Chapter 10, Problem 155AP

Interpretation Introduction

Interpretation:

The most probable speed and root mean square speed are to be compared and the value of the final temperature is to be calculated.

Concept Introduction:

The root mean square speed for a molecule can be calculated as:

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

Here, $u_{rms}$ represents the root mean square value of molecule, $R$ represents the gas constant, $T$ represents the temperature, and $M$ represents its molar mass.

The most probable speed can be calculated as:

$u_{mp}=\sqrt{\frac{2RT}{M}}.$

Here, $u_{mp}$ represents the most probable value of speed for the molecule, $R$ represents the gas constant, $T$ represents the temperature, and $M$ represents its molar mass.

Answer to Problem 155AP

Solution:

(a)

The root mean square speed is more than the probable speed for nitrogen at the given temperature.

(b)

The final temperature is $1200\text{ K}$.

Explanation of Solution

a)Nitrogen at $25°\text{C}$ of temperature

The most probable speed can be calculated as:

$u_{mp}=\sqrt{\frac{2RT}{M}}.$

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T\left(\text{in K}\right)=T\left(\text{in }°\text{C}\right)+\mathrm{273.15.}$

Substitute $25°\text{C}$ for $T\left(\text{in }°\text{C}\right)$:

$\begin{array}{c}T\left(\text{in K}\right)=\left[25°\text{C}+273.15\right]\text{ K}\\ =\text{298}\text{.15 K}\text{.}\end{array}$

The molar mass of nitrogen $M_{{\text{N}}_2}=28\text{ g}/\text{mol}\text{.}$

Convert gram to kilogram as:

$\begin{array}{c}28\text{ g}/\text{mol}=\left(28\cancel{\text{g}}/\text{mol}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\cancel{\text{g}}}\right)\\ =28\times {10}^{-3}\text{ kg}/\text{mol}\end{array}$ $\begin{array}{c}28\text{ g}/\text{mol}=\left(28\text{ g}/\text{mol}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =28\times {10}^{-3}\text{ kg}/\text{mol}\end{array}$

Substitute $28\times {10}^{-3}\text{ kg}/\text{mol}$ for $M_{{\text{N}}_2}$, $8.314\text{ J}/\text{K}\text{.mol}$ for $R$, and $\text{298}\text{.15 K}$ for $T$ in the above equation as:

$\begin{array}{c}{u}_{mp}=\sqrt{\frac{2\left(8.314\text{ J}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(\text{298}\text{.15 }\cancel{\text{K}}\right)}{28\times {10}^{-3}\text{ kg}/\cancel{\text{mol}}}}\\ =\sqrt{\frac{4957.6}{28\times {10}^{-3}}}\text{ m}/\text{s}\\ =\sqrt{177.06\times {10}^3}\text{ m}/\text{s}\\ =\sqrt{17.7\times {10}^4}\text{ m}/\text{s}\end{array}$ $\begin{array}{c}{u}_{mp}=\sqrt{\frac{2\left(8.314\text{ J}/\text{K}\text{.mol}\right)\left(\text{298}\text{.15 K}\right)}{28\times {10}^{-3}\text{ kg}/\text{mol}}}\\ =\sqrt{\frac{4957.6}{28\times {10}^{-3}}}\text{ m}/\text{s}\\ =\sqrt{177.06\times {10}^3}\text{ m}/\text{s}\\ =\sqrt{17.7\times {10}^4}\text{ m}/\text{s}\end{array}$

On taking the square root, the most probable speed for nitrogen is as:

$\begin{array}{c}{u}_{mp}=4.21\times {10}^2\text{ m}/\text{s}\\ =421\text{ m}/\text{s}\text{.}\end{array}$

The root mean square speed for a molecule can be calculated as:

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

Substitute $28\times {10}^{-3}\text{ kg}/\text{mol}$ for $M_{{\text{N}}_2}$, $8.314\text{ J}/\text{K}\text{.mol}$ for $R$, and $\text{298}\text{.15 K}$ for $T$ in the above equation as follows:

$\begin{array}{c}{u}_{rms}=\sqrt{\frac{3\left(8.314\text{ J}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(\text{298}\text{.15 }\cancel{\text{K}}\right)}{28\times {10}^{-3}\text{ kg}/\cancel{\text{mol}}}}\\ =\sqrt{\frac{7436.5}{28\times {10}^{-3}}}\text{ m}/\text{s}\\ =\sqrt{265.6\times {10}^3}\text{ m}/\text{s}\\ =\sqrt{26.5\times {10}^4}\text{ m}/\text{s}\end{array}$ $\begin{array}{c}{u}_{rms}=\sqrt{\frac{3\left(8.314\text{ J}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(\text{298}\text{.15 }\cancel{\text{K}}\right)}{28\times {10}^{-3}\text{ kg}/\cancel{\text{mol}}}}\\ =\sqrt{\frac{7436.5}{28\times {10}^{-3}}}\text{ m}/\text{s}\\ =\sqrt{265.6\times {10}^3}\text{ m}/\text{s}\\ =\sqrt{26.5\times {10}^4}\text{ m}/\text{s}\end{array}$

On taking the square root, the most probable speed for nitrogen is as:

$\begin{array}{c}{u}_{rms}=5.15\times {10}^2\text{ m}/\text{s}\\ =515\text{ m}/\text{s}\text{.}\end{array}$

Hence, the root mean square speed is more than the probable speed for nitrogen at given temperature.

b) Value of $T_2$

The most probable speed can be calculated as:

$u_{mp}=\sqrt{\frac{2RT}{M}}.$

From the figure,

At temperature $T_1=300\text{ K}$, the speed is $500\text{ m}/\text{s}$. (determined from graph)

At temperature $T_2$, the speed is $1000\text{ m}/\text{s}$. (determined from graph)

For temperature $T_1$, the most probable speed is as follows:

$u_{mp}=\sqrt{\frac{2R{T}_1}{M}}.$

For temperature $T_2$, the most probable speed is as:

$u_{mp\text{'}}=\sqrt{\frac{2R{T}_2}{M}}.$

Now, take the ratio of the most probable speed at temperature $T_1$ and $T_2$ as:

$\begin{array}{c}\frac{u_{mp}}{u_{mp\text{'}}}=\frac{\sqrt{\frac{2R{T}_1}{M}}}{\sqrt{\frac{2R{T}_2}{M}}}\\ =\sqrt{\frac{T_1}{T_2}}.\end{array}$

Substitute $500\text{ m}/\text{s}$ for $u_{mp}$, $1000\text{ m}/\text{s}$ for $u_{mp\text{'}}$, and $\text{300 K}$ for $T_1$ in the above equation as follows:

$\begin{array}{c}\frac{500\text{ m}/\text{s}}{1000\text{ m}/\text{s}}=\sqrt{\frac{\text{300 K}}{T_2}}\\ \frac{1}{2}=\sqrt{\frac{\text{300 K}}{T_2}}.\end{array}$

Now, on squaring both sides, solve as:

$\begin{array}{c}\frac{1}{4}=\frac{\text{300 K}}{T_2}\\ T_2=\text{300 K}\times 4\\ =1200\text{ K}\text{.}\end{array}$

Hence, the final temperatureis $1200\text{ K}$.