Chapter 13, Problem 46P

(a)

To determine

To Calculate: The average translational kinetic energy of the oxygen molecules.

Answer to Problem 46P

The average translational kinetic energy of the oxygen molecules $5.65\times {10}^{-21}\text{J}$ .

Explanation of Solution

Given:

At STP, the values of temperature and pressure are $273.15\text{K}$ and ${10}^5\text{Pa}$ .

The value of Boltzmann's constant, $\text{k=1}{\text{.38×10}}^{\text{-23}}\text{J/K}$

Formula used:

The average translational kinetic energy is,

  ${\text{KE}}_{\text{avg}}\text{=}\frac{\text{3}}{\text{2}}\text{kT}$

Calculation:

Now substitute the values in the given formula,

  $\begin{array}{l}{\text{KE}}_{avg}=\frac{3}{2}\text{kT}=\frac{3}{2}\times \left(1.38\times {10}^{-23}\text{J/K}\right)\times \left(273.15\text{K}\right)\\ =5.65\times {10}^{-21}\text{J}\end{array}$

Conclusion:

The average translational kinetic energy of the oxygen molecules is $5.65\times {10}^{-21}\text{J}$ .

(b)

To determine

To Calculate: The total translational energy.

Answer to Problem 46P

  $7.3\times {10}^3\text{J}$

Explanation of Solution

Given:

Number of moles of oxygen gas, $n_{o_2}=2.0\text{mole}$

Temperature is, $T=20°\text{C=}20+273=293\text{K}$

Calculation:

The total translational energy is equal to the product of the number of molecules and the average kinetic energy of gas molecules

  $\begin{array}{l}{\text{KE}}_{total}=N\left({\text{KE}}_{\text{avrg}}\right)=\text{N}\frac{\text{3}}{\text{2}}\text{kT}\\ =\left(\text{2}\text{.0}\text{mole}\right)\text{×}\left(\frac{\text{6}{\text{.023×10}}^{\text{23}}\text{molecules}}{\text{1}}\right)\text{×}\frac{\text{3}}{\text{2}}\left(\text{1}{\text{.38×10}}^{\text{-23}}\text{J/K}\right)\text{×}\left(\text{293K}\right)\\ =\text{7}{\text{.3×10}}^{\text{3}}\text{J}\end{array}$

Conclusion:

The total translational kinetic energy is $\text{7}{\text{.3×10}}^{\text{3}}\text{J}$ .