Van Der Waal's Gas Equation for an Adiabatic Process and Its Carnot

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Carnot equa- of a Waal’s der tion Van of obeys efficiency substance working explic- the whose relation obtained this pressure and use constant We itly volume. at constant heat at specific and the relates which hne h tt ftesse scle sadiabatic as called is system the of the state and which the system process changes forbids the thermodynamic boundary between the then energy the surroundings heat If of flow surroundings. the of its sys- form the and between the tem boundary are the across work transfer and energy heat thermodynamics, In Introduction 1 substance. working Carnot the the of of independent efficiency is the that engine shown also We cess. b a e alsgseuto o naibtcpoesadits and process adiabatic an for equation gas Waal’s der Van r a e alscntn n safactor a is Γ and constant Waal’s der Van are eta nvriyo aind,Trvrr6005 aind,Ind Tamilnadu, 005, 610 Tiruvarur Tamilnadu, of University Central γ P fteielgsfra daai pro- adiabatic an for gas ideal the of eateto hsc,Sho fBscadApidSciences, Applied and Basic of School Physics, of Department stepressure, the is ia .Kmr rvn .Bb n .Ponmurugan* M. and Babu P. Aravind Kumar, S. Kiran P + V antegn efficiency engine Carnot n V 2 2 stevolume, the is a ( V − nb ) Γ = n a 1 e etecagdbtentesse n h sur- the and system and the roundings between exchanged heat net rntwr efre ntesse.Teengine The system. the on performed efficiency work net or h nta tt ihn hnei nenlenergy to internal returns in system change ics the no of process, with part (∆ state cyclic another initial a the the In during reservoir cycle. cooler the to a rcs.Alha nie aeueo h mech- the ( of heat use converting make of isother- engines as anism heat called All is process. process system mal fixed a a such at then takes transfer temperature energy the If process. oradasalraon fheat of amount smaller a and voir efre ytewrigsbtnei nengine, an cycle in of substance be- heat part working operate a some the to During by said reservoirs. performed is two engine these the exchanges tween Therefore, which temper- substance reser- constant working heat. cold at the a maintained and and be atures, reservoir will hot which a voir, of surroundings. comprises its and This system the between flow energy be a will in state. system place initial or taking its engine to processes the returned which of in series manner a the cyclic is of state It the in system. change resultant any involving out η uigec ftepoess hr a eaheat a be may there processes, the of each During U Q = net ) o rmtefis a fthermodynam- of law first the from So, 0). = etabsorbed Heat − e Work Net η W Q sdfie s[,2 3], 2, [1, as defined is net H W sasre rmtehte reser- hotter the from absorbed is ,where 0, = net = = Q Q net H Q Q nt ok( work to in ) Q = − H net Q Q L H = 1 = − Q Q Q ia. H − L L − stetotal the is Q Q srejected is W Q H L L ,with- ), . sthe is (1) 1.1 Carnot Cycle 100% efficiency. Since it is fact of experience that some heat is always rejected to cooler reservoir, the Carnot proposed a thermodynamic cycle called as efficiency of actual engine is always less than the ideal Carnot cycle which is a set of equilibrium reversible and the Carnot engine [1, 2, 3]. processes any thermodynamic system can perform Considering a working substance in a Carnot en- [1, 2]. Initially the system or working substance in gine as an ideal gas, with no intermolecular inter- Carnot cycle is imagined to be in thermal equilib- actions, the gas satisfies the ideal gas equation of rium with a reservoir at lower temperature T . Four L state P V = nRT . Where P is the pressure, V is processes are then performed in the following order: the volume, T is the temperature of the system, n 1. A reversible adiabatic process is performed in is the number of moles and R is the universal gas such a direction that the temperature rises to constant. The thermal efficiency of a Carnot engine that of hotter reservoir temperature TH . How- whose working substance is an ideal gas is given by ever, no energy in the form of heat flows in to or [2, 3] out of the system. T η = 1 − L . (2) 2. The working substance is maintained in contact TH with the reservoir at T and a reversible isother- H Thus, a Carnot engine absorbing Q amount of heat mal process is performed in such a direction to H from the reservoir at higher emperature T and re- the extent that Q is absorbed from the reser- H H jecting Q amount of heat to the reservoir at lower voir. In this case the system is kept at a fixed L temperature T has an efficiency that is independent reservoir temperature. L of the nature of the working substance. The above 3. A reversible adiabatic process is performed in result is obtained by using the ideal gas equation a direction opposite to that of the first pro- of state for the isothermal process P V = constant cess with no heat exchange. This is done until and the ideal gas equation for the adiabatic process γ the temperature reaches TL, temperature of the P V = constant. Here, γ is the ratio of the specific cooler reservoir. heat capacity at constant pressure to the specific heat capacity at constant volume. Further studies using 4. A reversible isothermal process opposite to the various gas equations [4, 5, 6, 7] also showed that the direction of the second process is performed until efficiency of a Carnot cycle is independent of work- the working substance reaches the initial state ing substance and depends only on temperatures of and QL is rejected by the working substance. In reservoirs. this case the system is kept at a fixed reservoir temperature TL. 1.2 Van der Waal’s gas Thus an engine in Carnot cycle operates between By considering the intermolecular interactions, Van two reservoirs in a particular simple way. All the der Waal proposed the equation of state for a real gas absorbed heat enters the system at a constant high which is given by[2, 3, 5] temperature, namely that of a hotter reservoir. Also, all the rejected heat leaves the system at a constant n2a P + (V − nb)= nRT, (3) low temperature, that of a cooler reservoir. Since V 2 all processes are reversible, the Carnot engine is a reversible engine. An engine which operates in this where a and b are the Van der Waal’s constants. cycle is called as Carnot engine whose efficiency is al- There has been several studies for Van der Waal’s ways greater than any engines operated by the cycle equation of state [5, 6]. However, there is no specific other than Carnot cycle. Any hypothetical engine and direct studies available in undergraduate text operated in this cycle is said to be ideal if it gives books and also in literature for the equation of Van

2 der Waal’s gas subjected to an adiabatic process. In For an adiabatic process dQ = T dS = 0, hence this paper we explicitly find that the Van der Waal’s ∂P equation for an adiabatic process as nCV dT = −T dV. (10) ∂T V 2 n a Γ P + (V − nb) = constant, (4) Using Eq.(3), one can obtain V 2 ∂P nR where Γ is a factor which relates the specific heat at = . (11) ∂T V V − nb constant pressure and at constant volume. We use the above relation and obtained the relation between Therefore Eq.(10) becomes the specific heat capacity at constant pressure and at nR constant volume. As similar to ideal gase approach nCV dT = −T dV (12) we also used the above relation directly and showed V − nb that the efficiency of a Carnot cycle is independent 1 R dV dT = − . of the working substance and depends only on the T CV V − nb temperature of the reservoirs. Integrating the above equation we get R 2 Van der Waal’s gas equation ℓn T = − ℓn V − nb + ℓn z, (13) CV for an Adiabatic process where ℓnz is an integrating constant. Rearranging The entropy of pure substance can be considered as the above equation one can get a function of any two variable T and V as R T V − nb CV = z. (14) S = S(T, V ) (5) Combining Eq.(3) and Eq.(14) one can obtain Van ∂S ∂S dS = dT + dV. der Waal’s gas equation for an adiabatic process as ∂T V ∂V T 2 n a Γ Multiply both sides by T we get, P + (V − nb) = K, (15) V 2 ∂S ∂S T dS = T dT + T dV. (6) where Γ = R + 1 and K = nRz = a constant. It ∂T V ∂V T CV should be noted that, for real gas Γ =6 γ = CP of Since dQ = nCV dT and T dS = dQ for a reversible CV isochoric process, from the above equation one can an ideal gas, where CP is the specific heat capacity obtain at constant pressure. Thus, it would be interesting to obtain the relation between the CP and CV of the ∂S Van der Waal’s gas as follows. T = nCV , (7) ∂T V where CV is the specific heat capacity at constant vol- 2.1 Relation between CP and CV for ume. According to Maxwell’s third thermodynamic Van der Waal’s gas relation The internal energy of a given system can be con- ∂S ∂P = . (8) sidered as a function of any two variable T and V ∂V T ∂T V as Therefore Eq.(6) becomes U = U(T, V ) (16) ∂P ∂U ∂U T dS = nCV dT + T dV. (9) dU = dT + dV. ∂T V ∂T V ∂V T 3 Using the above equation, the first law of thermody- From Eq.(3) on can obtain namics in an infinitesimal form dQ = dU + dW with ∂P nR dW = −P dV , can be written as = , (27) ∂T V V − nb ∂U ∂U dQ = dT + dV − P dV. (17) ∂T V ∂V T ∂V 3 −1 = nRV V − nb G , (28) ∂T P In our study we have used the sign convention that 2 the work done on the system is taken as positive and where G = V 3nRT − 2n2a V − nb . the work done by the system is taken as negative. Therefore Eq.(24) can be written as For constant volume dV = 0, the above equation R becomes, CP − CV = , (29) fv ∂U nCV = . (18) where ∂T V 2na 2 fv = 1 − 3 V − nb . (30) Since dQ/dT = nCV for constant volume and V RT dQ/dT = nC for constant pressure, Eq.(17) can be n o P We have obtained the relation between the CP and rewritten as CV of Van der Waals gas. Divide Eq.(29) throughout ∂U by CV we get dQ = nCV dT + dV − P dV (19) ∂V T CP R Γ − 1 =1+ =1+ . (31) ∂U CV fvCV fv nCP dT = nCV dT + dV − P dV (20) ∂V T Thus we have finally obtained the relation between γ ∂U dV n(CP − CV )= − P . (21) of ideal gas and Γ of Van der Waal’s gas as ∂V T dT n o Γ − 1 From the first thermodynamic potential dU = T dS + γ =1+ . (32) fv P dV and using Eq.(17) and Eq.(8) at constant temperature one can obtain There has been few studies to find out the efficiency of the Carnot cycle whose working substance ∂U ∂S is different from ideal gases [5, 6, 7]. In what follows, = T + P (22) ∂V T ∂V T we employ simple approach to find out the efficiency ∂P of the Carnot engine whose working substance obeys = T + P. (23) ∂T V the Van der Waal’s equation of state. Then Eq.(21) becomes 3 Carnot engine efficiency for ∂P dV n(CP − CV ) = T . (24) ∂T V dT Van der Waal’s gas If V is a function of T and P , then change in V for Efficiency of the Carnot engine, η, is defined as the a constant pressure is ratio of the net work done to the heat absorbed in the Carnot cycle. As discussed earlier, this reversible cy- ∂V ∂V dV = dT + dP. (25) cle consist of four processes such as i) adiabatic com- ∂T P ∂P T pression ii) isothermal expansion iii) adiabatic expansion and iv) isothermal compression. In order to find dV ∂V out η, we calculate the total work done during the = . (26) Carnot cycle as follows. dT ∂T P 4 3.1 Work done in an adiabatic com- Substitute Eq.(36) and Eq.(37) in the above equa- pression tion, we get Γ−1 Γ−1 In this process, the volume changes from V1 to V2 nRTL(V1 − nb) = nRTH (V2 − nb) ≡ K. (39) (V1 > V2), the pressure changes from P1 to P2 and Therefore, the temperature from TL to TH . There is no heat 1−Γ exchange between the system and the surroundings. K(V1 − nb) = nRTL (40) 1−Γ The work done during this process W1 is given by, K(V2 − nb) = nRTH (41) V2 and hence the work done in an adiabatic compression W1 = − P dV. (33) becomes V1 Z 2 2 nR(TL − TH ) n a n a The Van der Waal’s equation for an adiabatic process W1 = − + . (42) − 2 1 obtained in Eq.(15) as 1 Γ V V

2 n a Γ 3.2 Work done in an isothermal ex- P + (V − nb) = K. (34) V 2 pansion In this process, the system undergoes volume expan- 2 K n a sion V2 to V3 and the pressure change from P2 to P = Γ − 2 . (35) (V − nb) V P3, while the temperature remains constant at TH . During this process the system absorbs QH amount Substitute Eq.(35) in Eq.(33) and integrating, we get of heat energy from the hot reservoir, then the work

V2 V2 done, K(V − nb)1−Γ n2a W1 = − − , V 3 1 − Γ V W2 = − P dV. (43) V1 V1 V2 Z 1−Γ 1−Γ 2 2 For a reservoir temperature TH , Eq.(3) can be written K(V1 − nb) − K(V2 − nb) n a n a W1 = − + . as, 1 − Γ V2 V1 2 nRTH n a The Van der Waal’s equation of state for the system P = − (44) V − nb V 2 in the initial state (P1, V1) at TL is Substitute the above equation in Eq.(43) and inte- 2 n a grating, we get P1 + 2 (V1 − nb)= nRTL (36) V 2 2 1 V2 − nb n a n a W2 = nRTH ℓn − + . (45) V3 − nb V3 V2 and for the system in the final state (P2, V2) at TH as From the Van der Waal’s equation of state (Eq.3), we n2a can relate the system in the initial state (P2, V2) and P2 + (V2 − nb)= nRT . (37) V 2 H final state (P3, V3) at a fixed temperature TH as 2 n2a n2a From the Van der Waal’s equation (Eq.34) for the P2 + (V2 − nb)= P3 + (V3 − nb). (46) V 2 V 2 adiabatic process, we can relate the system in the 2 3 initial state (P1, V1,TL) and final state (P2, V2,TH ) Substitute Eq.(46) in Eq.(38), then as n2a 1 2 −Γ 2 2 P + V (V3 − nb)(V1 − nb) Γ Γ 1 n a n a 2 = . (47) P1 + (V1 − nb) = P2 + (V2 − nb) . (38) n a 1−Γ 2 2 3 2 (V2 − nb) V V P + V 1 2 3

5 3.3 Work done in an adiabatic expan- 3.4 Work done in an isothermal com- sion pression

As similar to adiabatic compression, the heat ex- In this process, the pressure changes from P4 to P1, change in an adiabatic expansion is zero. As the the volume changes from V4 to V1 and QL amount of system expands from V3 to V4, the pressure changes heat energy rejected to a cold reservoir at constant from P3 to P4, and the temperature changes from TH temperature TL. Work done during this process W4 to TL, then the work done W3 during this process is is given by

given by, V 1 V4 W3 = − P dV. (48) W4 = − P dV. (56) V4 ZV3 Z Substitute Eq.(35) for an adiabatic process in the For a reservoir temperature TL, Eq.(3) can be written above equation, we get as, 2 1−Γ 1−Γ 2 2 nRTL n a K(V3 − nb) − K(V4 − nb) n a n a P = − 2 (57) W3 = − + . V − nb V 1 − Γ V4 V3 Substitute the above in equation in Eq.(56) and in- The Van der Waal’s equation of state for the system tegrating, we get in the initial state (P3, V3) at TH is 2 2 2 V4 − nb n a n a n a W4 = nRTLℓn − + . (58) P3 + (V3 − nb)= nRT (49) V1 − nb V1 V4 V 2 H 3 From the Van der Waal’s equation of state (Eq.3), we and for the system in the final state (P4, V4) at T as L can relate the system in the initial state (P4, V4) and 2 n a final state (P1, V1) at a fixed temperature TL as P4 + 2 (V4 − nb)= nRTL. (50) V 2 2 4 n a n a P4 + 2 (V4 − nb)= P1 + 2 (V1 − nb). (59) From the Van der Waal’s equation (Eq.34) for the V4 V1 adiabatic process, we can relate the system in the Substitute Eq.(59) in Eq.(51), then initial state (P3, V3,TH ) and final state (P4, V4,TL) 2 as n a − P3 + 2 Γ 2 2 V3 (V1 − nb)(V3 − nb) n a Γ n a Γ n2a = 1−Γ . (60) P3 + 2 (V3 − nb) = P4 + 2 (V4 − nb) . (51) P1 + 2 (V4 − nb) V V V1 3 4 Substitute Eq.(49) and Eq.(50) in the above equa- 3.5 Efficiency of the engine tion, we get Γ−1 Γ−1 The net work done Wnet for one complete cycle is 3 4 nRTH (V − nb) = nRTL(V − nb) ≡ K. (52) Wnet = W1 + W2 + W3 + W4. Therefore, the total work done for the Carnot cycle is given by Therefore,

1−Γ Wnet = wH + wL, (61) K(V3 − nb) = nRTH (53) 1−Γ K(V4 − nb) = nRTL (54) where and hence the work done in an adiabatic expansion V2 − nb wH = nRTH ℓn (62) becomes V3 − nb 2 2 nR(TH − TL) n a n a V4 − nb W3 = − + . (55) wL = nRTLℓn . (63) 1 − Γ V4 V3 V1 − nb

6 According to the first law of thermodynamics The above equation can be rewritten as Qnet − Wnet = ∆U, where Qnet = QH − QL is the V4−nb net heat energy exchange between the system and the TL ℓn V1−nb η = 1 − . (67) reservoir and ∆U is the change in the internal energy. V3−nb TH ℓn Here we have used the sign convention that the heat V2−nb energy flow in to the system is taken as positive and In order to simply the above equation, substitute the heat energy flow out of the system is taken as Eq.(60) in Eq.(47) and rearranging, we get negative. For a cyclic process, the change in internal 1−Γ 1−Γ 1−Γ 1−Γ energy ∆U is zero. Therefore, the net heat energy (V1 −nb) (V3 −nb) = (V2 −nb) (V4 −nb) exchange between the system and the reservoir in a 1−Γ 1−Γ given cycle is completely converted in to net work V3 − nb V4 − nb = . done which is given by V2 − nb! V1 − nb! So, Wnet = QH − QL. (64) V3 − nb V4 − nb = (68) 2 1 In the Carnot cycle, all the absorbed heat QH enters V − nb! V − nb! the system at a constant high temperature TH and all Therefore, Eq.(67) reduces to the rejected heat QL leaves the system at a constant low temperature TL. Hence, comparing Eq.(61) and TL 1 η =1 − . (69) Eq.(64), one can identify TH Thus, we used Van der Waal’s gas as a working substance and obtained the efficiency of the Carnot en- V2 − nb gine which is independent of the working substance. QH = wH = nRTH ℓn (65) V3 − nb

V4 − nb 4 Conclusion −QL = wL = nRTLℓn . (66) V1 − nb In summary, we have obtained the Van der

The efficiency of the engine obtained from Eq.(1) as Waal’s equation for the adiabatic process as n2a Γ P + V 2 (V − nb) = constant. Our result explic- − T ℓn V4 nb itly shows that Γ of the Van der Waal’s gas is differ- Q L V1−nb η =1 − L =1+ . ent from γ of ideal gas for adiabatic process. This − QH V2 nb TH ℓn V3−nb equation has been used directly in Carnot cycle for the adiabatic process and shown that the efficiency 1It should be noted that Q can also be obtained from other of the Carnot engine is independent of the working approaches. Using Eq.(23) and Eq.(27) for an isothermal pro- substance. For this calculation we have used the sim- cess, Eq.(17) becomes ple approach as similar to the ideal gas usually found ∂P nRT in the undergraduate text books. With this we find dQ = T dV = dV. ∂T V V − nb out the efficiency of the Carnot engine whose working substance obeys Van der Waal’s equation of state Integrating the above equation from the initial volume Vi to 2 the final volume V , one can obtain the amount of heat trans- n a f P + V 2 (V − nb) = nRT . Using the above two ferred between the system and the sorrounding as equations, we have also shown that the alternative Vf Van der Waal’s equation for an adiabatic process is − − Q = nRTℓn V nb . Γ 1 T (V − nb) = constant. We can also write the Vi 2 Γ n a 1−Γ above relation as P + V 2 T = constant. 7 *Corresponding author: [email protected]

References

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