Chapter 7, Problem 7.69QP

(a)

Interpretation Introduction

Interpretation: The root mean square speed and the wavelength of helium atom at the given temperature are to be calculated.

Concept introduction: Wavelength is defined as the equal distance between the two successive troughs and wave crests. It is denoted by a symbol lambda $\left(\lambda \right)$ and is calculated in the unit of meter or nanometer. The wavelength of photons emitted is inversely proportional to the frequency of photon.

To determine: The root mean square speed of helium atom at $500\text{K}$.

Answer to Problem 7.69QP

Solution

The root mean square speed is $\underset{\_}{1.765\times 10^{3}m/s}$.

Explanation of Solution

Explanation

Given

The temperature is $500\text{K}$.

The molar mass of helium is $4.0\text{g/mol}$.

The conversion of g to kg is done as,

$1\text{g}={10}^{-3}\text{kg}$

Therefore the conversion of $4.0\text{g}$ to kg is done as,

$4\text{g}=4\times {10}^{-3}\text{kg}$

The molar mass is $4\times {10}^{-3}\text{kg/mol}$

The speed of particles in gas phase is measured in terms of root mean square speed. The root mean square speed is given by the equation,

$C_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Where,

• $C_{\text{rms}}$ is the root mean square speed.
• $R$ is the universal gas constant $\left(8.314\text{J/mol}\text{.K}\right)$.
• $T$ is the temperature.
• $M$ is the molar mass.

Substitute the value of $R$, $T$ and $M$ in the above equation to calculate the root mean square speed of helium.

$\begin{array}{c}{C}_{\text{rms,He}}=\sqrt{\frac{3\times 8.314\text{J/mol}\text{.K}\times \text{500}\text{K}}{4.0\times {10}^{-3}\text{kg/mol}}}\\ =\sqrt{\frac{12471\text{J}}{4.0\times {10}^{-3}\text{kg}}}\\ =\sqrt{3117750\text{J/kg}}\\ =\underset{\_}{1.765\times 10^{3}m/s}\end{array}$

Since, $\text{J/kg}=\text{Nm/kg}=\text{kg}{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{kg}=\text{m/s}$.

Hence, the root mean square speed is $\underset{\_}{1.765\times 10^{3}m/s}$.

(b)

Interpretation Introduction

To determine: The wavelength of helium atom at $500\text{K}$.

Answer to Problem 7.69QP

Solution

The wavelength of helium atom is $\underset{\_}{5.521\times 10^{-31}nm}$.

Explanation of Solution

Explanation

Given

The mass of helium is $4\times {10}^{-3}\text{kg/mol}$

The wavelength is given by equation,

$\lambda =\frac{h}{mu}$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $u$ is the velocity $\left(\text{3}\times {\text{10}}^8\text{m/s}\right)$.
• $m$ is the mass of the particle in $\text{kg}$.
• $\lambda$ is the wavelength of a particle in nm.

Substitute the values of $h$, $u$ and $m$ in the above equation to calculate the wavelength of helium atom.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{.s}}{\left(4.0\times {10}^{-3}\text{kg}\right)\left(3\times {10}^8\text{m/s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{kg}{\text{.m}}^{\text{2}}\text{/s}}{12\times {10}^5\text{kg}\text{.m/s}}\\ =5.521\times {10}^{-40}\text{m}\end{array}$

The conversion of m to nm is done as,

$1\text{m}={10}^9\text{nm}$

Therefore, the conversion of $5.521\times {10}^{-40}\text{m}$ to nm is done as,

$\begin{array}{c}5.521\times {10}^{-40}\text{m}=5.521\times {10}^{-40}\times {10}^9\text{nm}\\ =\underset{\_}{5.521\times 10^{-31}nm}\end{array}$

Hence, the wavelength of helium atom is $\underset{\_}{5.521\times 10^{-31}nm}$.

Conclusion

1. a. The root mean square speed is $\underset{\_}{1.765\times 10^{3}m/s}$.
2. b. The wavelength of helium atom is $\underset{\_}{5.521\times 10^{-31}nm}$.