Thermodynamics - This time it really is adiabatic

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Homework Help Overview

The problem involves the adiabatic compression of 14 grams of nitrogen gas, starting at 20 degrees Celsius and 1 atmosphere of pressure, and being compressed to a pressure of 20 atmospheres. Participants are tasked with finding the final temperature, work done on the gas, compression ratio, heat transfer, and the pressure ratio, while also illustrating the process on a P-V graph.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate initial and final volumes using the ideal gas law and relationships for adiabatic processes. Some participants question the accuracy of their calculations, particularly regarding the use of units (Pascals vs. atmospheres) and its impact on results.

Discussion Status

Participants are actively discussing their calculations and assumptions, with some expressing uncertainty about the values obtained for final volume and temperature. There is recognition of the adiabatic nature of the process, and some participants are exploring the implications of this on heat transfer.

Contextual Notes

Participants note the complexity of the calculations and the potential for errors in unit conversion. The adiabatic condition implies that heat transfer is zero, which is a point of discussion regarding its implications for temperature changes during the process.

spherical23
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Homework Statement



14 grams of Nitrogen at 20 degrees Celsius in 1 atmosphere of pressure is compressed to a pressure of 20 atmospheres. What is its final temperature and what is the work done on the gas if the process is adiabatic? Obtain the compression ratio (i.e. Vmax/Vmin). How much heat comes in or leaves the gas? What is the ratio of Pmax/Pmin? Show this change on a P-V graph.

Homework Equations



PV = NkT

PinitialVinitial[tex]\gamma[/tex] = PfinalVfinal[tex]\gamma[/tex]

VinitialTinitial[tex]3/2[/tex] = VfinalTfinal[tex]3/2[/tex]

W = -[((PinitialVinitial[tex]\gamma[/tex])/(1-[tex]\gamma[/tex]))(Vfinal(1-[tex]\gamma[/tex]) - Vinitial(1-[tex]\gamma[/tex]))

The Attempt at a Solution



14 grams of Nitrogen = 1 atom

Tinitial = 20 degrees Celsius = 293 Kelvin

Tfinal = ?

Pinitial = 1 atm = 1.013 * 105 Pa (1 Pa = 1 N/m2)

~

PinitialVinitial = NkTinitial

(1.013 * 105 N/m2)Vinitial = 1(1.381 * 10-23 Nm/K)(293K)

Vinitial [tex]\approx[/tex] 4 * 10-26

~

PinitialVinitial[tex]\gamma[/tex] = PfinalVfinal[tex]\gamma[/tex]

(1.013 * 105 N/m2)(4 * 10-26) = (20 atm)(1.013 * 105 N/m2)Vfinal[tex]\gamma[/tex]

Vfinal = (4 * 10-26 m3) [tex]\sqrt[5]{[1.013 * 10<sup>5</sup> N/m<sup>2</sup>) / ((20 atm)(1.013 * 10<sup>5</sup> N/m<sup>2</sup>))]<sup>3</sup>}[/tex]

where [tex]\gamma[/tex] = 5/3

Vfinal [tex]\approx[/tex] 6.6 * 10-29 m3

Okay... I'm not done yet, but I am going to submit this to check out how my latex reference skills are working out. This is tedious.
 
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Oh holy crap, it is a disaster... well, the important parts can be made out. it would be nice to have a little chalk board we could draw on. Maybe I'll continue the problem after a while.. that is enough for now. Hopefully I will get better and faster at inputting this stuff.
 
I thought of a comment... under the fifth-root-radical it is suppose to be (# of Pascals in an atmosphere)/(20 atmospheres)(# of Pascals in an atmosphere) all cubed... because it is Pinitial / Pfinal. I did it with the Pascals because that helps me to work with my units. I think it should be the same outcome if I just use atmospheres because it seems like the # of Pascals in an atmosphere would cancel out anyway. Maybe I am putting in the calculator wrong, but I get different values when I do it with Pascals or atmospheres. It is a difference of like two orders of magnitude. That is turning out the be the most troublesome part of this problem for me. If anyone could figure out what Vfinal should be, that would be nice. :smile:
 
I went with 6.6*10-27 instead of 10-29 because the latter made way to much of a temperature increase. Other than that, I think I can move on from this point. I ended up getting that the change in heat coming in or leaving the system was zero. Does that make sense? I mean, it is adiabatic, so the change in heat has to be zero during the process, but there is adiabatic cooling. Should I worry about that here?
 

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