Chapter 19, Problem 19.15E

Interpretation Introduction

Interpretation:

The temperature required to have an rms-average speed of $200,400,600,800$ and $1000\text{m}/\text{s}$ for $\text{Cs}$ atoms is to be calculated. The pattern of the calculated temperatures is to be predicted.

Concept introduction:

A gas is made up of atoms or molecules that move with very high speeds. The kinetic energy of gases is very high. Every molecule or atom present in a gas can have different energies. Therefore, root mean square speed, most probable velocity, and mean velocity are calculated for a gas.

Answer to Problem 19.15E

The temperature required to have an rms-average speed of $200,400,600,800$ and $1000\text{m}/\text{s}$ is $213.15\text{K}$, $852.61\text{K}$, $1918.36\text{K}$, $3410.42\text{K}$ and $5328.79\text{K}$, respectively. The rate of increase in temperature is exponential to the rate of increase in root-mean-square speed of $\text{Cs}$ atoms.

Explanation of Solution

The root-mean-square speed is calculated by the formula,

$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$…(1)

Where,

• $R$ is the universal gas constant.

• $T$ is the temperature.

• $M$ is the molecular mass of a sample.

The molecular mass of $\text{Cs}$ is $0.1329\text{kg}/\text{mol}$.

The value of universal gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

The value of $v_{\text{rms}}$ is $200\text{m}/\text{s}$.

Substitute the value of universal gas constant, molecular mass and $v_{\text{rms}}$ in equation (1).

$\begin{array}{rll}200\text{m}/\text{s}& =& \sqrt{\frac{3\times 8.314\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}}\\ {\left(200\text{m}/\text{s}\right)}^2& =& \frac{24.94\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}\\ 40000{\text{m}}^2/{\text{s}}^2\times 0.1329\text{kg}/\text{mol}& =& 24.94\text{J}/\text{mol}\cdot \text{K}\times T\\ T& =& 213.15\text{K}\end{array}$

The value of $v_{\text{rms}}$ is $400\text{m}/\text{s}$.

Substitute the value of universal gas constant, molecular mass and $v_{\text{rms}}$ in equation (1).

$\begin{array}{rll}400\text{m}/\text{s}& =& \sqrt{\frac{3\times 8.314\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}}\\ {\left(400\text{m}/\text{s}\right)}^2& =& \frac{24.94\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}\\ 160000{\text{m}}^2/{\text{s}}^2\times 0.1329\text{kg}/\text{mol}& =& 24.94\text{J}/\text{mol}\cdot \text{K}\times T\\ T& =& 852.61\text{K}\end{array}$

The value of $v_{\text{rms}}$ is $600\text{m}/\text{s}$.

Substitute the value of universal gas constant, molecular mass and $v_{\text{rms}}$ in equation (1).

$\begin{array}{rll}600\text{m}/\text{s}& =& \sqrt{\frac{3\times 8.314\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}}\\ {\left(600\text{m}/\text{s}\right)}^2& =& \frac{24.94\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}\\ 360000{\text{m}}^2/{\text{s}}^2\times 0.1329\text{kg}/\text{mol}& =& 24.94\text{J}/\text{mol}\cdot \text{K}\times T\\ T& =& 1918.36\text{K}\end{array}$

The value of $v_{\text{rms}}$ is $800\text{m}/\text{s}$.

Substitute the value of universal gas constant, molecular mass and $v_{\text{rms}}$ in equation (1).

$\begin{array}{rll}800\text{m}/\text{s}& =& \sqrt{\frac{3\times 8.314\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}}\\ {\left(800\text{m}/\text{s}\right)}^2& =& \frac{24.94\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}\\ 640000{\text{m}}^2/{\text{s}}^2\times 0.1329\text{kg}/\text{mol}& =& 24.94\text{J}/\text{mol}\cdot \text{K}\times T\\ T& =& 3410.42\text{K}\end{array}$

The value of $v_{\text{rms}}$ is $1000\text{m}/\text{s}$.

Substitute the value of universal gas constant, molecular mass and $v_{\text{rms}}$ in equation (1).

$\begin{array}{rll}1000\text{m}/\text{s}& =& \sqrt{\frac{3\times 8.314\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}}\\ {\left(1000\text{m}/\text{s}\right)}^2& =& \frac{24.94\text{J}/\text{mol}\cdot \text{K}\times T}{0.1329\text{kg}/\text{mol}}\\ 1000000{\text{m}}^2/{\text{s}}^2\times 0.1329\text{kg}/\text{mol}& =& 24.94\text{J}/\text{mol}\cdot \text{K}\times T\\ T& =& 5328.79\text{K}\end{array}$

Therefore, the temperature required to have an rms-average speed of $200,400,600,800$ and $1000\text{m}/\text{s}$ is $213.15\text{K}$, $852.61\text{K}$, $1918.36\text{K}$, $3410.42\text{K}$ and $5328.79\text{K}$, respectively.

It is clear that the rate of increase in temperature is exponential to the rate of increase in root-mean-square speed of $\text{Cs}$ atoms.

Conclusion

The temperature required to have an rms-average speed of $200,400,600,800$ and $1000\text{m}/\text{s}$ is $213.15\text{K}$, $852.61\text{K}$, $1918.36\text{K}$, $3410.42\text{K}$ and $5328.79\text{K}$, respectively. The rate of increase in temperature is exponential to the rate of increase in root-mean-square speed of $\text{Cs}$ atoms.