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
Density = 0.7899 g/mL
mass = 28.7 g
Volume = ?
therefore:
D = m / V
0.7899 = 28.7 / V
V = 28.7 / 0.7899
V = 36.333 mL
It happens so because,
we know that evaporation is a surface phenomenon. If surface area is increased, the rate of evaporation increases. That's why when we pour hot water in a plate it cools faster.
