What Are the States of V and p in an Adiabatic Process?

Click For Summary
SUMMARY

The discussion focuses on the states of pressure (p) and volume (V) in the context of an adiabatic process, specifically using the internal energy equation U = (f/2) * N * k * T. The equation can also be expressed as U = (f/2) * p * V, leading to the differential form dU = (f/2)(Vdp + pdV). It is established that p and V represent the immediate states of the gas, which can vary throughout the process. The integration of dU requires careful consideration of whether p and V are constant or variable, particularly in processes like isobaric expansion.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly adiabatic processes
  • Familiarity with the ideal gas law and its applications
  • Knowledge of internal energy equations and their derivations
  • Basic calculus, specifically differentiation and integration techniques
NEXT STEPS
  • Study the implications of the first law of thermodynamics in adiabatic processes
  • Explore the concept of isobaric and isochoric processes in thermodynamics
  • Learn about the Maxwell relations and their applications in thermodynamic systems
  • Investigate the role of Boltzmann's constant in statistical mechanics and thermodynamics
USEFUL FOR

Students of thermodynamics, physicists, and engineers involved in energy systems or gas dynamics will benefit from this discussion, particularly those looking to deepen their understanding of adiabatic processes and internal energy calculations.

Niles
Messages
1,834
Reaction score
0

Homework Statement


I have the equation for the internal energy:

U = (f/2) * N * k * T, where f is the degrees of freedom, N is the number of molecules, k is Bolzmann's constant and T is the temperature in Kelvin.

This can be written as U = (f/2)*p*V using the ideal gas law. Differentiating this I get:

delta U = (f/2)*(delta_p*V + p*delta_V).

In this equation, I know what delta_p and delta_V are, but what about V and p? Are they the initial or final states?

Thanks in advance.
 
Physics news on Phys.org
I think you mean not delta but dU=(f/2)(Vdp+pdV). The values are those of the immediate states, and can vary depending on what you're doing to the gas. If you wanted to integrate dU to find the change in energy during a process, you would need to keep p and V inside the integrand unless they were constant, and this could make the integration complicated.

In practice, then, you wouldn't express dU this way unless p or V were constant (or a known function of other variables) during the process. For example, if you heat a gas while expanding from V1 to V2 in an isobaric process, then p is constant, dp=0, and U=(f/2)p(V2-V1).
 

Similar threads

Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
6K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
9K
  • · Replies 1 ·
Replies
1
Views
4K