Chapter 13, Problem 76GP

To determine

To Find: The number of air molecules and the number of moles.

Answer to Problem 76GP

The number of air molecules in a room of given dimensions is $\text{1}\text{.3}\times {\text{10}}^{\text{27}}$ and The number of moles that corresponds to the available molecules is $2160\text{ moles}$ .

Explanation of Solution

Given:

Length $l=6.5\text{ m}$ .

Width $b=3.1\text{ m}$ .

Height $h=2.5\text{ m}$ .

Temperature $T=22\text{°C }=295\text{ K}$ .

Pressure $P=1\text{ atm }=1.013\times {10}^5{\text{ N/m}}^{\text{2}}$ .

Boltzmann constant $k=1.38\times {10}^{-23}\text{ J/K}$ .

  $\begin{array}{l}\text{Total molecules }N=1.3\times {10}^{27}\\ \text{Number of moles per molecules = }6.02\times {10}^{23}\text{ moles / molecules}\end{array}$

Formula used:

For the determination of air molecules to the ideal gas law formula

  $\begin{array}{l}P\times V=N\times k\times T\\ N=\frac{P\times V}{k\times T}\end{array}$

  $P$=Pressure

  $V$ =volume

  $\eta$ =amount of substance

  $k$ =Ideal gas constant

  $T$ =temperature

For the no. of moles

The following formula is used to determine the total number of moles in the available molecules,

  $\text{Number of moles = }\frac{\text{Total molecules}}{\text{Number of moles / molecules}}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}N=\frac{1.013\times {10}^5\times (6.5\times 3.1\times 2.5)}{1.38\times {10}^{-23}\times 295}\\ N=1.3\times {10}^{27}\text{ molecules}\end{array}$

For the number of moles, substitute the given values in the above equation,

  $\begin{array}{l}\text{Number of moles = }\frac{1.3\times {10}^{27}}{6.02\times {10}^{23}}\\ \text{Number of moles = }2160\end{array}$

Conclusion:

Thus, the number of air molecules in a room of given dimensions is $1.3\times {10}^{27}$ and the number of moles that corresponds to the available molecules is $2160\text{ moles}$ .