[san1984]

The question goes like this. A fundamental fact of experience on which the second law of thermodynamics is based that the heat absorbed at any one temperature cannot be completely transformed in work without leaving some change in the system or its surroundings.


Consider now the following. When an ideal gas is allowed to expand isothemally against a piston, delta U = q + w =0. Thus in this case the work done by the gas is equal to the heat transferred from the reservoir to the gas, and the efficiency of turning into work is 100%. Explain why this is not a violation of the Second Law ?   

[xiankai]

http://en.wikipedia.org/wiki/Isothermal

according to it, since dU = dq + dw,

and dq = 0 (constant temperature)

dU = dw

and hence the change in internal energy must be converted entirely into work 

[san1984]

hmm... I think that being an isothermal process, q is not 0 in this case but instead change in internal energy is equal to zero and hence q = w.

But the main part of the question is still asking for why it is not a violation of the second law of thermodynamics in this scenerio.

[Winga]

The system becomes much random.

[Hunt]

I think you're right,  san1984, U = 3/2 RT for a gas, at least from KMT perspective. Perhaps this is the case because we're dealing with an ideal gas here?

[Yggdrasil]

The interpretation of the second law given in the OP applies only to engine cycles (i.e. thermodynamic processes which start and end at the same thermodynamic states).  Since an isothermal expansion is not a cyclic process, you can't apply this interpretation of the second law.

[san1984]

oic... now it makes more sense now.. 
tks !

[Donaldson Tan]

du = q + w
du = Cv.dT
q = T.ds
w = - p.dv
Cv.dT = T.ds - p.dV

Since this is an isothermal process, then dT = 0
T.ds = p.dv
ds = (p/T)dv

Since we are dealing with a perfect gas, then p/T = R/V
ds = (p/T)dv = (R/V)dv
=> ?S = R.ln(V_final/V_initial)

Since we are dealing with an expansion process, then V_final > V_initial
=> V_final/V_initial > 1
=> ln(V_final/V_initial) > 0
=> ?S > 0

The second law is not violated at all.

[Hunt]

The interpretation of the second law given in the OP applies only to engine cycles (i.e. thermodynamic processes which start and end at the same thermodynamic states).  Since an isothermal expansion is not a cyclic process, you can't apply this interpretation of the second law.

It makes sense now, but just out of curiousity , what exactly is the internal energy of a real gas? Is it a function of temperaure only?

[Donaldson Tan]

The internal energy of an ideal gas is a function of its temperature.

However, this is not true for real gases. This is reflected in the various equations of state.

The internal energy for the Van Der Waals' Equation of State:

[Hunt]

Thanks , Geodome