Question

A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass 5 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T 5 300 K? Take an average diameter of d 5 3.1 × 10210 m for the molecules in the air.

Answer 1

There are quite a bunch of typo errors in the question; here is the correct question below:

A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.

Answer:

9.57 × 10⁻⁷ atm

Explanation:

The mean free path ( λ ) can be illustrated by the equation:

λ =   $\frac{1}{\sqrt{2} \pi d{^2}N/V }$     ----------  (1)

N/V = $\frac{N_AP}{RT}$                      ------------- (2)

From the above relation, we can deduce that;

P=  $\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }$     -------------(3)

let I=  λ

From the above equations;

d= diameter of the atom

${N_A}$ = avogadro's constant

P= pressure

R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹

T= temperature

From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m

Therefore, we can determine the pressure using equation (3)

i.e

P=  $\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }$

=  $\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }$

=97.06 × 10⁻³ Pa   ×   $\frac{1atm}{1.01325*10^5Pa}$

=9.57 ×  10⁻⁷  atm

Therefore, the maximum residual gas pressure in the calculated space is; 9.57 ×  10⁻⁷  atm