Rotational/Translational Contributions?

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Rebecca Remple 1C
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Rotational/Translational Contributions?

Postby Rebecca Remple 1C » Sun Feb 09, 2020 5:56 pm

Hi all,

I am struggling to understand what is meant by rotational and translational contributions. Homework problem 4C.5 deals with these contributions of molecular motions. I understand how to add the two contributions together but do not understand the question conceptually. What do these contributions mean? Are we supposed to memorize their values (e.g. 3/2 or 5/2). What does "R" mean in this context? If anyone could explain this concept in a little more detail I would really appreciate it. Thank you so much much!

-Rebecca

VPatankar_2L
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Re: Rotational/Translational Contributions?

Postby VPatankar_2L » Sun Feb 09, 2020 9:05 pm

The values given indicate what the heat capacity of an ideal gas could be at constant volume and constant pressure. the main concept to know here is that the change in internal energy = (3/2)Rdelta T. You can find the heat capacity by dividing delta U by delta T. You can use the "How Do We Know?" box on page 265 to see that Cp= R + Cv when you divide delta H by delta T and substitute delta U + nRdeltaT for delta H. Knowing that Cv=3/2R, you can substitute this for Cv in Cp=R+Cv and get 5/2R.

Rebecca Remple 1C
Posts: 137
Joined: Wed Sep 18, 2019 12:16 am
Been upvoted: 1 time

Re: Rotational/Translational Contributions?

Postby Rebecca Remple 1C » Wed Feb 12, 2020 2:06 pm

VPatankar_2L wrote:The values given indicate what the heat capacity of an ideal gas could be at constant volume and constant pressure. the main concept to know here is that the change in internal energy = (3/2)Rdelta T. You can find the heat capacity by dividing delta U by delta T. You can use the "How Do We Know?" box on page 265 to see that Cp= R + Cv when you divide delta H by delta T and substitute delta U + nRdeltaT for delta H. Knowing that Cv=3/2R, you can substitute this for Cv in Cp=R+Cv and get 5/2R.

Thank you so much for your explanation! It makes a lot more sense now. Good luck on the midterm :)

-Rebecca


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