Textbook's Adiabatic Derivation

  • #1
cj
82
0
My textbook says, basically, for an adiabatic process...
  • All three variables in the ideal gas law—P, V, and T—change during an adiabatic process.
  • Let’s imagine an adiabatic gas process involving an infinitesimal change in volume dV and an accompanying infinitesimal change in temperature dT. The work done by the gas is W=PdV.
Question: if P, V, and T are all changing, should the infinitesimal work be: W=PdV+VdV and not simply W=PdV

I've seen pretty much the same treatment in other textbooks and online, and finally decided to try to find out hat I'm missing here. Any ideas?
 

Answers and Replies

  • #2
Quantum Defect
Homework Helper
Gold Member
495
116
My textbook says, basically, for an adiabatic process...
  • All three variables in the ideal gas law—P, V, and T—change during an adiabatic process.
  • Let’s imagine an adiabatic gas process involving an infinitesimal change in volume dV and an accompanying infinitesimal change in temperature dT. The work done by the gas is W=PdV.
Question: if P, V, and T are all changing, should the infinitesimal work be: W=PdV+VdV and not simply W=PdV

I've seen pretty much the same treatment in other textbooks and online, and finally decided to try to find out hat I'm missing here. Any ideas?
PV work is PdV. VdV is Volume^2, which is not work. Did you mean VdP instead? As Chester Miller has discussed in another recent thread, VdP is not work -- i.e. heating or cooling a gas in a closed, constant volume container does no work. The piston on the old train moves as the steam does work...

Perhaps you are bothered that P and V are both changing..?? This is just like any other integration of a non-constant function.
 
  • #3
cj
82
0
Yes and yes!

Yes, I did mean to type VdP rather than VdV. And yes, I can see how VdP is not work. Thank you very much, this is very helpful!!
 

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