A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N/m^2. During a

Q: Why is the distance between hits $2l$ and not just $l$?

The atom hits a *particular* wall. After hitting it, the atom must travel a distance $l$ to reach the opposite wall, bounce off, and travel another distance $l$ to return to the original wall. Thus, the total distance between consecutive hits on the *same* wall is $l + l = 2l$.

Q: Why do we use $M = 4 \times 10^{-3} \text{ kg/mol}$ instead of just $4$ in the temperature calculation?

In the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, all values must be in standard SI units to get the temperature in Kelvin. Since $v_{\text{rms}}$ is in m/s and $R$ is in J/mol-K, the molar mass $M$ must be in kg/mol. $4 \text{ g/mol}$ is equal to $4 \times 10^{-3} \text{ kg/mol}$.

Q: Why is the average kinetic energy $\frac{3}{2}kT$ and not $\frac{5}{2}kT$?

Helium is a noble gas, which means it is monoatomic. Monoatomic gases only have $3$ translational degrees of freedom and no rotational degrees of freedom. According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2}kT$, giving a total of $\frac{3}{2}kT$.

Q: Can I use $M = 4 \text{ g/mol}$ in the ideal gas equation $m = \frac{pVM}{RT}$?

Yes! If you use $M = 4 \text{ g/mol}$, the resulting mass $m$ will directly be in grams. If you use $M = 4 \times 10^{-3} \text{ kg/mol}$, the mass will be in kilograms ($0.0003 \text{ kg}$), which is equivalent to $0.3 \text{ g}$.

Q: Why did we assume the atom's speed is exactly $v_{\text{rms}}$?

The problem explicitly states that the atom is 'travelling with the root mean square speed'. This is a simplifying assumption provided by the question to allow us to connect the collision frequency directly to the macroscopic $v_{\text{rms}}$ of the gas.

Question

A cubical box of side $1	ext{ m}$ contains helium gas (atomic weight $4$) at a pressure of $100	ext{ N/m}^2$. During an observation time of $1	ext{ s}$, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make $500$ hits with a particular wall, without any collision with other atoms. Take, $R = \frac{25}{3}	ext{ J/mol-K}$ and $k = 1.38 	imes 10^{-23}	ext{ J/K}$.
(a) Evaluate the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
(c) Evaluate the total mass of helium gas in the box.

Sigma Percentile · Accepted Answer

Have you ever wondered what it would be like to track a single atom as it zips around inside a box? In this thrilling problem, we are going to do exactly that! We have a cubical box of side $1 	ext{ m}$ containing helium gas. We are given a fascinating piece of information: a single helium atom, travelling with the root mean square speed, hits a particular wall $500$ times in one second.

This might sound like a chaotic situation, but physics allows us to find order in the chaos. We are tasked with finding three things: the temperature of the gas, the average kinetic energy per atom, and the total mass of the helium gas in the box. Let's dive into the microscopic world and unravel this mystery step by step!

### Analyzing the Setup

Imagine you are standing inside this $1 	ext{ m}^3$ box. A helium atom flies past you, hits the right wall, bounces off, travels all the way to the left wall, bounces off again, and finally returns to hit the right wall.

To hit the **same** wall twice in a row, the atom must travel a total distance of $2l$, where $l$ is the side length of the cube.

We are told that the atom makes $500$ hits with this particular wall in $1 	ext{ s}$. This means the time interval $\Delta t$ between two consecutive hits on the same wall is:
$$ \Delta t = \frac{1}{500} 	ext{ s} $$

Since distance equals speed multiplied by time, we can relate the distance $2l$ to the root mean square speed $v_{	ext{rms}}$ and the time interval $\Delta t$:
$$ 2l = v_{	ext{rms}} \Delta t $$

Rearranging for $v_{	ext{rms}}$, we get:
$$ v_{	ext{rms}} = \frac{2l}{\Delta t} $$

Substituting $l = 1 	ext{ m}$ and $\Delta t = \frac{1}{500} 	ext{ s}$:
$$ v_{	ext{rms}} = \frac{2(1)}{1/500} = 1000 	ext{ m/s} $$

Wow! That atom is travelling at a blistering $1000 	ext{ m/s}$. Now that we have the speed, we can unlock the macroscopic properties of the gas.

### The Master Equation for Temperature

How does the microscopic speed of an atom relate to the macroscopic temperature of the gas? The Kinetic Theory of Gases provides the beautiful bridge between these two worlds. The root mean square speed is given by:
$$ v_{	ext{rms}} = \sqrt{\frac{3RT}{M}} $$

Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas. We want to find the temperature $T$, so let's square both sides and rearrange the equation:
$$ v_{	ext{rms}}^2 = \frac{3RT}{M} $$
$$ T = \frac{M v_{	ext{rms}}^2}{3R} $$

Helium is a light gas with an atomic weight of $4$, which means its molar mass $M$ is $4 	ext{ g/mol}$, or $4 	imes 10^{-3} 	ext{ kg/mol}$. We are given $R = \frac{25}{3} 	ext{ J/mol-K}$. Let's plug in the values:
$$ T = \frac{(4 	imes 10^{-3} 	ext{ kg/mol}) (1000 	ext{ m/s})^2}{3 \left( \frac{25}{3} 	ext{ J/mol-K} ight)} $$

Notice how the $3$ in the denominator beautifully cancels out!
$$ T = \frac{4 	imes 10^{-3} 	imes 10^6}{25} = \frac{4000}{25} = 160 	ext{ K} $$

We have successfully found the temperature of the gas! It is **$160 	ext{ K}$**.

### Unveiling the Kinetic Energy

Next, we need to evaluate the average kinetic energy per atom. Helium is a noble gas, which means it is monoatomic. For a monoatomic gas, the only form of kinetic energy is translational.

According to the equipartition of energy, each degree of freedom contributes $\frac{1}{2}kT$ to the energy. Since a monoatomic gas has $3$ translational degrees of freedom, the average kinetic energy per atom is:
$$ K_{	ext{avg}} = \frac{3}{2}kT $$

We already know the temperature $T = 160 	ext{ K}$, and the Boltzmann constant $k$ is given as $1.38 	imes 10^{-23} 	ext{ J/K}$. Let's substitute these values:
$$ K_{	ext{avg}} = \frac{3}{2} (1.38 	imes 10^{-23} 	ext{ J/K}) (160 	ext{ K}) $$
$$ K_{	ext{avg}} = 1.5 	imes 1.38 	imes 160 	imes 10^{-23} 	ext{ J} $$
$$ K_{	ext{avg}} = 331.2 	imes 10^{-23} 	ext{ J} = 3.312 	imes 10^{-21} 	ext{ J} $$

This tiny amount of energy is exactly what we would expect for a single atom!

### Final Calculation: The Total Mass

For the final part of our journey, we need to find the total mass of the helium gas in the box. Whenever we need to relate pressure, volume, temperature, and mass, the Ideal Gas Law is our best friend:
$$ pV = nRT $$

The number of moles $n$ is simply the total mass $m$ divided by the molar mass $M$. Substituting this into the equation:
$$ pV = \frac{m}{M} RT $$

We want to find the total mass $m$, so let's rearrange the equation:
$$ m = \frac{pVM}{RT} $$

Let's gather our known values. The pressure $p$ is $100 	ext{ N/m}^2$. The volume $V$ of a cube with side $1 	ext{ m}$ is $1 	ext{ m}^3$. The molar mass $M$ is $4 	ext{ g/mol}$ (we can keep it in grams to get the final mass in grams). The gas constant $R$ is $\frac{25}{3} 	ext{ J/mol-K}$, and the temperature $T$ is $160 	ext{ K}$.

Let's plug everything in:
$$ m = \frac{(100) (1) (4)}{\left( \frac{25}{3} ight) (160)} $$
$$ m = \frac{400}{\frac{4000}{3}} $$
$$ m = \frac{1200}{4000} = 0.3 	ext{ g} $$

And there we have it! The total mass of the helium gas in the box is just **$0.3 	ext{ g}$**.

By tracking a single atom, we were able to deduce the temperature, the kinetic energy, and even the total mass of the gas. This is the true power of physics—connecting the microscopic to the macroscopic in a seamless, elegant way. Keep visualizing, keep questioning, and you will master thermodynamics!

Visualized Solution

$l = 1 m, f = 500 hits/s$

$Δ t = \frac{1}{500} s$

$v_{rms} = 1000 m/s$

$T = \frac{M v _{rms}^{2}}{3 R}$

$T = 160 K$

$K_{avg} = \frac{3}{2} k T$

$K_{avg} = 3.312 \times 1 0^{- 21} J$

$m = \frac{p V M}{R T}$

$m = 0.3 g$

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation for Temperature

Unveiling the Kinetic Energy

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l=1 m,f=500 hits/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δt=5001​ s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vrms​=1000 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=3RMvrms2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=160 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kavg​=23​kT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kavg​=3.312×10−21 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=RTpVM​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=0.3 g