Equation apply to reversible only or not?

In summary, the equations Tds = du + pdv and du = dq + dw are valid for all processes, but dw = -pdv only applies to reversible cases. To prove du = Tds + fdx, the process must be reversible. For the question of whether Cp is independent of pressure, it is done at a constant pressure and can be shown to be independent of pressure. However, in general, Cp is dependent on pressure.
  • #1
siresmith
5
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??
 
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  • #2
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
 
  • #3
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 [itex]p[/itex] constant, then of course the pressure dependence of [itex]C_p[/itex] will not appear! In general, [itex]C_p[/itex] is dependent on pressure. It can be shown by using Maxwell relations that for constant [itex]T[/itex],

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

where [itex]\alpha[/itex] is the volumetric coefficient of thermal expansion.
 

1. Is the equation only applicable to reversible processes?

Yes, the equation is specifically designed for reversible processes and may not accurately represent irreversible processes.

2. What is the difference between reversible and irreversible processes?

Reversible processes are those that can be reversed and do not result in any net change in the system, while irreversible processes result in a net change in the system.

3. Can the equation be used for all types of equations?

No, the equation is only applicable to thermodynamic equations that involve reversible processes.

4. Why is the equation only applicable to reversible processes?

The equation is based on the assumption that the system is always in thermal equilibrium, which is only true for reversible processes.

5. Can the equation be used to predict the behavior of irreversible processes?

No, the equation is not accurate for irreversible processes and cannot be used to predict their behavior.

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