Adiabatic process thermodynamics help

In summary, the problem asks for the final temperature of an ideal diatomic gas that is adiabatically compressed from 75 K to 40% of its original volume. Using the equations delta U = Q - W, W = the integral of PdV, and PVgamma = constant, and the knowledge that an ideal diatomic gas has a constant internal energy U = 5/2nKT = 5/2nRT, the final temperature can be solved for using the adiabatic equation TVgamma-1 = constant. This equation can be derived by combining pVgamma = constant with pV/T = constant, and the value of gamma can be found in most textbooks. Therefore, the problem can
  • #1
uchicago2012
75
0

Homework Statement


An ideal diatomic gas at 75 K is adiabatically compressed to 40% its original volume. What is its final temperature?

Homework Equations


delta U = Q - W
W = the integral of PdV
PVgamma = constant
U = 5/2nKT = 5/2nRT (diatomic, ideal gas)

The Attempt at a Solution


So I'm having trouble because I don't know the pressure. I'm sure you can solve it without the pressure, but I didn't really see how to. Since it's an adiabatic process, Q= 0 so delta U = -W. I tried to write equations separately for U and W and set them equal to the negative of one another, but that ended badly, since the two equations were almost identical and everything, including my final temperature, divides out. Also, I wasn't sure if I can just solve for the problem using 1 mole as the amount present, but that's what I tried to do.
 
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  • #2


You need the form of the adiabatic equation with T and V in it.
Do you have this in your notes or book?
It is derived by combining the one you mention, pVγ = constant, with pV/T = constant. [by division of one by the other to eliminate p]
 
  • #3


uchicago2012 said:
PVγ = constant

Since you know this equation, you should probably know this one, too:

TVγ-1 = constant.

And you should know (or at least have in your textbook) the value of γ. Therefore, you can solve this problem fairly easily.
 

FAQ: Adiabatic process thermodynamics help

What is an adiabatic process in thermodynamics?

An adiabatic process in thermodynamics is a process where no heat is exchanged between the system and its surroundings. This means that the system is isolated and there is no transfer of thermal energy.

What is the first law of thermodynamics and how does it relate to adiabatic processes?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In an adiabatic process, the first law still applies, but since no heat is exchanged, the change in internal energy of the system is equal to the work done on the system.

How is the temperature affected in an adiabatic process?

In an adiabatic process, the temperature of the system can change due to the work done on the system. For example, if the system is compressed, the temperature will increase due to the work being done on the system. Similarly, if the system expands, the temperature will decrease due to the work done by the system.

What is the difference between adiabatic and isothermal processes?

In an adiabatic process, no heat is exchanged between the system and its surroundings, while in an isothermal process, the temperature of the system remains constant. This means that in an isothermal process, heat is constantly being exchanged to maintain the temperature, while in an adiabatic process, there is no heat exchange and the temperature can change.

How is the entropy affected in an adiabatic process?

In an adiabatic process, the entropy of the system remains constant since there is no heat exchange. This means that the disorder or randomness of the system does not change. However, if work is done on the system, the entropy may increase due to the increase in internal energy.

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