Question

Modern vacuum pumps make it easy to attain pressures of the order of${\mathbf{10}}^{\mathbf{-}\mathbf{13}}$atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of $\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}$atm and an ordinary temperature of 300.0 K, how many molecules are present in a volume of$\mathbf{1}\mathbf{.}\mathbf{0}\mathit{c}{\mathit{m}}^{\mathbf{3}}$? (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

Short answer

1. $2.201\times {10}^6$molecules are present in a volume of$1.00c{m}^3$
2. $2.446\times {10}^{19}$ molecules are present at temperatures of 300K and 1.00atm pressure.

Step-by-step solution

Ideal Gas Equation

Given data:

$$\begin{gathered} P=9\times {10}^{-14}\times 1.013\times {10}^5 \\ V=1\times {10}^{-6}{m}^3 \\ R=8.314J/mol \\ T=300K \end{gathered}$$

The ideal gas equation is

$$\begin{gathered} PV=nRT \\ n=\frac{PV}{RT} \\ n=\frac{9\times {10}^{-14}\times 1.013\times {10}^5\times 1\times {10}^{-6}}{8.314\times 300} \\ n=3.665\times {10}^{-18}mol \end{gathered}$$

No. of moles

No. of moles

$$\begin{gathered} N=n{N}_A \\ =3.665\times {10}^{-18}\times 6.022\times {10}^{23} \\ =2.201\times {10}^{-18}\times 6.022\times {10}^{23} \\ =2.201\times {10}^{6}molecules \end{gathered}$$

Therefore,$2.201\times {10}^6$molecules are present in a volume of $1.00c{m}^3$.

Step 3:

Molecules present at the same temperature but at 1.00atm pressure;

Assuming$p_{1}V=n_{1}RT$as the first case and $p_{2}V=n_{2}RT$as the second case

Comparing both the cases:

$$\begin{gathered} \frac{p_{1}V}{p_{2}V}=\frac{n_{1}RT}{n_{2}RT} \\ \frac{p_1}{p_2}=\frac{n_1}{n_2} \\ \frac{p_1}{p_2}=\frac{{\displaystyle \raisebox{1ex}{$N_1$}\!\left/ \!\raisebox{-1ex}{$N_A$}\right.}}{{\displaystyle \raisebox{1ex}{$N_2$}\!\left/ \!\raisebox{-1ex}{$N_A$}\right.}}=\frac{N_1}{N_2} \end{gathered}$$

Now the equation and putting the values for no. of molecules

$$\begin{gathered} N_2=\frac{N_1{p}_2}{p_1} \\ =\frac{2.201\times {10}^6\times 1}{9\times {10}^{-14}} \\ =2.446\times {10}^{19}molecules \end{gathered}$$

Hence, $2.446\times {10}^{19}$molecules are present at temperatures of 300K and 1.00atm pressure.