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

In summary, the conversation is discussing how to find the work done by a gas that is expanding adiabatically and reversibly. The first attempt to solve the problem using the equation dw=-pexdV does not give the correct solution. The second attempt, using du=dq+dw and du=nCv,mdt, requires finding the final temperature, which can be done using Tf=(Vf/Vi)^(1/c). There is confusion about the value of c, and the solution guide does not provide a clear explanation. The conversation also touches on using the specific gas constant for CO2 and finding the specific heat capacities for CO2.
  • #1
speny83
17
0

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 don't understand why i can't 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 which 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 don't 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 wouldn't Cp,m just be 7R/2?


now I am so confused i don't 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
 
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  • #2
For R you need to use the specific gas constant for CO2. Which is 0.1889 [itex]\frac{kj}{kg K}[/itex]
 
  • #3
Miaq- I am 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
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
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
I have no idea what your getting at. And I am 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
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.
 

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

1. What are C(v,m) and C(p,m) and how are they related?

C(v,m) and C(p,m) are specific heat capacities, which measure the amount of heat required to raise the temperature of a unit mass of a substance by 1 degree (in Kelvin). C(v,m) refers to specific heat at constant volume, where the volume of the substance remains constant during the heating process. C(p,m) refers to specific heat at constant pressure, where the pressure of the substance remains constant during the heating process. The two are related by the equation C(p,m) = C(v,m) + R, where R is the gas constant.

2. What is the difference between adiabatic and reversible work?

Adiabatic work refers to work done on or by a system without any heat exchange with the surroundings. This means that there is no change in the internal energy of the system. Reversible work, on the other hand, refers to work done on or by a system while maintaining thermal equilibrium with the surroundings. This allows for the possibility of heat exchange and a change in internal energy.

3. How do C(v,m) and C(p,m) relate to adiabatic and reversible work?

For an ideal gas, the relationship between specific heat capacities and work is given by the equation W = C(p,m)(T2-T1) = C(v,m)(T2-T1), where T2 and T1 are the final and initial temperatures, respectively. This shows that the amount of work done is the same for both constant pressure and constant volume processes, as long as the temperature change is the same. However, in real-world situations, the specific heat capacities and work done may vary due to factors such as heat loss or gain.

4. How is the first law of thermodynamics related to C(v,m) and C(p,m)?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. This law is reflected in the equations for specific heat capacities, as the heat added to a substance is equal to the change in its internal energy and the work done on the substance. Therefore, the first law of thermodynamics is closely related to the concept of specific heat capacities and their role in heat and work transfer.

5. What factors can affect the values of C(v,m) and C(p,m)?

The values of specific heat capacities can vary depending on the substance, temperature, and pressure of the system. Other factors that can affect the values include the presence of impurities, phase changes, and the type of process being studied (e.g. adiabatic or reversible). In real-world situations, the values of specific heat capacities may also be influenced by external factors such as heat loss or gain and changes in the surroundings.

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