Chapter 21, Problem 21.8OQ

(i)

To determine

The factor by which the average kinetic energy of the molecules changes.

Answer to Problem 21.8OQ

Option (b), by a factor of $3$.

Explanation of Solution

Given info: The initial temperature of an ideal gas is $300\text{K}$ and the final temperature of an ideal gas is $900\text{K}$.

Formula to calculate the initial average kinetic energy of the gas molecules is,

${\left(K_{\text{i}}\right)}_{\text{avg}}=\frac{3}{2}{k}_{\text{B}}{T}_{\text{i}}$ (1)

Here,

$k_{\text{B}}$ is the Boltzmann constant.

$T_{\text{i}}$ is the initial temperature of an ideal gas.

Formula to calculate the final average kinetic energy of the gas molecules is,

${\left(K_{\text{f}}\right)}_{\text{avg}}=\frac{3}{2}{k}_{\text{B}}{T}_{\text{f}}$ (2)

Here,

$k_{\text{B}}$ is the Boltzmann constant.

$T_{\text{f}}$ is the initial temperature of an ideal gas.

Divide equation (2) by equation (1).

$\begin{array}{c}\frac{{\left(K_{\text{f}}\right)}_{\text{avg}}}{{\left(K_{\text{i}}\right)}_{\text{avg}}}=\frac{\frac{3}{2}{k}_{\text{B}}{T}_{\text{f}}}{\frac{3}{2}{k}_{\text{B}}{T}_{\text{i}}}\\ =\frac{T_{\text{f}}}{T_{\text{i}}}\end{array}$

Substitute $300\text{K}$ for $T_{\text{i}}$ and $900\text{K}$ for $T_{\text{f}}$ in above equation.

$\begin{array}{c}\frac{{\left(K_{\text{f}}\right)}_{\text{avg}}}{{\left(K_{\text{i}}\right)}_{\text{avg}}}=\frac{900\text{K}}{300\text{K}}\\ \frac{{\left(K_{\text{f}}\right)}_{\text{avg}}}{{\left(K_{\text{i}}\right)}_{\text{avg}}}=3\\ {\left(K_{\text{f}}\right)}_{\text{avg}}=3{\left(K_{\text{i}}\right)}_{\text{avg}}\end{array}$ (3)

Conclusion:

The average kinetic energy of the gas molecules increase by a factor of $3$. Hence the option (b) is correct.

The average kinetic energy of the gas molecules increase by a factor of $3$ that is contradictory to option (a). Hence the option (a) is incorrect.

The average kinetic energy of the gas molecules increase by a factor $3$ that is contradictory to option (c). Hence the option (c) is incorrect.

The average kinetic energy of the gas molecules increase by a factor of $3$ that is contradictory to option (d). Hence the option (d) is incorrect.

The average kinetic energy of the gas molecules increase by a factor of $3$ that is contradictory to option (e). Hence the option (e) is incorrect.

(ii)

To determine

The factor by which the rms speed of the gas molecules changes.

Answer to Problem 21.8OQ

Option (c) a factor of $\sqrt{3}$.

Explanation of Solution

Given info: The initial temperature of an ideal gas is $300\text{K}$ and the final temperature of an ideal gas is $900\text{K}$.

Formula to calculate the initial rms speed for the molecules of gas is,

${\left(v_{\text{rms}}\right)}_{\text{i}}=\sqrt{\frac{3R{T}_{\text{i}}}{M}}$ (4)

Here,

$R$ is the universal gas constant.

$M$ is the molar mass of the gas.

Formula to calculate the final rms speed for the molecules of gas is,

${\left(v_{\text{rms}}\right)}_{\text{f}}=\sqrt{\frac{3R{T}_{\text{f}}}{M}}$ (5)

Divide equation (5) by equation (4).

$\begin{array}{c}\frac{{\left(v_{\text{rms}}\right)}_{\text{f}}}{{\left(v_{\text{rms}}\right)}_{\text{i}}}=\frac{\sqrt{\frac{3R{T}_{\text{f}}}{M}}}{\sqrt{\frac{3R{T}_{\text{i}}}{M}}}\\ {\left(v_{\text{rms}}\right)}_{\text{f}}=\sqrt{\frac{T_{\text{f}}}{T_{\text{i}}}}\left({\left(v_{\text{rms}}\right)}_{\text{i}}\right)\end{array}$

Substitute $300\text{K}$ for $T_{\text{i}}$ and $900\text{K}$ for $T_{\text{f}}$ in above equation.

$\begin{array}{c}{\left(v_{\text{rms}}\right)}_{\text{f}}=\sqrt{\frac{900\text{K}}{300\text{K}}}{\left(v_{\text{rms}}\right)}_{\text{i}}\\ =\sqrt{3}{\left(v_{\text{rms}}\right)}_{\text{i}}\end{array}$

Conclusion:

The rms speed of the gas molecules increase by a factor of $\sqrt{3}$. Hence the option (c) is correct.

The rms speed of the gas molecules increase by a factor of $\sqrt{3}$ that is contradictory to option (a). Hence the option (a) is incorrect.

The rms speed of the gas molecules increase by a factor of $\sqrt{3}$ that is contradictory to option (b). Hence the option (b) is incorrect.

The rms speed of the gas molecules increase by a factor of $\sqrt{3}$ that is contradictory to option (d). Hence the option (d) is incorrect.

The rms speed of the gas molecules increase by a factor of $\sqrt{3}$ that is contradictory to option (e). Hence the option (e) is incorrect.

(iii)

To determine

The factor by which the average momentum changes.

Answer to Problem 21.8OQ

Option (c) a factor of $\sqrt{3}$.

Explanation of Solution

Given info: The initial temperature of an ideal gas is $300\text{K}$ and the final temperature of an ideal gas is $900\text{K}$.

Formula to calculate the initial average kinetic energy of the gas molecules is,

${\left(K_{\text{i}}\right)}_{\text{avg}}=\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{i}}$

Here,

$m_0$ is the mass of one molecule of gas.

${\left(v_{\text{avg}}\right)}_{\text{i}}$ is the initial average velocity of the gas molecules.

Formula to calculate the final average kinetic energy of the gas molecules is,

${\left(K_{\text{f}}\right)}_{\text{avg}}=\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{f}}$

Here,

${\left(v_{\text{avg}}\right)}_{\text{f}}$ is the final average velocity of the gas molecules.

From equation (3), the relation between the final and average kinetic energy is given as,

${\left(K_{\text{f}}\right)}_{\text{avg}}=3{\left(K_{\text{i}}\right)}_{\text{avg}}$ (6)

Substitute $\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{f}}$ for ${\left(K_{\text{f}}\right)}_{\text{avg}}$ and $\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{i}}$ for ${\left(K_{\text{i}}\right)}_{\text{avg}}$ in above equation.

$\begin{array}{c}\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{f}}=3\left(\frac{1}{2}{m}_0{\left(v_{\text{avg}}^2\right)}_{\text{i}}\right)\\ {\left(v_{\text{avg}}^2\right)}_{\text{f}}=3{\left(v_{\text{avg}}^2\right)}_{\text{i}}\\ {\left(v_{\text{avg}}\right)}_{\text{i}}=\sqrt{3}{\left(v_{\text{avg}}\right)}_{\text{f}}\end{array}$ (7)

Formula to calculate the average momentum of a molecule is,

$p_{\text{avg}}=-2m_0{v}_{\text{avg}}$

From the above equation the average momentum of a molecule that undergoes in a collision with particular wall is directly proportional to the average speed of a gas molecule. But from the equation (7), the final average speed of a gas molecule increases by a factor of $\sqrt{3}$ so the final average momentum of a molecule is also increase by a factor of $\sqrt{3}$ times as that of the initial average momentum of a molecule.

Conclusion:

The average momentum of a molecule is increase by a factor of $\sqrt{3}$. Hence the option (c) is correct.

The average momentum of a molecule is increase by a factor of $\sqrt{3}$ that is contradictory to option (a). Hence the option (a) is incorrect.

The average momentum of a molecule is increase by a factor of $\sqrt{3}$ that is contradictory to option (b). Hence the option (b) is incorrect.

The average momentum of a molecule is increase by a factor of $\sqrt{3}$ that is contradictory to option (d). Hence the option (d) is incorrect.

The average momentum of a molecule is increase by a factor of $\sqrt{3}$ that is contradictory to option (e). Hence the option (e) is incorrect.

(iv)

To determine

The factor by which the rate of collision of molecules changes.

Answer to Problem 21.8OQ

Option (c) a factor of $\sqrt{3}$.

Explanation of Solution

Given info: The initial temperature of an ideal gas is $300\text{K}$ and the final temperature of an ideal gas is $900\text{K}$.

Formula to calculate the average rate of collision of molecules with the walls is,

$\Delta t_{\text{avg}}=\frac{2d}{v_{\text{avg}}}$

From the above equation the time required for the collision of molecules is inversely proportional to the average speed of the gas molecules but from equation (7), the final average speed of a gas molecule increases by a factor of $\sqrt{3}$, hence the average rate of collision of molecules with the walls decreases by a factor of $\sqrt{3}$. So the change in the factor for the rate of collision is $\sqrt{3}$.

Conclusion:

The rate of collision of molecules with walls change by a factor of $\sqrt{3}$. Hence the option (c) is correct.

The rate of collision of molecules with walls change by a factor of $\sqrt{3}$ that is contradictory to option (a). Hence the option (a) is incorrect.

The rate of collision of molecules with walls change by a factor of $\sqrt{3}$ that is contradictory to option (b). Hence the option (b) is incorrect.

The rate of collision of molecules with walls change by a factor of $\sqrt{3}$ that is contradictory to option (d). Hence the option (d) is incorrect.

The rate of collision of molecules with walls change by a factor of $\sqrt{3}$ that is contradictory to option (e). Hence the option (e) is incorrect.

(v)

To determine

The factor by which the pressure of gas changes.

Answer to Problem 21.8OQ

Option (b) a factor of $3$.

Explanation of Solution

Given info: The initial temperature of an ideal gas is $300\text{K}$ and the final temperature of an ideal gas is $900\text{K}$.

Formula to calculate the pressure of $N$ molecules of an ideal gas is,

$P=\frac{2}{3}\left(\frac{N}{V}\right)\left(K_{\text{avg}}\right)$

From the above pressure of gas is directly proportional to the average kinetic energy of the gas molecules but from equation (6), the final average kinetic energy of gas molecules increases by a factor of $3$, hence the aver pressure of $N$ molecules of an ideal gas increase by a factor of $3$. So the change in the factor for the pressure of gas is $3$.

Conclusion:

The pressure of gas change by a factor of $3$. Hence the option (b) is correct.

The pressure of gas change by a factor of $3$ that is contradictory to option (a). Hence the option (a) is incorrect.

The pressure of gas change by a factor of $3$ that is contradictory to option (c). Hence the option (c) is incorrect.

The pressure of gas change by a factor of $3$ that is contradictory to option (d). Hence the option (d) is incorrect.

The pressure of gas change by a factor of $3$ that is contradictory to option (e). Hence the option (e) is incorrect.