The mass of a helium atom is 6.64 10-27 kg, while the mass of an argon atom is 6.63 10-26 kg. Find the average speed of helium atoms at 490 K.
The correct answer and explanation is :
To find the average speed of helium atoms at 490 K, we can use the following formula from the kinetic theory of gases:
[
v_{\text{avg}} = \sqrt{\frac{3kT}{m}}
]
Where:
- ( v_{\text{avg}} ) is the average speed of the gas particles,
- ( k ) is the Boltzmann constant (( 1.38 \times 10^{-23} \, \text{J/K} )),
- ( T ) is the temperature in Kelvin (490 K),
- ( m ) is the mass of a helium atom.
Step 1: Substituting values
Given:
- Mass of helium atom, ( m = 6.64 \times 10^{-27} \, \text{kg} ),
- Temperature, ( T = 490 \, \text{K} ),
- Boltzmann constant, ( k = 1.38 \times 10^{-23} \, \text{J/K} ).
Now substitute these into the formula:
[
v_{\text{avg}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 490}{6.64 \times 10^{-27}}}
]
Step 2: Simplifying the expression
First, calculate the numerator:
[
3 \times 1.38 \times 10^{-23} \times 490 = 2.0286 \times 10^{-20}
]
Now divide the numerator by the mass:
[
\frac{2.0286 \times 10^{-20}}{6.64 \times 10^{-27}} = 3.05 \times 10^{6}
]
Finally, take the square root of this value:
[
v_{\text{avg}} = \sqrt{3.05 \times 10^{6}} \approx 1745 \, \text{m/s}
]
Conclusion:
The average speed of helium atoms at 490 K is approximately 1745 m/s.
Explanation:
In the kinetic theory of gases, the average speed of a gas molecule depends on both the temperature and the mass of the molecule. The formula shows that at higher temperatures, gas particles move faster. Conversely, lighter molecules (like helium) tend to move faster than heavier molecules (like argon) at the same temperature.
In this problem, helium atoms are lighter than argon atoms, and at 490 K, the helium atoms will have an average speed of around 1745 m/s. The result is consistent with the fact that lighter gases like helium exhibit higher speeds compared to heavier gases, as the kinetic energy of gas molecules is inversely proportional to their mass.