The vibrational modes of molecular nitrogen are 'frozen out' at r...

Question

The vibrational modes of molecular nitrogen are 'frozen out' at room temperature but become active at temperatures above ≈1500 K. The temperature in the combustion chamber of a jet engine can reach 2000 K, so an engineering analysis of combustion requires knowing the thermal properties of materials at these temperatures. What is the expected specific heat ratio γ for nitrogen at 2000 K?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to calculate or find the theoretical ratio of specific heats of the mysterious gas that we have. In this problem statement, we'll have a mysterious diatomic gas with a strong covalent bond requiring a high temperature to experience vibrational mode. The vibration modes are experienced at temperatures greater than or equal to Kelvin. And suppose the gas enters a combustion chamber of an auto cycle engine that reaches the ignition temperatures of Kelvin. We're being asked to find the theoretical ratio of specific heats of the mysterious gas at 2400 Kelvin. The options given are a 7/2 B 9/ C 9/7 and D seven. So let's consider the mysterious diatomic gas to be ideal. We need to note that the diatomic gas molecules can store a considerable amount of energy as they rotate around two perpendicular axis at constant volume. The specific heat capacity CV is going to be given by seven divided by two, multiplied by R where R is the universal ideal gas constant. Therefore, the change in the thermal energy or delta E T H based on this CV is going to equals to seven divided by two multiplied by N R delta T. We need to remember that when the temperature changes at constant pressure, the thermal energy of the gas rises and the energy must also be put into doing work to expand the gas. Therefore, work will equals to P multiplied by delta V or at the same time equals to N multiplied by R multiplied by delta T. This implies that the total energy that has to be added into the system to increase the temperature is going to be equals by or is going to be given by Q equals to delta E T H plus W or substituting this two equation right here into our Q formula. Q will then be equal to seven divided by two N R delta T for the uh delta E T H plus the W which is just N R delta T. So that will give us a Q value to be equal to nine divided by two of an N R delta TT or nine divided by two multiplied by N multiplied by R and multiplied by delta T. So when we factor out the number of moles, and we can see that C P equals to CV plus R which will then be equal to seven divided by two R plus R, which is equals to nine divided by two multiplied by R. Therefore, in this case, we can conclude that the C P divided by CV or the C P divided by CV ratio or the theoretical ratio of the specific heats will then be equal to nine divided by two R divided by seven divided by two R. We can cross out the R S and the twos to then get the C P divided by CV of nine divided by seven. So that will be the final answer to this particular practice problem. So the theoretical ratio or specific heat of the mysterious gas at 2400, Kelvin is going to equals to nine divided by seven. And that will correspond to option C. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on our website and that'll be it for this one. Thank you.

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