Deriving the Isentropic Relations for Gasses 

I'm not sure I fully understand your question. For an irreversible process, dS > dQ/T. Is there a specific thermodynamic case or system configuration you are curious about?

@@ArlingtonMatrix If dq < 0 then ds = dq/T + entropy production could be zero for non reversible processes

​@@klingeron5929 If we look at the second law of thermodynamics in difference form, we would have S_2 - S_1 = Delta(Q) / T. We can compute Delta(Q) from the first law of thermodynamics. S_2 - S_1 is basically your entropy production from state 1 to state 2. Entropy is only generated because heat transfer occurs generating entropy. Once you have heat transfer in or out of the system, the process is irreversible because you cannot restore it to its original state without somehow stealing heat back from somewhere (or supplying heat, depending on the direction of the heat transfer). For a refrigeration cycle as an example, you can have an apparent "negative" entropy because of the direction of heat transfer, where the total entropy of the system + the surroundings increases. Regardless, such a system would not be isentropic. For such cases, you need to modify the isentropic relations to consider the entropy change. To include entropy relations, you have to introduce that calculation for dQ so that S_2 - S_1 is not zero. In the video here, I've made the assumption S_2 = S_1 for an isentropic system, so you would have to modify the equations to yield slightly different relations that depend on entropy. For example, (P_2 v_2^(gamma))/(P_1 v_1^(gamma)) = exp((S_2 - S_1)/C_v), which yields a ratio of pressure and volume for an entropy-generating process (which is also valid for an isentropic process, because the ratio is 1 when S_2-S_1 = 0). In this description, C_v is the specific heat at constant volume, and gamma is the ratio of specific heats.

That is an interesting relationship, but not quite what I was thinking about. Your difference form is missing the entropy production, delta sigma, for irreversible processes. For irreversible non adiabatic processes, S2 - S1 could very well equal zero and as such be isentropic. Of course this would imply negative delta Q since entropy production is always greater or equal to zero. The final relations like p*v^n still hold, which can be shown with the the Gibbs equations, but they only hold for the final state and not inbetween. So you wouldnt be able to compute things like the work integral. The fact that non adiabatic irreversible processes can also be isentropic is largely ignored, which is quite frustrating.

@@klingeron5929 Work is computed as the integral PdV or VdP, not from entropy. Entropy is a state variable, so you can compute S2 and S1 regardless of what happens in between (we define it for a reversible path that satisfies the equilibrium states). Work can be computed from the first law of thermodynamics: dE = dQ - dW (or dE = dQ + dW depending on which sign conventions are being used). If we instead define dS = dQ/T + S_gen (where S_gen takes care of the inequality from dS >= dQ/T, this definition could be used for non-equilibrium systems), we might be tempted to re-arrange the equations to get dS - S_gen = dQ/T. If the process is "isentropic", then dS = 0, but this would then produce -S_gen = dQ/T, which would indicate a decrease in entropy; this can't occur unless we have work or mass transfer to the surroundings. **However**: that isn't the equation you use to compute the pressure-volume relationship. S_gen simply states that useful work has been lost to the surroundings. Computing pressure-volume relationships and the like requires you to consider the state relationships (i.e. state 2 and state 1, or S_2 - S_1 locally, we might have S_2 - S_1 less than zero, but we understand that the system + the surroundings must have an increase in entropy). If useful work is lost, then it would be reflected in the first law of thermodynamics by dE = dQ - dW. Unfortunately my research for the past few years hasn't intersected with this subject matter so I'm not able to provide any better insight without sifting through my old notes. At the moment I can't really devote time to such endeavors, but if I should ever receive funding for more videos (or if I am asked to make educational or training content as part of my job) I'll definitely dedicate a video to this particular question.

Ah, I think I've written these incorrectly. The correct way would be to compute between 2 states as State_2 - State_1, but I've written State_1 - State_2. The derivation is essentially the same and you should get the same results (albeit with the subscripts changed). Here's an article with the correct equations: www.grc.nasa.gov/www/k-12/airplane/compexp.html

The process is not isobaric (otherwise ln(P2/P1) = 0). I suppose another derivation here would consider dU = -P dV and dH = V dP where dU = Cv dT and dH = Cp dT such that the ratio, Gamma, equals Cp/Cv = -V dP / P dV which tells us that P * V^{Gamma} = Constant; following from this, we confirm the properties in the video are also correct and the process followed here is not an isobaric one.

@@ArlingtonMatrix But just to say that dH = cp*dT you must have already assumed that cp = cv + R since dH = dU + nRT = ncvdT + nRT. And just that dH = cp*dT indicates it's constant pressure.

@@MrYahya0101 For an ideal gas, internal energy and enthalpy are both functions of temperature. The way the specific heat quantities are defined is such that Cv = dU/dT and Cp = dH/dT. "Constant Pressure" and "Constant Volume" refers to the process by which these quantities are measured, the use of Cp for enthalpy doesn't indicate constant pressure, it just means that dH = (dH/dT)*dT; if you wanted to MEASURE these specific heats, you would do so at constant-volume/pressure as indicated by the subscript. So using the definition of enthalpy: dH = dU + d(PV) where dH = Cp dT and dU = Cv dT such that Cp dT = Cv dT + d(PV) which, based on ideal gas law, is the same as writing Cp dT = Cv dT + R dT, or Cp = Cv + R.

@@ArlingtonMatrix Ohh I see. Thank you so much! Just to make sure I understand, even if I see dQ = n*cv*dT, doesn't mean the heat transfer is happening at constant volume, just that the specific heat constant is measured at constant volume?

The equations given here apply to only reversible processes. An irreversible process is one which generates entropy, and the generation of entropy produces additional changes in these properties (for example: Cp*ln(T1/T2) is no longer equal to R*ln(P1/P2) due to entropy). If you consider an alternative derivation using dS = dQ/T, you see a similar result whereby you have to specify dS = 0 to recover the isentropic gas relations.

@@ArlingtonMatrix But there seems to be no reasons cp*ln(T1/T2) shouldn't equal R*ln(P1/P2) in adiabatic irreversible process. You only specify that dQ = 0 in your derivation, so it should apply for all adiabatic processes right? Also, I wanted to ask about the enthalpy equation you wrote: dH = dU + d(PV). Some seem to define it as dH = dU + PdV, which is correct? If defined as dH = dU + d(PV), then doesn't dH always equal dQ (since dU = dQ - d(PV))?

@@MrYahya0101 In an irreversible process we generate losses (or perform irreversible work on the environment); If we perform work, then we have to add an additional term to account for that work and thus change the energy balance of the system. The derivation here assumes irreversible, and as such does not show terms relating to irreversible processes (for example, friction or viscous dissipation from the boundary). The definition of enthalpy as dH = dU + PdV is correct for an isobaric process (where V dP = 0); U measures internal energy of a system, whereas H measures the total energy (including energy related to volume); you can define enthalpy as H = U + PV, such that dH = dU + d(PV) = dU + V dP + P dV

@@ArlingtonMatrix Thanks for clarifying the reversibility part. if enthalpy is as you say dH = dU + VdP + PdV, then wouldn't it always equal dQ? Since dU = dQ - d(PV). I see people writing dU = dQ - P*dV but then why would you add (V*dP + P*dV) to the enthalpy equation? It's like saying the system is doing P*dV work, and now we will add P*dV to eliminate it which makes sense but why add V*dP then? Who is doing the V*dP? If the system is doing V*dP then it should be included in dU = dQ - P*dV - V*dP, right?

​@@MrYahya0101 In the dU equation, the expression PdV represents "boundary work" (work performed by pressure on a changing volume boundary) - this is included in the dU equation as the "work" term and accounts for changes in the system when the boundary is allowed to move; the expression VdP, however, relates to process flow work caused by a changing pressure on volume V. Internal energy is a sum of heat transfer and thermodynamic work done by the system on its surroundings, but at constant volume, there's no work done on the surroundings - this may be thought of as WORK = FORCE * DISPLACEMENT: if there's no displacement, there's no work. It is important here to understand the difference between Internal Energy and Enthalpy: Internal Energy is a property of the system relating to the energy contained *within* the system (minus work performed by the system, PdV), whereas enthalpy is the sum of Internal Energy and the *work required* to achieve its state. The "true" definition of enthalpy (insofar as to define a value) is H = U + PV, whereas internal energy is U = Q - W, where W is an actual 'work' quantity. Generally, however, we're not interested in the raw value of enthalpy (which is kindof-sortof meaningless), but rather the change in enthalpy which is of interest in thermodynamic systems. Enthalpy is a difficult concept to grasp and even more difficult to explain, however it simplifies many thermodynamic computations so the convenience of enthalpy is worth the headache of understanding it. If you're still having difficulties, I might recommend asking at chemistry.stackexchange.com as they are better equipped to explain these concepts (the youtube comments section leaves much to be desired when it comes to readability)