Equation apply to reversible only or not?

  • Thread starter Thread starter siresmith
  • Start date Start date
  • Tags Tags
    Apply Reversible
AI Thread Summary
The equations Tds = du + pdv and du = dq + dw are valid for all processes, but the relationship dw = -pdv holds true only for reversible processes. To prove du = Tds + fdx without knowing if the process is reversible, one must consider the system's response under quasistatic conditions. When work is done on a wire, it expands, and reversing the process leads to a shrinkage, suggesting reversibility. Regarding heat capacity at constant pressure (Cp), it is calculated under constant pressure conditions, making it effectively independent of pressure, although Cp does generally depend on pressure in other contexts. Understanding these principles clarifies the application of thermodynamic equations in various scenarios.
siresmith
Messages
5
Reaction score
0
Equation apply to reversible only or not??

My thinking must be faulty somewhere, but I can't work out what's gonig on...

The equation

Tds = du + pdv and

du = dq + dw

are supposed to be valid for all processes right?

yet dw = -pdv only for reversible cases, yes?

So how do you prove that du = Tds + fdx

given that fdx = dw (for a stretched wire) if you don't know if its a reversible process?

As fdx = dw can only be equated with the -pdv if its reversible.

[or similarly, ydA in place of fdw for the surface tension of a droplet. Are you just supposed to know these are reversible?]

Secondly, is Cp, 'heat capacity at constant pressure' independent of pressure?

Anybody know the answers??
 
Physics news on Phys.org
Firstly, I agree with what was stated on your first two questions. Those equations do hold for all processes and dw = -pdv is only true given you know it is reversible. The problem, I believe arises since you assume you don't know it is reversible. To answer this think about how your system in this case the wire will respond to certain quasistatic conditions.

According to your equation fdx = dw, if work is done on the wire, it should expand/get stretched. Now slowly reverse the process. Slowly do less work and the wire should begin to shrink. This process seems to me to be reversible.

For the question regarding Cp, it is done at a constant pressure, such that dp = 0. This allows one to calculate it by using Cp = dH/dT (partial derivative). In such a case, it can be shown to be independent of pressure.

Hope this will help.
Cheers CB
 
Col.Buendia said:
For the question regarding Cp, it is done at a constant pressure, such that dp = 0. This allows one to calculate it by using Cp = dH/dT (partial derivative). In such a case, it can be shown to be independent of pressure.

If you're keeping p constant, then of course the pressure dependence of C_p will not appear! In general, C_p is dependent on pressure. It can be shown by using Maxwell relations that for constant T,

\left(\frac{\partial C_p}{\partial p}\right)_T=-VT\left[\alpha^2+\left(\frac{\partial \alpha}{\partial T}\right)_p\right]

where \alpha is the volumetric coefficient of thermal expansion.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top