Question

Oxygen $\left(O_2\right)$has a molar mass of $\mathbf{32}\mathbf{.}\mathbf{0}\frac{\mathbf{9}}{\mathbf{m}\mathbf{o}\mathbf{l}}$ . What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of 300 K ; (b) the average value of the square of its speed; (c) the root-mean-square speed; (d) the momentum of an oxygen molecule travelling at this speed? (e) Suppose an oxygen molecule travelling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 m on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule’s velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules travelling at this speed are necessary to produce an average pressure of ? (h) Compute the number of oxygen molecules that are contained in a vessel of this size at 300 K and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

Short answer

(a) The average translational kinetic energy of an oxygen molecule at a temperature of 300 K is $6.21x{10}^{-21}J$.

(b) The average value of the square of its speed is $2.34\times {10}^5\frac{m^2}{s^2}$

(c) The root-mean-square speed is $484\frac{m}{s}$

(d) The momentum of an oxygen molecule travelling at this speed is $2.57\times {10}^{-23}\mathrm{kg}·\frac{m}{s}$

(e) The average force the molecule exerts on one of the walls of the container is $1.24x{10}^{-19}\mathrm{N}$.

(f) The average force per unit area is $1.24x{10}^{-17}\mathrm{Pa}$.

(g) The number of oxygen molecules travelling at this speed is $8.17\times {10}^{21}molecules$

(h) The number of oxygen molecules that are contained in a vessel of this size at 300 K and atmospheric pressure is role="math" localid="1668177595278" $2.45\times {10}^{22}molecules$.

(i) The discrepancy in the values is because the average velocities of individual molecules are equal to $\frac{1}{3}$ of the average velocity of the molecules striking the walls.

Step-by-step solution

A concept:

Momentum is the result of a particle's mass and velocity.

A gas that strictly abides by the ideal-gas law: a gas in which there is typically no molecular attraction.

Identification of the given data:

The molar mass of oxygen is $M=32.0\frac{9}{\mathrm{mo}}$

Side length of cubical vessel is 0.10 m

The temperature is 300 K .

The Boltzmann constant is$k=1.38\times {10}^{-23}\frac{J}{molecules·K}$

(a) The average translational kinetic energy of an oxygen molecule at a temperature of 300 K :

The average translational kinetic energy is expressed as,

$$\begin{gathered} K=\frac{1}{2}m{\left(v^2\right)}_{av} \\ =\frac{3}{2}kT \end{gathered}$$

Here, K is the kinetic energy, m is the mass, v is the velocity, k is the Boltzmann constant and T is the temperature.

Substitute the values in the above equation, and you have

$\begin{array}{r}\mathrm{K}=\frac{3}{2}\left(1.38\times {10}^{-23}\frac{\mathrm{J}}{\text{molecule}\cdot \mathrm{K}}\right)\left(300\mathrm{K}\right)\\ =6.21\times {10}^{-21}\mathrm{J}\end{array}$

(b) The average value of the square of its speed:

The mass has to be calculated from the molar mass,

$\begin{array}{r}\mathrm{m}=\frac{\mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}}\\ =\frac{32.0\times {10}^{-3}\frac{\mathrm{kg}}{\mathrm{mol}}}{6.022\times {10}^{23}\frac{\text{molecules}}{\mathrm{mol}}}\\ =5.314\times {10}^{-26}\frac{\mathrm{kg}}{\text{molecule}}\end{array}$

From part (a), solve for the velocity,

$\begin{array}{r}{\left({\mathrm{v}}^2\right)}_{\mathrm{av}}=\frac{2\left(6.21\times {10}^{-21}\mathrm{J}\right)}{\mathrm{m}}\\ =\frac{2\left(6.21\times {10}^{-21}\mathrm{J}\right)}{5.314\times {10}^{-26}\frac{\mathrm{kg}}{\text{molecule}}}\\ =2.34\times {10}^5\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}\end{array}$

(c) The root-mean-square speed

The root mean square speed is expressed as,

$\begin{array}{r}{\mathrm{v}}_{\mathrm{ms}}=\sqrt{{\left({\mathrm{v}}^2\right)}_{\mathrm{av}}}\\ =\sqrt{2.34\times {10}^5\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}}\\ =484\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Thus, the kinetic energy is $2.34\times {10}^5\frac{m^2}{s^2}$, the average squared velocity is and the rms speed is $484\frac{m}{s}$.

(d) The momentum of an oxygen molecule travelling at the calculated speed.

The momentum of the oxygen molecule is expressed as,

$p=m{v}_{rms}$

Substitute $5.314\times {10}^{-26}\frac{kg}{molecule}$ for $484\frac{m}{s}$and $v_{rms}$ for in the above equation.

$\begin{array}{r}p=\left(5.314\times {10}^{-26}\frac{\mathrm{kg}}{\text{molecule}}\right)\left(484\frac{\mathrm{m}}{\mathrm{s}}\right)\\ =2.57\times {10}^{-23}\mathrm{kg}\cdot \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

(e) The average force the molecule exerts on one of the walls of the container: 

The collision time between the walls is,

$\begin{array}{r}t=\frac{0.20\mathrm{m}}{v_{rms}}\\ =\frac{0.20\mathrm{m}}{484\frac{\mathrm{m}}{\mathrm{s}}}\\ =4.13\times {10}^{-4}\mathrm{s}\end{array}$

The change in momentum can be expressed by taking into consideration that the velocity changes direction,

$\begin{array}{r}\mathrm{\Delta p}={\mathrm{mv}}_{\mathrm{rms}}-\left(-{\mathrm{mv}}_{\mathrm{rms}}\right)\\ =2{\mathrm{mv}}_{\mathrm{ms}}\\ =2\left(2.57\times {10}^{-23}\mathrm{kg}\cdot \frac{\mathrm{m}}{\mathrm{s}}\right)\\ =5.14\times {10}^{-23}\mathrm{kg}\cdot \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Now, according to the second law of Newton, the force is equal to the rate of change of momentum.

$\begin{array}{r}\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\\ =\frac{\mathrm{\Delta p}}{\mathrm{\Delta t}}\\ =\frac{5.14\times {10}^{-23}\mathrm{kg}\cdot \frac{\mathrm{m}}{\mathrm{s}}}{4.13\times {10}^{-4}\mathrm{s}}\\ =1.24\times {10}^{-19}\mathrm{N}\end{array}$

(f) The average force per unit area:

First calculate the area as below.

$\begin{array}{l}A={(sidelength)}^2\\ ={(0.10m)}^2\end{array}$

Pressure is nothing but force per unit area. Mathematically,

$P=\frac{F}{A}$

Substitute $1.24x{10}^{-19}\mathrm{N}$for F and ${(0.10\mathrm{m})}^2$ for A in the above equation.

So, the pressure exerted by a single molecule is,

$\begin{array}{r}P=\frac{\left(1.24\times {10}^{-19}\mathrm{N}\right)}{(0.10\mathrm{m}{)}^2}\\ =1.24\times {10}^{-17}\mathrm{Pa}\end{array}$

Thus, the momentum is $2.57\times {10}^{-23}\mathrm{kg}·\frac{m}{s}$, the force exerted on the walls of the container is $1.24x{10}^{-19}\mathrm{N}$ and the pressure is $1.24x{10}^{-17}\mathrm{Pa}$.

(g) The number of oxygen molecules travelling at the calculated speed:

From part (f), the pressure exerted is$1.24x{10}^{-17}\mathrm{Pa}$by a single molecule. The atmospheric pressure is given in question as,

$1atm=1.013\times {10}^{5}Pa$

Thus, the number of molecules is,

$\begin{array}{r}\mathrm{n}=\frac{1.013\times {10}^5\mathrm{Pa}}{1.24\times {10}^{-17}\frac{\mathrm{Pa}}{\text{molecule}}}\\ =8.17\times {10}^{21}\text{molecules}\end{array}$

(h) The number of oxygen molecules travelling at the calculated speed:

The equation of state for one mole of the gas is,

PV = RT

PV = NkT

Here, P is the pressure, V is the volume of the vessel, N is the number of oxygen molecules, k is the Boltzmann constant and T is the temperature.

Solve for N and substitute all the values,

$\begin{array}{r}N=\frac{PV}{kT}\\ =\frac{\left(1.013\times {10}^5\mathrm{Pa}\right)(0.10\mathrm{m}{)}^3}{\left(1.38\times {10}^{-23}\frac{\mathrm{J}}{\text{molecule}\cdot \mathrm{K}}\right)\left(300\mathrm{K}\right)}\\ =2.45\times {10}^{22}\text{molecules}\end{array}$

(i) The number of oxygen molecules that are contained in a vessel of this size at   and atmospheric pressure. Explain the discrepancy in the values.

From part (g) and (h), it is concluded that,

$\frac{n}{N}\approx 3$

This discrepancy arises because the average velocities of individual molecules are equal to $\frac{1}{3}$ of the average velocity of the molecules striking the walls.