Chapter 14, Problem 117P

(a)

To determine

The gas having larger rms speed.

Answer to Problem 117P

The gas having larger rms speed is $\underset{\_}{\text{Nitrogen}}$.

Explanation of Solution

Given that the volume of nitrogen is $3.0\text{L}$, the volume of oxygen is $5.0\text{L}$, the temperature of both gases is $20\text{°C}$, and the pressure of both gases is $1.0\text{atm}$.

The average kinetic energy of the gaseous molecules in a container is proportional to the temperature. In the given situation, both the gases are at same temperature. Hence the average kinetic energy of both gases will be same.

Since the average kinetic energy is directly proportional to the mass of the molecule, the less massive molecule will have high rms speed in order to have same average kinetic energy.

Conclusion:

Among nitrogen and oxygen, nitrogen molecules are less massive. Thus, it will have larger rms speed than oxygen at same temperature.

Therefore, the gas having larger rms speed is $\underset{\_}{\text{Nitrogen}}$.

(b)

To determine

The temperature at which the oxygen gas will have same rms speed as nitrogen.

Answer to Problem 117P

The temperature at which the oxygen gas will have same rms speed as nitrogen is $\underset{\_}{62\text{°C}}$.

Explanation of Solution

Given that the temperature of nitrogen gas is $20\text{°C}$ ($293.15\text{K}$).

Write the expression for the rms speed of a gas molecule.

$v_{\text{rms}}=\sqrt{\frac{3kT}{m}}$ (I)

Here, $v_{\text{rms}}$ is the rms speed, $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass.

Rewrite equation (I) for ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ molecules.

$v_{\text{rms}\left({\text{N}}_2\right)}=\sqrt{\frac{3k{T}_{\left({\text{N}}_2\right)}}{m_{\left({\text{N}}_2\right)}}}$ (II)

$v_{\text{rms}\left({\text{O}}_2\right)}=\sqrt{\frac{3k{T}_{\left({\text{O}}_2\right)}}{m_{\left({\text{O}}_2\right)}}}$ (III)

Since the rms speeds are equal, equate the right hand sides of equations (II) and (III) and solve for $T_{\left({\text{O}}_2\right)}$.

$\begin{array}{c}\sqrt{\frac{3k{T}_{\left({\text{N}}_2\right)}}{m_{\left({\text{N}}_2\right)}}}=\sqrt{\frac{3k{T}_{\left({\text{O}}_2\right)}}{m_{\left({\text{O}}_2\right)}}}\\ T_{\left({\text{O}}_2\right)}=\frac{m_{\left({\text{O}}_2\right)}}{m_{\left({\text{N}}_2\right)}}{T}_{\left({\text{N}}_2\right)}\end{array}$ (IV)

Conclusion:

Substitute $\left(2\times 15.9994\text{u}\right)$ for $m_{\left({\text{O}}_2\right)}$, $\left(2\times 14.0067\text{u}\right)$ for $m_{\left({\text{N}}_2\right)}$, and $293.15\text{K}$ for $T_{\left({\text{N}}_2\right)}$ in equation (IV) to find $T_{\left({\text{O}}_2\right)}$.

$\begin{array}{c}{T}_{\left({\text{O}}_2\right)}=\frac{\left(2\times 15.9994\text{u}\right)}{\left(2\times 14.0067\text{u}\right)}\left(293.15\text{K}\right)\\ =334.9\text{K}\\ =\left(334.9-273.15\right)\text{°C}\\ =62\text{°C}\end{array}$

Therefore, the temperature at which the oxygen gas will have same rms speed as nitrogen is $\underset{\_}{62\text{°C}}$.

(c)

To determine

The amount of heat that must flow into or out of the container of oxygen to change its temperature from $20\text{°C}$ to $62\text{°C}$.

Answer to Problem 117P

Heat of $\underset{\_}{180\text{J}}$ must flow $\underset{\_}{\text{into}}$ the container of oxygen to change its temperature from $20\text{°C}$ to $62\text{°C}$.

Explanation of Solution

Given that the volume of oxygen is $5.0\text{L}$, the temperature of oxygen gas is $20\text{°C}$ ($293.15\text{K}$) , and the pressure of the gas is $1.0\text{atm}$, and the final temperature of the gas is $62\text{°C}$ ($335.15\text{K}$)

Write the expression for the molar specific heat at constant volume of an ideal diatomic gas.

$C_{\text{v}}=\frac{5}{2}R$ (I)

Here, $C_{\text{v}}$ is the molar specific heat at constant volume, $R$ is the universal gas constant.

Write the ideal gas law equation for the gas.

$PV=nR{T}_{\text{i}}$ (II)

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, and $T_{\text{i}}$ is the initial temperature.

Solve equation (II) for $n$.

$n=\frac{PV}{R{T}_{\text{i}}}$ (III

Write the expression for the heat energy required to change the temperature of the gas.

$Q=n{C}_{\text{v}}\Delta T$ (IV)

Here, $Q$ is the heat, , and $\Delta T$ is the change in temperature.

Use equation (I) and (III) in (IV) and expand the term $\Delta T$.

$\begin{array}{c}Q=\left(\frac{PV}{R{T}_{\text{i}}}\right)\left(\frac{5}{2}R\right)\left(T_{\text{f}}-T_{\text{i}}\right)\\ =\frac{5PV}{2T_{\text{i}}}\left(T_{\text{f}}-T_{\text{i}}\right)\end{array}$ (V)

Here, $T_{\text{f}}$ is the final temperature.

Conclusion:

Substitute $1.0\text{atm}$ for $P$, $5.0\text{L}$ for $V$, $335.15\text{K}$ for $T_{\text{f}}$, and $293.15\text{K}$ for $T_{\text{i}}$ in equation (V) to find $Q$.

$\begin{array}{c}Q=\frac{5\left(1.0\text{atm}\right)\left(5.0\text{L}\right)}{2\left(293.15\text{K}\right)}\left(335.15\text{K}-293.15\text{K}\right)\\ =\frac{5\left(1.0\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(5.0\text{L}\times \frac{1{\text{m}}^3}{1000\text{L}}\right)}{2\left(293.15\text{K}\right)}\left(335.15\text{K}-293.15\text{K}\right)\\ =180\text{J}\end{array}$

Therefore, heat of $\underset{\_}{180\text{J}}$ must flow $\underset{\_}{\text{into}}$ the container of oxygen to change its temperature from $20\text{°C}$ to $62\text{°C}$.