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Average atomic and molecular speeds $\left(v_{\text { rms }}\right)$ are large, even at low temperatures. What is $v_{\text { rms }}$ for helium atoms at $5.00 \mathrm{K},$ just one degree above helium's liquefaction temperature?
$v_{r m s}=177 \mathrm{m} / \mathrm{s}$
Physics 101 Mechanics
Chapter 13
Temperature, Kinetic Theory, and the Gas Laws
Temperature and Heat
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so the expression for the root means squared velocity or the root means squared speed would be equal to the square root of three times bolts, man constant times the temperature divided by the mass. In this case, the mass would be equal to the Moler mass divided by ah, the number of Adams. Permal. So this is always gonna be avocados number. And so the mass of a single hydrogen single helium adam would be 4.0 to six times 10 to the negative third kilograms. Permal, divided by we could say 6.2 times 10 to the 23rd more. And so we have that the masses equaling 6.6 five times 10 to the negative 27 kilograms for a single helium atom. And so the root means squared speed would be equaling three times. Ah bolts. One constant, 1.38 times 10 to the negative. 23rd Jules per kelvin. This would be multiplied by five Calvin, and this would be divided by 6.65 times 10 to the negative 27th kilograms. And we find that the root means squared speed is equaling one point 76.4 meters per second. This would again before helium. This would be our final answer. That is the end of the solution. Thank you.
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