The average kinetic energy of an ideal-gas atom or molecule is (3...

Question

The average kinetic energy of an ideal-gas atom or molecule is (3/2)kT, where T is the Kelvin temperature (Chapter 18). The rotational inertia of the H2 molecule is 4.6 * 10^-48 kg•m^2. What is the value of T for which (3/2)kT equals the energy separation between the l = 0 and l = 1 energy levels of H2? What does this tell you about the number of H2 molecules in the l = 1 level at room temperature?

Accepted Answer

Hi, everyone. In this practice problem, we will have to recall the expression of three divided by two multiplied by K T where T is the temperature. In Kelvin describes an ideal Gasso average kinetic energy per molecule. We're being asked to find the temperature at which the energy difference between the I equals to zero and I equals to one levels of the 02 molecule is equal to the average kinetic energy. The moment of inertia of the 02 molecule is 1.94 times to the power of negative 46 kg meters squared. And we were also being asked to comment if there will be enough excited electrons at room temperature. The actions given are a 277 Kelvin with no excited electrons. B 277 Kelvin with a substantial number of excited electrons. C 2.77 Kelvin with no excited electrons. And lastly, D 2.77 Kelvin with a substantial number of excited electrons. So we wanna re recall that the rotational energy level or E L E L is going to be equals to L multiplied by L plus one multiplied by H bar squared divided by two I L is the rotational quantum number. Each bar is essentially H or the plants constant divided by two pi which will be the constant of 1.55 times 10 to the power of negative 34 seconds. And I is the moment of inertia. The change in energy or delta E will be equals to E one minus E zero where the, for E one L will equals to one and for E zero L will equals to zero. So we can uh use this formula here to calculate E one and E zero. So for E one, um our E L will be equals to one multiplied by one plus one multiplied by H bar squared divided by two I minus zero, multiplied by zero plus one multiplied by H bar squared divided by two I which will essentially be equals to just zero. So then our delta E will then equal to just H bar squared divided by I just like. So, so it is given that the change in energy will equals to the average kinetic energy. So essentially delta E will be equal to the expression of three divided by two, multiplied by K multiplied by T where K is the Boltzmann constant and D is the temperature. So from the first, uh from the equation that we have here, which is H bar squared divided by I, we can substitute that into our equation here. So that three divided by two K T will be equal to H bar squared divided by I. So now what we are interested to find is an expression for the temperature. So let's rearrange this equation so that we get an equation for T, which will then be equals to two multiplied by H bar squared divided by three K I. This is the final equation that we will have for the temperature. And now we want to actually substitute all of our information that we know into this equation. So T will then equals to two multiplied by H bar squared, which is going to be 1.55 times 10 to the power of negative 34 to second squared. All of that is going to be divided by three multiplied by K, which is going to be the Boltzmann constant, which is going to be 1.38 times 10 to the power of negative divided by Kelvin and multiplied that with the I which is given in the problem statement to be the moment on of inertia for the molecule which is 1.94 times 10 to the power of four negative 46 kram meters squared. And then calculating all of this, we will get the temperature value to then be equal to 0.2772 times 10. Kelvin, which will be equals to 2. Calvin just like So, so at any temperature higher than 2.77 Kelvin, as we keep on increasing the temperature, the molecules will get more energy to excite more and more electrons. So essentially at room temperature or at 300 Kelvin, which is going to be much higher than 2.77 Kelvin, there will be a substantial amount of excited electrons, substantial number of excited electrons just like. So, so there we have it the temperature at which the energy difference between the al equal to zero and all equals to one levels of the molecule is equal to the average kinetic energy is going to be 2.77 Kelvin. And also at room temperature, there will be a substantial number of excited electrons in our molecule. So that will be it for this video. Those two answers will correspond to option D which will be the answer to this practice problem. So if you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics. But other than that, that will be it for this video. Thank you so much for watching. Bye bye.

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