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C(v,m) vs C(p,m) and adiabatic, reversible work

  • Thread starter speny83
  • Start date
  • #1
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Homework Statement


A sample of carbon dioxide, mass 2.45g, is allowed to expand reversible & adiabaticly from 500cm3 to 3.00 dm3
What is the work done by the gas

Homework Equations


dw=-pexdV
du=dq+dw
du=nCv,mdt


The Attempt at a Solution



At first i just wanted to say that because dw=-pexdV and pex=p=[itex]\frac{nrt}{v}[/itex] so w=-nrt∫dV/V....w=-nrt*ln(vf/vi)
however this doesent give me the correct sollution, but i dont understand why i cant say that.

my second thought was to say that because du=dq+dw and dq=0 for adiabatics that it would make du=dw and with du=nCv,mdt so w=nCv,mΔT i could go about it this way, but i need to find the final temp, as an aside: is the reason i cannot use charles law here because work is being done so p is not constant?

ok so i need final T wich i can get by Tf=([itex]\frac{Vf}{Vi}[/itex])1/c

my notes say that c=Cvm/R and, this is not part of the notes, Cvm for CO2 (because its linear) is just 5R/2 right?

however the solution guide goes into all this c=[itex]\frac{Cvm}{R}[/itex]=[itex]\frac{Cpm-R}{R}[/itex]=3.463

but they dont explain any numbers...if you try to reverse calc it (3.463*8.3145)+8.3145 you get 37 somethig which doesent coordinate to anything either....also wouldnt Cp,m just be 7R/2?


now im so confused i dont know what to do

I moved on and worked some other problems (involving CO2) where i had to find [itex]\gamma[/itex] being Cpm/Cvm doing this using Cpm=7R/2 and Cvm=5R/2 worked just fine so I am inclined to believe that the above just doesent make a lick of sense
 
Last edited:

Answers and Replies

  • #2
47
0
For R you need to use the specific gas constant for CO2. Which is 0.1889 [itex]\frac{kj}{kg K}[/itex]
 
  • #3
17
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Miaq- Im not sure what you mean by that comment. As far as i can see R=8.3145J/kmol is the only R value that would apply...Ive never heard of each gas having their own unique R. and for that matter the solution guide (which does a horrible job explaining the solution) uses the 8.3...
 
  • #4
17
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Miaq- i see your edit addition to your first comment. however that number doesn't appear to resolve anything in the problem, nor is it something we have covered in the first two chapters of this silly book they gave me. but thanks for suggesting a solution.
 
  • #5
47
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The specific gas constant is just R divided by the gas's molar mass. I guess the m is for CO2's molar mass which is 44. Does dividing 37 by 44 give you the right Cp?
 
  • #6
17
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I have no idea what your getting at. And Im sorry but its just getting me more confused trying to use something that i truly believe doesent belong in this problem. Also, I really want to understand where my though processess went wrong, not to find some abstract way of getting to an answer.

Thanks for giving it a shot
 
  • #7
47
0
Sorry about that. I was using the tables in my thermo book that list Cp and Cv. I worked out c using those Cp and Cv values and the specific gas constant I mentioned earlier and got 3.46. I kinda assumed that maybe you had those tables also.
 

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