
p = 0.195 Pa
The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa
The mean free path of argon at 20°C becomes comparable to the diameter of spherical vessel at 0.195 Pascal.
Given the following data:
- Temperature = 20°C to K = [tex]273.15+20[/tex] 293.15 K.
- Volume of vessel = 100 [tex]cm^3[/tex].
- [tex]\sigma = 0.36 \; nm^2[/tex]
Scientific data:
- Boltzmann's constant, k = [tex]1.38 \times 10^{-23}\;J/K[/tex]
- Mean free path of argon = [tex]5.76 \times 10^{-23}\;m[/tex]
How to calculate the pressure.
First of all, we would determine the diameter of the spherical vessel. Mathematically, the volume of a sphere is given by this formula:
[tex]V = \frac{1}{6} \pi d^3[/tex]
Making d the subject of formula, we have:
[tex]d=(\frac{6V}{\pi} )^{\frac{1}{3} }\\\\d=(\frac{6\times 100}{3.142} )^{\frac{1}{3} }\\\\d=(\frac{600}{3.142} )^{\frac{1}{3} }\\\\d=(190.96)^{\frac{1}{3} }[/tex]
d = 5.76 centimeter.
Now, we can determine the pressure by using this formula:
[tex]P=\frac{kT}{\lambda \sigma} \\\\P=\frac{1.38 \times 10^{-23} \times 293.15}{0.36 \times 10^{-9} \times 5.76 \times 10^{-23} }[/tex]
P = 0.195 Pascal.
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