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At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at 20.0$^\circ$C? ($Hint$: Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element. The molar mass of H$_2$ is twice the molar mass of hydrogen atoms, and similarly for N$_2$.)

$3800^{\circ} \mathrm{C}$

Thermal Properties of Matter

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University of Washington

Hope College

University of Sheffield

McMaster University

in this problem, we're gonna talk about the root mean square speed off gas molecules. Eso the root, mean square speed off a gas that has Mueller mass M and a temperature T is equal to the square root of three times the ideal gas. Constant are times the temperature teeth divided by the molar mass M. And in our problem, we have hi hydrogen at a temperature off 20 degrees Celsius and are going to find this is actually a two. So it's the either gen molecule, the hydrogen atoms. Uh, and our goal is to find what should be the temperature off the nitrogen molecules such that it route meet its root mean square speed is equal to the root mean square speed off the H two model s o. The speed of the nitrogen is three r times the temperature of the nitrogen divided by the Mueller mask the nitrogen, and this must be equal to three times our times the temperature of the hydrogen divided by the Mueller mass of the I. So the three R's cancel out and even the square roots vanished. I'm gonna actually tease to have cheese equal to zero times M h i to develop. I am too. No notice that zero is 20 degrees Celsius. I want to convert this to Calvin, so I'm gonna some 173 Kelvin in order to obtain 293. Hello. So the temperature t it's 293 Calvi. Time to the molar mass of the hydrogen. And that is 2 g per mall. Because each hydrogen atom, um, has a massive has a motor, massive 2 g from all. And then we divided Bye. The Mueller mass of the nitrogen, which is 28 g from all because each nitrogen molecule has a more massive 14 webs remote, so cheap is equal to 4102. Calvin, I'm gonna subtract 273 Celsius in order to obtain three 100 from sorry, 3800 29 degrees Celsius, which is the answer to our exercise