dG Vdp (T = const.) Second Tds equation for simple, compressible substance ... Tds dh vdP=− dh vdP ds TT or =− (7-24, 26) 7-25 and 7-26 are valid for reversible and irreversible processes. C. It is valid when all This problem has been solved! First of all you want to know about what is entropy. It is the amount of heat energy contained in a substance and degree to which it is ordered is... Also known as macroscopic property. For a closed system, the general relation is δ Q ≤ T d S, as is illustrated by the Clausius Theorem ( http://en.wikipedia.org/wiki/Clausius_theorem ). The law of conservation of energy states that the total … Working out problems is a necessary and important aspect system's internal energy, while dq and dw, respec- tively, represent any heat absorbed and any work done by that system. E = internal energy (arising from molecular motion - primarily a function of temperature) + kinetic energy + potential energy + chemical energy. (B) For an irreversible process TdSdU+pdV. If the composition of system does not change, then dU=TdS-pdV . By differentiating this function and and combining with dU= TdS -PdV + ∑μdn for the open system You will get what you wish, namely: dG= -SdT +VdP +∑μdn. According to First Law of Thermodynamics we know that, dQ = dU + pdV From Second law of Thermodynamics, dQ = TdS The Increase of Entropy Principle Entropy change of a closed system during an irreversible process is greater that the ... Entropy 8 TdS = dU + PdV or, per unit mass Tds = du + Pdv This is called the first Gibbs equation. Transcribed Image Text: Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)? : D. It is equivalent to 1^{st} law, ... 9 Explanation: T d S=d U+p d V This equation holds good for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the both. The differential expression of Helmholtz free energy: dA = dU - TdS - SdT dU = - pdV + TdS dA = - pdV - SdT (2.4) The change of Helmholtz free energy in an isothermal reversible process is equal to the work. Since the path of a reversible process consists of a series of equilibrium states, the boundary work involved in a reversible process shall be given by (7.5). This is true only for a reversible (quasi-static) process. Find quality x (=0.86). δQ = dU + δW. This is because U, T, S, p, and V are all functions of ... inserting dU = TdS – pdV ⇒ ** dH = TdS + Vdp ** The natural variables for H are then S and p Question: Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)? T = temperature. Solution outline: Find final state. P = pressure. Please derive an expression of dU in terms of dP and %3D - dV. d U = T d S − p d V is not always valid. Furthermore, the reason it's not generally valid isn't because of the distinction between reversible and irreversible processes. There's an easy workaround for that*. • Differentiating the combined 1st and 2nd laws, dU = TdS - PdV by V at constant T we have Using Maxwell’s equation: this is valid for any system, we have not used any approximation. [mirror download link : https://goo.gl/o24NN ] Solving problems in school work is the exercise of mental faculties, and examination problems are usually picked from the problems in school work. The internal energy u = u (T,v) B. The equation dQ=dU+pdW holds good for any process undergone by a closed stationary system. Because it is the relation between properties which are independent to path. Download Solution PDF Share on Whatsapp India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank (B) TdS = dU+pdV This equation holds good for any process reversible or irreversible, undergone by a closed system. path independent). Entropy 2. The four Gibbsian relations for a unit mass are. The equation that relates partial derivatives of properties of p, v, T, and s of a compressible fluid are called Maxwell relations. Q6. Ans. ∆S total =∆S system + ∆S ... system TdS system –[dU system + PdV system] ≥0 This relation is in terms of the system variables alone. Q7. The internal energy of the air leaving is 90 kJ/kg greater than that of air entering. Examples here include disordered systems (like glasses, or indeed almost any real physical system at low T). This is valid for all cycles, reversible or irreversible. W < 0 when work done by the system, energy lost. KE + PE. the entropy of a closed system that is not in equilibrium increases with time reaching its maximum value when equilibrium Δq ts; 10.3 thermodynamical potentials of the system: internal emergy du = tds - pdv, in varriables s, p v t; potential entropy h = u + pv or dh = d (u + pv) = du + pdv + vdp = tds + vdp in variables s, p, v, t TdS= dU −(dWin)rev (11.5) The work term in a reversible process of a simple compressible system is the boundary work only. Thermodynamics FOURTH EDITION, , M. David Burghardt Hofstra University James A. Harbach U.S. Consider a closed system interacting dQ amount of energy with the surrouding. C. It is valid only for an ideal gas . Which of the following relations is valid only for a reversible processes undergone by a closed system? However when both equations (of 1st and 2nd law) merged to give the equation, all path dependent functions got eliminated. This is because U, T, S, p, and V are all functions of ... From dU = TdS – pdV ⇒ T dU pdV dS + = At constant T, dU=0 ⇒ T pdV dS T = For an ideal gas, pV = nRT ⇒ V nRdV dS c. first law for a closed system in out 21 ... Tds du Pdv= + b. Veja grátis o arquivo Termodinamica enviado para a disciplina de Termodinâmica Categoria: Aula - 20 - 42871428 (B) For an irreversible process TdSdU+pdV. Merchant Marine Academy =t. e.g. (C) It is valid only for an ideal gas. (a) SQ= dU+ SW (b) Tds = dU + pdV (c) Tds = dU + SW (d) 8Q = dU+ pdV This problem has been solved! dG Vdp (T = const.) (a) dU = PdV+TdS (b) VdP+TdS (c) VdP-SdT (d) -SdT-PdV B5. Which of the following equations is NOT true for a reversible process? If the system is closed but there are chemical reactions, they can vary. An open system of constant composition B. Hope this helps. Sharing is caring! According to First Law of Thermodynamics we know that, dQ = dU + pdV From Second law of Thermodynamics, dQ = TdS Find Study Resources . The boundary work of a closed system is δW rev = PdV (3) Substituting equations (1) and (3) into equation (2) gives dU = TdS- PdV TdS = dU + PdV or Tds = du +Pdv (4) where s = entropy per unit mass Equation (4) is known as the first relation of Tds, or Gibbs equation. The pressure in the pipe at section S1 (elevation: 10m) is 50kPa. V = volume. Tds = du + pdv is derived from first and second laws of Thermodynamics, assuming reversible pdv work. However when both equations (of 1st and 2nd l... See the answer Number 4 only please a) true b) false Answer: a You don't see a Q or a W in the Tds equations. Share Improve this answer $dU = TdS-pdV$ is not always valid. Furthermore, the reason it's not generally valid isn't because of the distinction between reversible and ir... Dr. M. Medraj MECH6661 lecture 5/14 Maxwell Equations: Example 2 Considering V = V(P,T): and differentiating by T at constant V we have: since ⇒ If we are to include other types of work, then dU = TdS − PdV + (δω −) rev (Eq. Yes, it is absolutely true that for reversible p V work the condition p e x t = p holds. ... TdS = dU + pdV TdS = dH – Vdp. Originally Answered: How Tds=dU + PdV is applicable to reversible as well as irreversible process when PdV is defined only for a closed reversible work ? In equation TdS = dU + PdV , temperature, entropy, internal energy, pressure and volume all are properties of system. Properties of system are point functions and independent of path. dQ = dU + dW is true for all process , reversible or irreversible , for any phase , solid, liqiud or gas , however under given condition that we ap... It follows that (see the derivation ofthe first law for a closed system undergo a cycle the differential is an exact different and is a property of the system (i.e. There's an easy workaround for that*. First Law of Thermodynamics (VW, S & B: 2.6) There exists for every system a property called energy. Considering the relationship Tds=dU+pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which of the statement is correct. 1. General equation of energy conservation from 1st law of thermodynamics is given by $Q = dE + $W But as soon as we talk about closed system, dE beco... In under-damped system, amplitude of vibration decreases slowly in exponential manner. So the equation only applies to differentially separated thermodynamic equilibrium states. (d) 49. TdS= dU + PdV This equation holds good for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path. Second, TD is often applied to systems that are only in thermodynamic equilibrium ‘for all practical purposes’ (FAPP). [GATE-2007] (a) δQ = dU + δW (b) TdS = dU + pdV (c) TdS = dU + δW (d) δQ = dU + pdV 56. Irreversibilities are factors which makes a process irreversible. Thus if these factors will not be present process will be reversible External irr... Piston+Cylinder ⇒ Closed system ⇒ dU =( Q + W )dt For the heat exchange : Qdt = TdS (macroscopic definition) For the work Wdt = - PdV (W = Fdx) ⇒dU = TdS - PdV This is the fundamental property relationship. (From the definition of entropy, the d Q in d S = d Q T is for reversible paths between the inital and final states.) But when I checked on the net, it says d U = T d S − p d V is always true for any type of process. S10. (b) Find the ratio … The pextdV term includes the external pressure on the system and the change in volume of the system and implies that the system pressure is only slightly different from the external pressure. Additionally, from the second law of thermodynamics, in terms of entropy, we know that the heat transferred is given by: δ Q = T d S δ Q = T d S . (a) δQ=dU+PdV (b) δQ=TdS (c) TdS=dU+PdV (d) None of these. But, if you want to move between two equilibrium states that are separated by a … dQ = dU+ PdV (2) where the bar indicates an imperfect differential (i.e., it is not the differential of some hypothetical function Q). For a system of variable composition, the internal energy depends on a) entropy b) volume c) moles d) all of the mentioned Answer: d Clarification: If some substance is added to the system, then energy of the system increases. General equation of energy conservation from 1st law of thermodynamics is given by $Q = dE + $W But as soon as we talk about closed system, dE becomes dU i.e $Q = dU + $W And when we talk about only PdV work in a closed system i.e for quasi static or reversible process , the equation further modified to $Q = dU + pdV (a) Q = dU + W (b) Q = dU + pdV (c) Tds = dU + W (d) Tds = dU + pdV Ans: (b) 41. entropy S TdS = dU + PdV U;V internal energy U dU = TdS PdV S;V enthalpy H H = U + PV dH = TdS + VdP S;P Helmholtz free energy F F = U TS dF = PdV SdT T;V the Gibbs free energy G G = H TS dG = VdP SdT T;P The \Natural Variables" de ne the boundary conditions for which the potential is most useful. Maxwell's thermodynamic relations are valid for a thermodynamic system in equilibrium. Substituting this in the above expression for d U d U, we get: This expression is valid for a closed system doing volume work 8. A. any system, any process. for reversible and irreversible process equation is same TdS=PdV+dU but for reversible process W=P.dV and dS=dQ/T mean dQ=T.dS but for irreversible... a) true b) false Answer: a Clarification: When a closed stationary system undergoes a process, this equation holds true. of the system has a unique ground state (as defined in quantum mechanics), and this is not always so. Remember that the mechanical work done is given by dW = PdV. The equation DU = Tds - PdV is applicable to infinitesimal changes occuring in A. The equation d U = T d S − P d V describes the mutual variations in dU, dS, and dV between two closely neighboring thermodynamic equilibrium states. For an irreversible process TdS \gt dU + pdV . Cooling water in the compressor jackets absorbs heat from the air at the rate of 58 kW. The equation Tds=du+Pdv describes the unique relationship between ds, du, and dv between these two thermodynamic equilibrium states for the material. From energy equation for a closed stationary system (a reversible process) δQW dU int rev int rev out−δ =, δ=QTdS int rev δWPdV int rev out, = Thus TdS dU PdV= + Or per unit mass Tds du Pdv= + Called Gibbs equation From the definition of enthalpy hu Pv= + →=+ +dh du Pdv vdP Tds dh Pdv vdP Pdv=− − +( ) Thus Tds dh vdP= − and See the answer Show transcribed image text Since all the variables of this equation are state properties, so does not matter be a reversible or not, In dq=Tds , heat is a path function, Andi... In the case of open or closed system, there are two ways 1. Specific volume happens to be between sat’d liquid and sat’d vapor value- hence must be wet. A closed system of constant composition C. An open system with changes in composition D. A closed system with changes in composition Answer: Option D Solution (By Examveda Team) The equation TdS=dU+pdV holds good for a) reversible process b) reversible process c) both of the mentioned d) none of the mentioned Answer: c Explanation: This equation holds good for any process undergone by a closed system since it is a relation among properties which are independent of the path. If we hold the volume constant, this leads to an expression for temperature as a partial derivative of the entropy with respect to the internal energy. (C) It is valid only for an ideal gas. This ... Tds = du + Pdv 2. This is a relation between properties and is always true (D) δQ = dU+ pdV This equation holds good for a closed system when only pdv work is present. Unformatted text preview: 20.110/5.60 Fall 2005 Lecture #8 page 1Fundamental Equations, Equilibrium, Free Energy, Maxwell Relations • Fundamental Equations relate functions of state to each other using 1st and 2nd Laws 1st law with expansion work: dU = đq - pextdV need to express đq in terms of state variables because đq is path dependent Use 2nd law: đqrev = TdS For a … (a) 8Q = dU +8W (c) TdS = dU + SW (b) TdS = dU + … 15. This is the First Law of Thermodynamics: that the internal energy of an isolated system is constant. Rather, it's because it only applies to closed systems. The differential statement of the first law, $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ is valid for all states---but not all processes---for t... If dU is the change in internal energy of the system and pdV is work done by the system against pressure p due to volume change dV. (A) It is applicable only for a reversible process. The law of conservation of energy states that the total … Note that the above differential expressions of H, F and G are valid for a reversible process in a closed system that is restricted to only P–V work. 5 WS2002 5 Chemical Potential • Up to now we have discussed changes in closed systems in homogeneous materials • Terms can be added to the definitions of U, H, F, and G to deal with changes in composition • From these equations, it can be shown that: • The chemical potential is a measure of the propensity of a constituent of a system to undergo change • At equilibrium, the … In such Closed system, constant S and p (9) An unconstrained system at constant entropy and pressure will spontaneously evolve into a state that minimizes enthalpy. The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, distinguishing three kinds of transfer of energy, as heat, as thermodynamic work, and as energy associated with matter transfer, and relating them to a function of a body's state, called internal energy.. by School by Literature Title by Subject Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)? QUESTION: 2. Find specific entropy of wet mixture as weighted average. This principle is known as the enthalpy minimum principle.It states that For an unconstrained, closed system at constant pressure and entropy, the equilibrium state will be that state which has the minimum enthalpy. A. A2A 1st law with expansion work: [math]dU = dq_r - p_edV [/math] Use 2nd law: [math] dq_r = TdS [/math] For a reversible process: [math]p_e = p , a... It is valid when all processes in the system are reversible and only PV work is done. Energy transferred to the system. (A) It is applicable only for a reversible process. Defines a useful property called "energy". 2. Solution: TdS= dU + PdV This equation holds good for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path. We shall generally consider closed systems but ones that … The equation dQ=dU+pdV holds good for a) open system b) closed system c) both of the mentioned d) none of the mentioned Answer: b (a) Q = dU + W (b) Q = dU + pdV (c) Tds = dU + W (d) Tds = dU + pdV Ans: (b) 41. Indeed, taking into account that A = 0 and p is constant, then TdS = dU + pdV = ... valid for a system within which a single chemical reaction occurs, represented by . From energy equation for a closed stationary system (a reversible process) δQW dU int rev int rev out−δ =, δ=QTdS int rev δWPdV int rev out, = Thus TdS dU PdV= + Or per unit mass Tds du Pdv= + Called Gibbs equation From the definition of enthalpy hu Pv= + →=+ +dh du Pdv vdP Tds dh Pdv vdP Pdv=− − +( ) Thus Tds dh vdP= − and Piston+Cylinder ⇒ Closed system ⇒ dU =( Q + W )dt For the heat exchange : Qdt = TdS (macroscopic definition) For the work Wdt = - PdV (W = Fdx) ⇒dU = TdS - PdV This is the fundamental property relationship. Evaluate ∆S for systems and surroundings. ∆U = q + W. For an isolated system, ∆U = 0. Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy? If dU is the change in internal energy of the system and pdV is work done by the system against pressure p due to volume change dV. Sol. At Section S2 (elevation: 12m) the pressure is 20kPa and velocity is 2ms-1.Density of water is 1000kgm-3 and acceleration due to gravity is … 6 lesson 18 VI. Upvote Reply Answer this doubt An electric heater uses natural convection to heat water w K 0.6 mK using a rod of 1 cm diameter and 0.65 m length at 110 volt. If we consider an isobaric process, dP = 0, then dH is the heat transferred to the system, or dH = C P dT. Remember that the mechanical work done is given by dW = PdV. An electric heater uses natural convection to heat water w K 0.6 mK using a rod of 1 cm diameter and 0.65 m length at 110 volt. Thermodynamics Miscellaneous. Q2. The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, distinguishing three kinds of transfer of energy, as heat, as thermodynamic work, and as energy associated with matter transfer, and relating them to a function of a body's state, called internal energy.. Tds = dh – vdP Derivation: d U = δ Q + δ W d U = δ Q + δ W. We know that the work done on a system, δ W δ W, is given by: δ W = − P d V δ W = − P d V . (b) TdS = dlJ + pdV (c) TdS = dU + ðw (d) ðQ = dlJ + pdV Telegram - MECHA UNACADEMY xoombook@hotmail.com TO GET MECHA MAX DISCOUNT (a) 56. du = (sign)Tds + (sign)Pdv du = +Tds - Pdv To write the Maxwell relations we need to concentrate on the direction of the arrows and the natural variables only. 3.105 in text However, since dU is an exact differential, its value is path … Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)? Another way of writing it is d S = δ Q / T + d S i r r where d S i r r is the entropy change due to irreversibilitiy in the closed system transformation. du = Tds – Pdv; dh = Tds + vdP; df = - Pdv – sdT; dg = -sdT + vdP We could expand dS from the equation above to explore its temperature and pressure dependence, but the dV term has been derived for the special case of reversible process, it only contain state properties of the system. If there is other work in addition to P – V work, TdS system –[dU B. If both the arrows pointing in the same direction, there is no need to change the sign, otherwise the equation should carry a negative sign. It is wrong to say 'total heat' or 'heat content' of a closed system, because heat or work is not a property of the system. Constraints that prevent the flow of energy, volume, or matter between the sub-systems are called "internal" constraints. It is valid when all processes in the system are reversible. and work on closed system of fixed mass). Ans (c) Sol. The entropy-change for any closed system which undergoes an irreversible adiabatic process (a) may be positive, negative or zero (b) may be positive or zero but not negative (c) is always positive (d) is always negative. H = U + PV ⇒ dH = dU + d(PV) = dU + PdV + VdP Substitution of the fundamental property relationship: S = entropy. Consider a closed system interacting dQ amount of energy with the surrouding. Valid for all processes, reversible or irreversible, open system or closed system (b) dQ dU dW Every process but closed system (c) dQ dU PdV Closed and reversible (Quasi-static) (d) dQ TdS reversible only (e) TdS dU PdV Valid for all process and … Internal Energy = total energy of all particles. Thermodynamic identity: dU = TdS - PdV. The Tds Relations for Open System The definition of enthalpy gives h = u + Pv 5 Dependence of Gibbs energy on temperature and pressure dG dH d(TS) dH TdS SdT ... G H TS dH dU pdV Vdp dG dU pdV Vdp TdS SdT Closed system performing volume work: dU S V dG S pdV pdV p S SdT dG Vdp SdT dG T (p = const.) The relation, TdS = dU + PdV, is valid for. Tds = du + pdv is derived from first and second laws of Thermodynamics, assuming reversible pdv work. process, the equation is always correct and valid for a closed (no mass transfer) system, even in the presence of an irreversible process. H = U + PV ⇒ dH = dU + d(PV) = dU + PdV + VdP Substitution of the fundamental property relationship: U = internal energy. Tds = du + pdv is derived from first and second laws of Thermodynamics, assuming reversible pdv work. However when both equations (of 1st and 2nd law) merged to give the equation, all path dependent functions got eliminated. You don't see a Q or a W in the Tds equations. So all the terms there in (like T, S, p, v) are point functions. process, the equation is always correct and valid for a closed (no mass transfer) system, even in the presence of an irreversible process. by differential: dH = dU + pdV +Vdp The natural variables of H are S and p represented as H(S,p) and dU =TdS − pdV ∴dH =TdS − pdV + pdV +Vdp ∴dH =TdS +Vdp The last equation is the fundamental equation for H and for a closed system in which only pV work, and since H is a state function: dS p dH T ∴ = S dp dH V Thermodynamics. In a small reversible change of the system, dU = TdS - PdV. We traditionally limit dw to the isobaric, isothermal case, writing dw = Tds = du + Pdv, where ds and dv, respectively, represent any infini- tesimal change to the system's entropy and volume, entire system or is a function of position which is continuous and does not vary rapidly over microscopic distances, except possibly for abrupt changes at boundaries between phases of the system; examples are temperature, pressure, volume, concentration, surface tension, and viscosity. i.e., TdS- pdV= CdT denotes the energy of thermal motion within the system. This is a concise statement of both the First and the Second laws in one equation. d U = T d S − p d V is not always valid. by School by Literature Title by Subject Furthermore, the reason it's not generally valid isn't because of the distinction between reversible and irreversible processes. The four elemental equations are in the first image down the page. 5 0 dt dH. Previous Question Next Question q > 0 when E transferred to the system, and vice versa. TdS = dU + … Final pressure is 100kPa. Review: Combine 1st and 2nd Laws The first law:dU đq đw For a reversiblechange in a closedsystem of constant composition, and in the absence of any additional non-PV work, rev đwdw PdV rev đqdq TdS Second Law for reversible heat flow Thus for a reversiblechange in a closed system, dU TdS PdV Eq. 4. This is because when we started developing the relations between entropy, internal energy, work , we started with reversible process or internally... We shall generally consider closed systems but ones that are composite of two or more simple systems. (a) Compute the rate of shaft work input to the air compressor in kW. 1; (Valid for all thermodynamic cycles, reversible or irreversible, including ref. This expression is valid for a closed system doing volume work 8. dU=δq+δw =δq +δwrev =TdS−PdV where the equality of the first and second lines derives from U being a state function, and the final line comes from the definition of reversible dS and reversible work. 5. For an irreversible process, TdS > dU+ pdV It is valid only for an ideal gas It is equivalent to 1st law, for a reversible process Correct Option: D The equation is valid for between reversible or inversible process of a closed system. For any two processes (reversible or irreversible) connecting two equilibrium states that are described by the state variables $U,T,S,V,p$ it is al... "d" denotes the total differential of the associated quantity. In a small reversible change of the system, dU = TdS - PdV. 5 Dependence of Gibbs energy on temperature and pressure dG dH d(TS) dH TdS SdT ... G H TS dH dU pdV Vdp dG dU pdV Vdp TdS SdT Closed system performing volume work: dU S V dG S pdV pdV p S SdT dG Vdp SdT dG T (p = const.) So all the terms there in (like T, S, p, v) are point functions. Ans. TdS = du + PdV This equation is applicable for any process reversible or irreversible, undergone by a closed system since it is a relation among properties that are independent of the path. (b) For an adiabatic process in a closed system, AS cannot be negative (c) For a process in an isolated system, AS cannot be negative (d) For an adiabatic process in a closed system, AS must be zero B4. ∆S for solids and liquids (incompressible substances) In equation TdS = dU + PdV , temperature, entropy, internal energy, pressure and volume all are properties of system. Properties of system are poin... F is useful for xed volume system in 6.1 Properties Relations for Homogeneous Phases From the 1st law and 2nd law of Thermodynamics, dWrev PdV (for reversible process) dU dQrev dWrev dQrev TdS dU TdS PdV Although this eqn. TdS=dU+PdV This equation consists of only point functions, so it is valid for all processes and also for perfect gas. Considering the relationship Tds=dU+pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which of the statement is correct. Find Study Resources . Which of the following relations is valid only for a reversible processes undergone by a closed system? In a closed system and in the absence of reaction, they remain invariant. cycles) ... For closed system (irreversible); For adiabatic closed system (irreversible); Entropy generation; 1) S gen. ≥ 0 (always) ... TdS = dU + PdV or Tds = du + Pdv First TdS or Gibbs equations Since h = u + Pv dh = du + Pdv + vdP Tds Yes, for an ideal gas U does not depend on V, TdS and the term −pdV can be combined into an independent term. When is the fundamental property relation du - Tds - Pdv valid for a closed system in material equilibrium? Find s1 and s2 from tables. A smooth pipe of diameter 200mm carries water. If we consider the function H = U + PV, then dH = TdS - PdV + PdV + VdP, or dH = TdS + VdP.