Answer:
The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa
Explanation:
<u>Step 1</u>: 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
<u>Step2 </u>: 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
Answer:
Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.
Hope this helped and if it did, please give my answer a brainliest.
Answer:
68 g
Explanation:
Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =
= 136 g/mol
m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g
n(C10H16) - number of moles of C10H16
M(C10H16) - molar mass of C10H16
