# Reversible Processes

1. Nov 13, 2013

### mateomy

I'm working from Callen ch. 4:

Consider a monatomic ideal gas in a cylinder fitted with a piston.The walls of the cylinder and the piston are adiabatic. The system is initially in equilibrium, but the external pressure is slowly decreased. The energy change of the gas in the resultant expansion dV is dU = -PdV . Show, from equation 3.34, that dS = 0, so that the quasi-static adiabatic expansion is isentropic and reversible.

Equation 3.34 is defined as:
$$S = Ns_o + NRln\left[\left(\frac{U}{U_o}\right)^{c} \left(\frac{V}{V_o}\right) \left(\frac{N}{N_o}\right)^{-(c+1)}\right]$$

I'm completely lost as to what this question is asking me. I see the dS = 0 and I think, 'Well they were just talking about thermodynamic configuration space and for dS to be zero..." and then I try to take derivatives. But, I don't know with respect to what, the volume? I'm not looking for answers (not that anybody should be), I just want some guidance on how to tackle this problem.

2. Nov 13, 2013

### qbert

I think the approach should be something like:
S = S(U,V) => dS = dS/dU dU + dS/dV dV
now use what you know about "monatomic ideal gasses"
and the fact dU =-P dV to show the terms in dS cancel.

3. Nov 13, 2013

### mateomy

I'll give that a shot. Thank you for the pointer.