Change in entropy adiabatic expansion

Question

I think that an adiabatic expansion of a gas should cause the entropy to increase. On the other hand we have for adiabatic processes that $dQ = 0$ and therefore $dS= 0$, which is why I thought that adiabatic processes are always isentropic. But somehow this adiabatic expansion of an ideal gas does not fit into this scheme, as it is quite obvious that this cannot be a reversible process. So how is this reconcilable with $dS=0$?

Answer 1

I guess you refer to the free expansion of a gas, which is an irreversible process. During free expansion, no work is done by the gas. The gas goes through states of no thermodynamic equilibrium before reaching its final state, which implies that one cannot define thermodynamic parameters as values of the gas as a whole. For example, the pressure changes locally from point to point, and the volume occupied by the gas (which is formed of particles) is not a well defined quantity. For that reason the standard equation $dS=dQ/T$ cannot be used because is not well defined. In such a case there is a change in entropy. For a calculation of this change you cah ckeck this link.

Answer 2

Irreversible free expansion of a gas in adiabatic condition is not isentropic. There is an increase of entropy. The equation of dS=dQ/T is not an accurate equation, the actual equation should be dS=dQ(reversible)/T. For dQ of irreversible the equation should be changed into the clausius inequality form which is dS> dQ (irreversible)/T. Under this clausius inequality equation, even dQ is zero in the adiabatic condition, and the right hand side equation is equal to zero, the left side dS is greater than zero which means that the dS entropy is increasing. You may want to ask why the equation should be changed to this clausius inequality, then, you should read on the derivation of this clausius inequality which is a bit of long story. To elucidate the clausius inequality briefly (and thus incompletely), we start from the concept of internal energy, U (I assume you know what is internal energy, if not the story will be longer). The equation of internal energy is U = Q (heat) + (-W, Work) (Work is flow in different direction, though not neccesarily, so negative is placed, not because of minus out the work). However, U is a state function (i assume you know what is state function), so whether the condition is reversible or irreversible, the U should be the same. But W and Q are not state function and they are condition depending. From the Carnot experiment, we know that the amount of reversible work, W(R), is greater > than the amount of irreversible work, W(IR) (I assume you had read about Carnot cycle and understand why reversible work> irreversible work). Thus, as U is a state function and reversible U(R) is equal to irreversible U(IR), U(R) = U(IR), to equate the condition in the aspect of Q and W; We must have reversible heat, Q(R) is also greater than irreversible heat Q(IR) so that when Q(R)+[-W(R)] = Q(IR)+[-W(IR)]. The original equation is dS = dQ(R)/T. As dQ(R)> dQ(IR), the equation for reversible heat and irreversible heat cannot be equal. dQ(R)/T > dQ(IR)/T; Thus, dS> dQ(IR)/T which is the clausius inequality.

Answer 3

Total entropy only increases when the system undergoes an irreversible process. that is when the process occurs spontaneously. we know intuitively that free expansion of gas is an irreversible reaction because, from our experience, we have never witnessed the gas flowing back into its starting point. therefore, the entropy of this system plus the surrounding would increase. if we wanted to find the increased value of the system's entropy, we should look for a reversible path. because the system's entropy is a state function, the change in system's entropy depends solely on the initial and final states of the reaction. if we can find the path that has identical initial and final points and calculated the heat that is involved in the reversible path, we can calculate the entropy of the system.

Answer 4

I think the question here is really how adiabatic is defined. I did not find a clear answer to that in the literature, but in my opinion, the only definition that makes sense is $\delta Q_\mathrm{rev}=0$, meaning no heat exchange with the surrounding. If this is the case, there can be still generation of heat due to entropy production (conversion of work into heat internally), so that $TdS=\delta Q_\mathrm{irr}$. This is why in the adiabatic expansion case, the entropy can increase.

Answer 5

A different way to look at what user65081 have already said in their answer:
energy conservation requires that $$dU = \delta Q +\delta W,$$ i.e., the energy change of the system is due to the heat transferred and the work done. Clausius/Gibbs definition of entropy is that in equilibrium process $$\delta Q=TdS,$$ which gives us $$dU=TdS -pdV$$ (where I assume an ideal gas for clarity, although it is not essential for the argument.)

However, if no thermal equilibrium is assumed, we only have Clausius inenquality, $$\delta Q \leq TdS,$$ so that the differential form of the first law is not really a strict equality: $$dU\leq TdS - pdV.$$

Related:
Entropy for irreversible adiabatic process

Answer 6

dQ – dW = dU

dQ = dU + dW

where dU = mcvdT and dW = pdV for closed system substitute dU and dW in the above equation, we get

dQ = mCvdT + p*dV

We know that dQ = T dS (from entropy), then above equation simplifies to

T dS = mCvdT + p*dV

dS= (mCv dT+pdV)/T On integrating the above equation between the limits 1 and 2 and using ideal gas equation, we get,

S2-S1=mCvln(T2/T1)+mR(ln V2/V1)..