Chapter 20, Problem 64PQ

(a)

To determine

The number of molecules.

Answer to Problem 64PQ

The number of molecules is $\underset{\_}{1.07\times {10}^{23}\text{molecules}}$.

Explanation of Solution

Write the equation of ideal gas law.

  $PV=N{k}_{B}T$                                                                                               (I)

Here, $P$ is the pressure, $V$ is the volume, $N$ is the number of molecules, $k_B$ is the Boltzmann constant and $T$ is the temperature.

Write the expression for the volume.

  $V=\frac{4}{3}\pi r^3$                                                                                         (II)

Here, $r$ is the radius.

Rearrange the expression from equation (I) to express in terms of number of molecules.

  $N=\frac{P\left[\frac{4}{3}\pi r^3\right]}{k_{B}T}$                                                                                                    (III)

Conclusion:

Substitute $1.013\times {10}^5\text{Pa}$ for $P$, $10.0\text{cm}$ for $r$, $1.38\times {10}^{-23}\text{J/K}$ for $k_B$ and $15\text{°C}$ for $T$ in equation (III) to find $N$.

  $\begin{array}{c}N=\frac{\left(1.013\times {10}^5\text{Pa}\right)\left[\frac{4}{3}\pi {\left\{\left(10.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}}^3\right]}{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(15\text{°C+273}\text{.15}\right)}\\ =\frac{\left(1.013\times {10}^5\text{Pa}\right)\left[\frac{4}{3}\pi {\left(0.100\text{m}\right)}^3\right]}{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(\text{288}\text{.15K}\right)}\\ =1.07\times {10}^{23}\text{molecules}\end{array}$

Thus, the number of molecules is $\underset{\_}{1.07\times {10}^{23}\text{molecules}}$.

(b)

To determine

The average kinetic energy.

Answer to Problem 64PQ

The average kinetic energy is $\underset{\_}{5.96\times {10}^{-21}\text{J}}$.

Explanation of Solution

Write the equation of average kinetic energy.

  $K_{av}=\frac{3}{2}{k}_{B}T$                                                                                                             (IV)

Here, $K_{av}$ is the average kinetic energy.

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k_B$ and $15\text{°C}$ for $T$ in equation (IV) to find $K_{av}$.

  $\begin{array}{c}{K}_{av}=\frac{3}{2}\left(1.38\times {10}^{-23}\text{J/K}\right)\left(15\text{°C+273}\text{.15}\right)\\ =\frac{3}{2}\left(1.38\times {10}^{-23}\text{J/K}\right)\left(\text{288}\text{.15K}\right)\\ =5.96\times {10}^{-21}\text{J}\end{array}$

Thus, the average kinetic energy is $\underset{\_}{5.96\times {10}^{-21}\text{J}}$.

(c)

To determine

The rms speed of oxygen.

Answer to Problem 64PQ

The rms speed of oxygen is $\underset{\_}{474\text{m/s}}$.

Explanation of Solution

Write the equation for the rms speed.

  $v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$                                                                                                            (V)

Here, $v_{rms}$ is the rms speed, $m$ is the molar mass.

Conclusion:

Substitute $32\left(1.66\times {10}^{-27}\text{kg}\right)$ for $m$, $1.38\times {10}^{-23}\text{J/K}$ for $k_B$ and $15\text{°C}$ for $T$ in equation (V) to find $v_{rms}$.

  $\begin{array}{c}{v}_{rms}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(15\text{°C}+273.15\right)}{\left[32\left(1.66\times {10}^{-27}\text{kg}\right)\right]}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(\text{288}\text{.15}\text{K}\right)}{\left[32\left(1.66\times {10}^{-27}\text{kg}\right)\right]}}\\ =474\text{m/s}\end{array}$

Therefore, the rms speed of oxygen is $\underset{\_}{474\text{m/s}}$.