Compute

1. The number of moles
2. The number of molecules in$\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{cm}}^3$of an ideal gas at a pressure of$\mathbf{100}\mathbf{Pa}$and a temperature of$\mathbf{220}\mathbf{K}$.

1. Pressure,$p=100\text{\hspace{0.17em}pascal}$
2. Volume,$V=1{\text{\hspace{0.17em}cm}}^3=1\times {10}^{-6}{\text{m}}^3$
3. Gas constant,$R=8.31\frac{\text{J}}{\text{mol}\cdot \text{K}}$
4. Temperature,$T=220\text{\hspace{0.17em}K}$
5. Avogadro’s number,$N_A=6.02\times {10}^{23}{\text{\hspace{0.17em}mol}}^{-1}$

In this problem, use the ideal gas law. This law holds for any single gas or for any mixture of different gases, i.e., for n total number of moles in the mixture. Ideal gas law gives the equation$\mathbf{\text{pV=nRT.}}$Also, the total number of molecules N can be calculated as$\text{N=n,}$where${\text{N}}_{\text{A}}$denotes Avogadro’s number.

Formula is as follow:

$pv=nRT$

Here,pis the absolute pressure, n is the number of moles of gas present, Tis the temperature in Kelvin, and R is the gas constant.

To calculate the number of moles n, considering ideal gas law,

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

Substitute the values and solve as:

$$\begin{gathered} n=\frac{\left(100\right)(1\times {10}^{-6})}{\left(8.31\right)\left(220\right)} \\ n=5.47\times {10}^{-8}\text{mol} \end{gathered}$$

Hence,the number of molesare$5.47\times {10}^{-8}\text{mol}$

Consider the number of molecules is given by:

$N=n{N}_A$

Substitute the values and solve as:

$N=3.29\times {10}^{16}\text{molecules}$

Hence,the number of molecules in ${\text{1\hspace{0.17em}cm}}^{\text{3}}$of an ideal gas at a pressure of $\text{100\hspace{0.17em}Pa}$and a temperature of$\text{220\hspace{0.17em}K}$are$3.29\times {10}^{16}$