Chapter 5, Problem 5.82P

(a)

Interpretation Introduction

Interpretation:

The root-mean-square speed of $\text{He}$ at $0°\text{C}$ is to be calculated. Also, the root-mean-square speed of $\text{He}$ at $\text{30 }°\text{C}$ is to be calculated.

Concept introduction:

The expression to calculate the root-mean-square speed is as follows:

$u_{rms}=\sqrt{\frac{3RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

Answer to Problem 5.82P

The root-mean-square speed of $\text{He}$ at $0°\text{C}$ is $1.30\times {10}^3\text{ m/s}$ and the root-mean-square speed of $\text{He}$ at $\text{30 }°\text{C}$ is $1.37\times {10}^3\text{ m/s}$.

Explanation of Solution

The formula to convert $°\text{C}$ to Kelvin is as follows:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (1)

Substitute $\text{0 }°\text{C}$ for $T_{\text{°C}}$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=0+273.15\\ =273.15\text{ K}\end{array}$

The expression to calculate the root-mean-square velocity of $\text{He}$ is as follows:

$u_{rms}=\sqrt{\frac{3RT}{M}}$ (2)

Substitute the value $273.15\text{ K}$ for $T$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for $R$ and $0.004003\text{ kg/mol}$ for $M$ in the equation (2).

$\begin{array}{c}{u}_{rms}=\sqrt{\frac{3\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(273.15\text{ K}\right)}{\left(0.004003\text{ kg/mol}\right)}\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)}\\ =1.3042\times {10}^3\text{ m/s}\\ \approx 1.30\times {10}^3\text{ m/s}\end{array}$

Substitute $\text{30 }°\text{C}$ for $T_{\text{°C}}$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=30+273.15\\ =303.15\text{ K}\end{array}$

Substitute the value $303.15\text{ K}$ for $T$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for $R$ and $0.004003\text{ kg/mol}$ for $M$ in the equation (2).

$\begin{array}{c}{u}_{rms}=\sqrt{\frac{3\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(303.15\text{ K}\right)}{\left(0.004003\text{ kg/mol}\right)}\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)}\\ =1.374\times {10}^3\text{ m/s}\\ \approx 1.37\times {10}^3\text{ m/s}\end{array}$

Conclusion

The root-mean-square speed of $\text{He}$ at $0°\text{C}$ is $1.30\times {10}^3\text{ m/s}$. And the root-mean-square speed of $\text{He}$ at $\text{30 }°\text{C}$ is $1.37\times {10}^3\text{ m/s}$.

(b)

Interpretation Introduction

Interpretation:

The root-mean-square speed of $\text{He}$ and $\text{Xe}$ at $\text{30 }°\text{C}$ are to be compared.

Concept introduction:

The expression to calculate the root-mean-square speed is as follows:

$u_{rms}=\sqrt{\frac{3RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

$\frac{\text{Rate A}}{\text{Rate B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}$

Here, $M_{\text{B}}$ is the molar mass of gas $\text{B}$.

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.

Answer to Problem 5.82P

The root-mean-square speed of $\text{He}$ and $\text{Xe}$ at $\text{30 }°\text{C}$ is $5.73$

Explanation of Solution

The formula to convert $°\text{C}$ to Kelvin is:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (1)

Substitute $\text{30 }°\text{C}$ for $T_{\text{°C}}$ in equation (1)

$\begin{array}{c}{T}_{\text{K}}=30+273.15\\ =303.15\text{ K}\end{array}$

The expression to calculate the root-mean-square velocity of $\text{Xe}$ is as follows:

$u_{rms}=\sqrt{\frac{3RT}{M}}$ (2)

Substitute the value $303.15\text{ K}$ for $T$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for $R$ and $0.004003\text{ kg/mol}$ for $M$ in the equation (2).

$\begin{array}{c}{u}_{rms}=\sqrt{\frac{3\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(303.15\text{ K}\right)}{\left(0.13113\text{ kg/mol}\right)}\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)}\\ =239.913\times {10}^3\text{ m/s}\end{array}$

The expression to compare the root-mean-square speed of $\text{He}$ with that of $\text{Xe}$ is as follows:

$\frac{\text{Rate of He}}{\text{Rate of Xe}}=\frac{u_{rms}\text{ of He}}{u_{rms}\text{of Xe}}$ (3)

Substitute the value $1.3740\times {10}^3\text{ m/s}$ for $u_{rms}$ of $\text{He}$ and $239.913\text{ m/s}$ for $u_{rms}$ of $\text{Xe}$ in the equation (3).

$\begin{array}{c}\frac{\text{Rate He}}{\text{Rate Xe}}=\frac{1.3740\times {10}^3\text{ m/s}}{239.913\text{ m/s}}\\ =5.727076\\ \approx 5.73\end{array}$

Conclusion

The root-mean-square speed $\text{He}$ and $\text{Xe}$ at $\text{30 }°\text{C}$ is $5.73$.

(c)

Interpretation Introduction

Interpretation:

The average kinetic energy of $\text{He}$ at $\text{30 }°\text{C}$ is to be calculated. Also, the average kinetic energy of $\text{Xe}$ at $\text{30 }°\text{C}$ is to be calculated.

Concept introduction:

The expression to calculate the average kinetic energy is as follows:

$E_{\text{k}}=\frac{1}{2}\left(\text{m}\right){\left(u\right)}^2$

Here, $E_{\text{k}}$ is the average kinetic energy, $m$ is the mass and $u$ is the speed.

The kinetic energy is directly proportional to the temperature.

Answer to Problem 5.82P

The average kinetic energy of $\text{He}$ at $\text{30 }°\text{C}$ is $3.78\times {10}^3$. And also the average kinetic energy of $\text{Xe}$ at $\text{30 }°\text{C}$ is $3.78\times {10}^3\text{ J/mol}$.

Explanation of Solution

The expression to calculate the average kinetic energy of $\text{He}$ is as follows:

$E_{\text{He}}=\frac{1}{2}\left(\text{m}\right){\left(u\right)}^2$ (4)

Substitute the value $1.3740\times {10}^3\text{ m/s}$ for the speed of $\text{He}$ and $0.004003\text{ g/mol}$ for the mass of $\text{He}$ in the equation (4).

$\begin{array}{c}{E}_{\text{He}}=\frac{1}{2}\left[\left(0.004003\text{ kg/mol}\right){\left(1.3740\text{ m/s}\right)}^2\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)\right]\\ =3778.58\text{ J/mol}\\ =3.78\times {10}^3\text{ J/mol}\end{array}$

The expression to calculate the average kinetic energy of $\text{Xe}$ is as follows:

$E_{\text{Xe}}=\frac{1}{2}\left(\text{m}\right){\left(u\right)}^2$ (5)

Substitute the value $239.913\times {10}^3\text{ m/s}$ for the speed of $\text{Xe}$ and $0.1313\text{ g/mol}$ for the mass of $\text{Xe}$ in the equation (5).

$\begin{array}{c}{E}_{\text{Xe}}=\frac{1}{2}\left[\left(0.1313\text{ Kg/mol}\right){\left(239.913\text{ m/s}\right)}^2\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)\right]\\ =3778.70\text{ J/mol}\\ \text{= }3.78\times {10}^3\text{ J/mol}\end{array}$

Conclusion

The average kinetic energy of $\text{He}$ at $\text{30 }°\text{C}$ is $3.78\times {10}^3$. And also the average kinetic energy of $\text{Xe}$ at $\text{30 }°\text{C}$ is $3.78\times {10}^3\text{ J/mol}$.

(d)

Interpretation Introduction

Interpretation:

The average kinetic energy per molecule of $\text{He}$ at $\text{30 }°\text{C}$ is to be calculated.

Concept introduction:

The expression to calculate the average kinetic energy is as follows:

$E_{\text{k}}=\frac{1}{2}\left(\text{m}\right){\left(u\right)}^2$

Here, $E_{\text{k}}$ is the average kinetic energy, $m$ is the mass and $u$ is the speed.

The kinetic energy is directly proportional to the temperature.

Answer to Problem 5.82P

The average kinetic energy per molecule of $\text{He}$ at $\text{30 }°\text{C}$ is $6.27\times {10}^{-21}\text{ J/atom}$.

Explanation of Solution

The expression to calculate the average kinetic energy of $\text{Xe}$ is as follows:

$E_{\text{Xe}}=\frac{1}{2}\left(\text{m}\right){\left(u\right)}^2$ (6)

Rearrange the equation (6) in per molecule by dividing ${\text{N}}_{\text{A}}$.

$E_{\text{Xe}}=\frac{1}{2}\frac{\left(\text{m}\right){\left(u\right)}^2}{{\text{N}}_{\text{A}}}$

Substitute the value $239.913\times {10}^3\text{ m/s}$ for the speed of $\text{Xe}$ and $0.1313\text{ g/mol}$ for the mass of $\text{Xe}$ in the equation (5).

$\begin{array}{c}{E}_{\text{Xe}}=\frac{1}{2}\frac{\left[\left(0.1313\text{ Kg/mol}\right){\left(239.913\text{ m/s}\right)}^2\left(\frac{\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{\text{J}}\right)\right]}{\left(6.022\times {10}^{23}\text{ atoms}\right)}\\ =6.2746\times {10}^{-21}\text{ J/atom}\\ \approx 6.27\times {10}^{-21}\text{ J/atom}\end{array}$

Conclusion

The average kinetic energy per molecule of $\text{He}$ at $\text{30 }°\text{C}$ is $6.27\times {10}^{-21}\text{ J/atom}$.