Chapter 1, Problem 1.1P

To determine

Absolute pressure (in Pa) for air.

Answer to Problem 1.1P

Absolute pressure (in Pa) for air is $4.0\text{ Pa}$.

Explanation of Solution

Given Information:

Temperature of gas is $\text{20}\text{°}\text{C}$

Number of molecules is ${10}^{12}$ per cubic millimeter.

The mass of single molecule of air calculated as

$\begin{array}{c}\text{m }=\frac{\text{Molecular}\text{weight}}{\text{Avogadro's}\text{number}}\\ =\frac{\text{28}\text{.97}{\text{mol}}^{\text{-1}}}{\text{6}\text{.023E23}\text{molecules/g}\text{.mol}}\\ =\text{4}\text{.81E-23}\text{g}\end{array}$

Then the density of air including 10¹² molecules per mm³ is, in SI units,

$\begin{array}{c}\text{ρ}=\left({\text{10}}^{\text{12}}\frac{\text{molecules}}{{\text{mm}}^{\text{3}}}\right)\left(\text{4}\text{.81E-23}\left(\frac{\text{g}}{\text{molecule}}\right)\right)\\ =\text{4}\text{.81E-11}\frac{\text{g}}{{\text{mm}}^{\text{3}}}\\ =\text{4}\text{.81E-5}\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

Conclusively, from the perfect gas law, at $20\text{ °C }293\text{ K}$, we get the pressure:

$\begin{array}{c}\text{p}=\text{ρRT}\\ =\left(\text{4}\text{.81E-5}\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right)\left(\text{287}\left(\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}\text{K}}\right)\right)\left(\text{293}\text{}\text{K}\right)\\ =\text{4}\text{.0}\text{}\text{Pa}\end{array}$

Conclusion:

Therefore, the absolute pressure (in Pa) for air is $4.0\text{ Pa}$.