Question

Question: The best laboratory vacuum has a pressure of about,$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{}\mathit{a}\mathit{t}\mathit{m}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}\mathbf{}\mathit{p}\mathit{a}\mathit{s}\mathit{c}\mathit{a}\mathit{l}$. How many gas molecules are there per cubic centimeter in such a vacuum at 293 k?

Short answer

Answer

Total of gas molecules$25\frac{\text{molecules}}{{\text{cm}}^3}$are there per cubic centimeter in such a vacuum at 293 k .

Step-by-step solution

Determine the concept 

By using ideal gas equation, i.e. pv = nRT we can find the number of mole of gas and then find the total number of gas molecule per cubic centimeter at 293 k .

Formula is as follow:

pv = nRT

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

Determine the number of gas molecules in a vacuum at 293 k

Initially, find the number of mole of the gas by using ideal gas equation,

$$\begin{gathered} pv=nRT \\ n=\frac{pv}{RT} \end{gathered}$$

Substitute the given values and solve:

$$\begin{gathered} n=\frac{1.01\times {10}^{-13}\times 1\times {10}^{-6}}{8.31\times 293} \\ n=4.1\times {10}^{-23}\text{mol} \end{gathered}$$

Use this value to find the total number of gas molecules. The formula is,

$N=n{N}_A$

Here, N_A is Avogadro’s number.

$N=4.1\times {10}^{-23}\times 6.02\times {10}^{23}$

$N=25\frac{\text{molecules}}{{\text{cm}}^3}$

Hence, $25\frac{\text{molecules}}{{\text{cm}}^3}$gas molecules are there per cubic centimeter in such a vacuum at 293 k.