At what temperature is the root-mean-square speed of nitrogen mol...

Question

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°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 H2 is twice the molar mass of hydrogen atoms, and similarly for N2.)

Accepted Answer

everyone in this problem. We have oxygen and chlorine K. These are gas elements in the periodic table. And were asked to determine the temperature when the root mean squared speed of chlorine is the same as the root mean squared speed of oxygen at room temperature. Okay. We're given the molar mass of oxygen and the molar mass of chlorine. Okay. So let's recall that the speed root mean square speed V. R. M. S. Is equal to the square root of three K. T. Over little M. Which can also be written as three R. T. over again. Alright, so we're giving information about the molar mass. Big M. We don't have any information about the mass. So let's go ahead and use that second um formula. And what we want is we want the speed root mean squared speed V. Rms of oxygen. Okay. Equal to the root mean square word speed of chlorine. Okay. That's what we wanna know. We want to know the temperature when those are equal. Alright, so writing up the equation for both. This means that we want the square root of three are the temperature and we're gonna have subscript O for the temperature that we're talking about with the oxygen molecules. Okay, divided by the molar mass of oxygen is equal to the square root of three. Are the temperature When we're talking about chlorine. Tcl provided by the molar mass of chlorine. Okay. And we want to know the temperature when this occurs. The temperature for chlorine, gas when this occurs. So we're looking for this T. C. L. We know tina. We're told the temperature of oxygen is at room temperature, we know the molar mass of oxygen. We know the molar mass of chlorine. K. R. Is r gas constant. And actually when we work this out that's gonna cancel. And we're not even gonna need to plug in that value. Okay so let's try to stall for tcl. Okay, squaring both sides, we get three R. T. O. Divided by M. O. Is equal to three. R. T. C. L. Divided by M. C. L. three are we can divide by three are and then multiplying we're gonna get tcl. The temperature we're looking for is going to be equal to t not I'm M. C. L. Divided by M. So this temperature we're looking for is going to be the initial temperature or the temperature that oxygen was at divided or times of ratio of their molar masses. Okay. Alright so our temperature, we want to write this in Calvin. Okay, temperature we're given is 25°C. So we're gonna have 25 plus 273 Calvin. And then the molar mass we're taking the ratio of the molar masses. So as long as the units of both of those molar masses are the same, we're good. Okay so both of them are given in grams per mole. So we can leave them in grams per mole. So we're gonna have 70.90 g per mole. Okay the unit of or sorry the molar mass of chlorine. And then we have divided by the molar mass of oxygen which is 32 g per day. And so the unit g per mole will cancel because it's the same. We're gonna be left with T. C. L. is equal to .26 Calvin. Okay, subtracting 273 to get us back 2°C. We're going to get 387.26°C. Okay. And so the temperature that we need for those root mean squared speeds to be the same is 387°C. Okay, so that is going to correspond with answers see here. Alright. I hope this video helped. Thanks everyone for watching. See you in the next one.

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