Question 53: Thermodynamics & KTG

Question

A cubical box of side $$2 \mathrm{~m}$$ contains helium gas. It was observed that in a time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, made 250 hits with one of the walls, without any collision with other atoms. The average kinetic energy of the helium gas is Take $$R=\frac{25}{3} \mathrm{~J} / \mathrm{mol}-\mathrm{K}$$ and $$\mathrm{kB}=1.38 	imes 10^{-23} \mathrm{JK}{ }^{-1}$$

Accepted Answer

LetÃ¢ÂÂs break down the problem step by step.
Determine the particle's speed from the collision data:
The cubical box has a side length of $$L = 2\,	ext{m}.$$
An atom moving along one edge (perpendicular to a pair of parallel walls) bounces back and forth between two walls.
To hit the same wall again, the atom must travel from one wall to the opposite wall and back, covering a distance of
$$	ext{Distance} = 2L = 2 + 2 = 4\,	ext{m}.$$
If the atom makes 250 hits with the wall in 1 second, then the time between successive hits on that wall is
$$\Delta t = \frac{1}{250}\,	ext{s} = 0.004\,	ext{s}.$$
The relationship between distance, speed, and time is
$$\Delta t = \frac{4}{v_{m rms}}.$$
Solving for the root-mean-square speed, we have
$$ v_{m rms} = \frac{4}{0.004} = 1000\,	ext{m/s}.$$
Relate the rms speed to the temperature:
For a monatomic ideal gas,
$$v_{m rms} = \sqrt{\frac{3k_B T}{m}},$$
where:
Ã¢ÂÂ¢ $$k_B = 1.38 	imes 10^{-23}\,	ext{J/K}$$ is BoltzmannÃ¢ÂÂs constant, and
Ã¢ÂÂ¢ $$m$$ is the mass of one helium atom.
The mass of one helium atom is obtained from its molar mass. For helium, the molar mass is approximately $$0.004\,	ext{kg/mol}.$$
Using AvogadroÃ¢ÂÂs number $$N_A \approx 6.02 	imes 10^{23}\,	ext{mol}^{-1},$$
$$ m = \frac{0.004}{6.02 	imes 10^{23}} \approx 6.64 	imes 10^{-27}\,	ext{kg}.$$
Rearranging the formula for $$v_{m rms},$$ we have:
$$ v_{m rms}^2 = \frac{3k_B T}{m} \quad\Longrightarrow\quad T = \frac{m\,v_{m rms}^2}{3k_B}.$$
Plugging in the values:
$$ T = \frac{6.64 	imes 10^{-27}\,	ext{kg} 	imes (1000\,	ext{m/s})^2}{3 	imes 1.38 	imes 10^{-23}\,	ext{J/K}}.$$
Calculate the numerator:
$$ 6.64 	imes 10^{-27} 	imes 10^6 = 6.64 	imes 10^{-21}\,	ext{J}.$$
And the denominator:
$$ 3 	imes 1.38 	imes 10^{-23} = 4.14 	imes 10^{-23}\,	ext{J/K}.$$
Then,
$$ T \approx \frac{6.64 	imes 10^{-21}}{4.14 	imes 10^{-23}} \approx 160\,	ext{K}.$$
Find the average kinetic energy per atom:
The average kinetic energy of a gas atom in a monatomic ideal gas is given by
$$	ext{Avg KE} = \frac{3}{2}k_BT.$$
Substituting the values:
$$	ext{Avg KE} = \frac{3}{2} 	imes 1.38 	imes 10^{-23}\,	ext{J/K} 	imes 160\,	ext{K}.$$
First, calculate:
$$1.38 	imes 10^{-23} 	imes 160 \approx 2.208 	imes 10^{-21}\,	ext{J}.$$
Then,
$$\frac{3}{2} 	imes 2.208 	imes 10^{-21} \approx 3.312 	imes 10^{-21}\,	ext{J}.$$
Thus, the average kinetic energy of the helium atoms is approximately
$$3.31 	imes 10^{-21}\,	ext{J},$$
which corresponds to Option B.
Final Answer: Option B is correct.

Answer

$$
82.8 	imes 10^{-21} \mathrm{~J}
$$

Answer

$$
3.31 	imes 10^{-21} \mathrm{~J}
$$

Answer

$$
82.8 	imes 10^{-19} \mathrm{~J}
$$

Answer

$$
1 	imes 10^{-21} \mathrm{~J}
$$

Thermodynamics & KTG

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