Question

Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{-15}\, atm$ at room temperature $(300\, K)$. Taking $R = 8.3\, JK^{-1} \,mole^{-1}, 1\, atm = 10^5\, Pa$ and $N_{\text{Avogadro}} = 6 \times 10^{23} mole^{-1}$, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

• A. $0.2 \,\mu m$
• B. $0.2 \,mm$
• C. $0.2\, cm$
• D. $0.2\, nm$

Answer 1

The Correct Option is B

 To determine the mean distance between molecules of gas in an evacuated vessel, we start by considering the ideal gas equation and some basic conceptual understanding related to molecular spacing.

Given:

• Pressure, \(P = 4.0 \times 10^{-15} \, \text{atm}\)
• Room Temperature, \(T = 300 \, \text{K}\)
• Gas constant, \(R = 8.3 \, \text{J K}^{-1}\text{mole}^{-1}\)
• 1 atm = \(10^5 \, \text{Pa}\)
• Avogadro's number, \(N_{\text{Avogadro}} = 6 \times 10^{23} \, \text{mole}^{-1}\)

First, we need to convert the pressure from atm to pascals:

\(P = 4.0 \times 10^{-15} \times 10^5 \, \text{Pa} = 4.0 \times 10^{-10} \, \text{Pa}\)

Using the ideal gas equation, we have:

\(PV = nRT\)

For a single molecule of gas, we consider the gas constant per molecule \((k = \frac{R}{N_{\text{Avogadro}}})\):

\(k = \frac{8.3}{6 \times 10^{23}} \, \approx \, 1.38 \times 10^{-23} \, \text{J/K}\)

The ideal gas equation for a single molecule becomes:

\(PV = NkT \Rightarrow V = \frac{kT}{P}\)

Substituting the values, we can calculate the volume occupied by a single molecule:

\(V = \frac{1.38 \times 10^{-23} \times 300}{4.0 \times 10^{-10}} \approx 1.035 \times 10^{-21} \, \text{m}^3\)

The mean distance \(d\) between the molecules is the cube root of the volume \((d = V^{1/3})\):

\(d = (1.035 \times 10^{-21})^{1/3} \approx 1.01 \times 10^{-7} \, \text{m}\)

After converting the distance into a more suitable unit:

\(d = 1.01 \times 10^{-7} \, \text{m} = 0.1 \, \text{mm}\)

Considering approximations and relevant order of magnitudes, the mean distance is of the order of \(0.2 \, \text{mm}\) which matches the option:

Correct Answer:

$0.2 \,mm$