Maxwell relations
and
RidlResidual properties
1 B.Srinivas ChEngg GVP CE Maxwell relations and Residual properties
Why they have been defined and how do we use them in thermodynamics.
Towards the end of the class I am going to show you with three examples how residual ppproperty can be used to obtain ∆U, ∆H and ∆S for a gas obeying the
Van der Waals EOS.
2 B.Srinivas ChEngg GVP CE Till date we have seen non flow systems, flow systems and entropy calculations.
You would have noticed that everytime we were given P and T and asked to calculate ∆U, ∆H and ∆S.
You would start wondering why cant we measure ∆U, ∆H and ∆S
just like we measure P and T and solve the ppyppgroblem by skipping
these calculations of ∆U, ∆H and ∆S.
3 B.Srinivas ChEngg GVP CE The answer is simple: we just do not have measuring
instruments to measure ∆U, ∆H and ∆S as of today.
So we measure P and T with instruments and then thermodynamics gives us the relevant equations to estimate ∆U, ∆H and ∆S.
4 B.Srinivas ChEngg GVP CE Take the case of an ideal gas undergoing an isothermal reversible process.
We measure P and T at the initial and final states and then thermodynamics gives us the necessary equations to calculate ∆U, ∆HHand and ∆S.
UCTV
H CTP
TV22 SCV ln R ln TV11 constant Cp and Cv
TP22 SCp ln R ln TP11
5 B.Srinivas ChEngg GVP CE What about more complicated cases of when the gas is not
following the ideal gas behavior and say it is following the
van der Waals EOS ? Then how do the equations look like ?
How do we estimate ∆U, ∆H and ∆S.
6 B.Srinivas ChEngg GVP CE 7 B.Srinivas ChEngg GVP CE 8 B.Srinivas ChEngg GVP CE 9 B.Srinivas ChEngg GVP CE How entropy changes with P at const T, this ibliiis a problematic quantity since we cannot measure this quantity. But Maxwells reltilation rep laces this quantity with another relation which can be easily obtained .
10 B.Srinivas ChEngg GVP CE This integrand now involves all measurable properties !!
11 B.Srinivas ChEngg GVP CE 12 B.Srinivas ChEngg GVP CE 13 B.Srinivas ChEngg GVP CE 14 B.Srinivas ChEngg GVP CE Another application of Maxwells eqn in throttling process
15 B.Srinivas ChEngg GVP CE 16 B.Srinivas ChEngg GVP CE 17 B.Srinivas ChEngg GVP CE 18 B.Srinivas ChEngg GVP CE Cubic EOS like vdW, PR, SRK or any other EOS can predict the thermodynamic properties of liquids as well as gases if the molecules are non polar like hydrocarbons. But they predict very poorly for polar molecules like alcohols, acids etc.
19 B.Srinivas ChEngg GVP CE 20 B.Srinivas ChEngg GVP CE 21 B.Srinivas ChEngg GVP CE 22 B.Srinivas ChEngg GVP CE 23 B.Srinivas ChEngg GVP CE 24 B.Srinivas ChEngg GVP CE 25 B.Srinivas ChEngg GVP CE 26 B.Srinivas ChEngg GVP CE Since ∆H is a state function we calculate this in a two step process.
Approach 1:
27 B.Srinivas ChEngg GVP CE 28 B.Srinivas ChEngg GVP CE 29 B.Srinivas ChEngg GVP CE Approach 2:
30 B.Srinivas ChEngg GVP CE 31 B.Srinivas ChEngg GVP CE 32 B.Srinivas ChEngg GVP CE 33 B.Srinivas ChEngg GVP CE In order to get over this problem we solve the problem in a three step process.
Step 1 and 3 are isothermal Step 2 is iso bar ic btbut a t low pressure that the gas behaves like an ideal gas and we can use the idea l gas Cp values !!
34 B.Srinivas ChEngg GVP CE 35 B.Srinivas ChEngg GVP CE 36 B.Srinivas ChEngg GVP CE 37 B.Srinivas ChEngg GVP CE 38 B.Srinivas ChEngg GVP CE 39 B.Srinivas ChEngg GVP CE 40 B.Srinivas ChEngg GVP CE For a gas obeying vanderWaal EOS
41 B.Srinivas ChEngg GVP CE Two moles of a gas are confined in a piston-cylinder. Initially, the tempera ture is 300 K an d the pressure is 1 bar. The gas is compresse d isothermally to 5 bar. The gas has Cp=(7/2)R. Find Q, W, ΔU, ΔH and ΔS for the gas if: a) The gas is an ideal gas b) The gas follows the van der Waals equation of state, with a=5*106 bar cm6/mol2 and b=30 cm3/mol RT a P 2 VV VbV 22RT a W n PdV n dV PV VbV 2 Z VV11 RT
Vb2 aa a n RT ln HZRTR (1) VbV121 V V a U R V Vb SRZRR ln ln V 42 B.Srinivas ChEngg GVP CE 43 B.Srinivas ChEngg GVP CE 44 B.Srinivas ChEngg GVP CE 45 B.Srinivas ChEngg GVP CE 46 B.Srinivas ChEngg GVP CE 47 B.Srinivas ChEngg GVP CE 48 B.Srinivas ChEngg GVP CE 49 B.Srinivas ChEngg GVP CE 50 B.Srinivas ChEngg GVP CE 51 B.Srinivas ChEngg GVP CE 52 B.Srinivas ChEngg GVP CE Methane enters a turbine at T = 600 K and P = 10 bar and leaves at T = 400 K and P = 2 bar. How much work is produced for each mole of gas? Use the following data: van der Waals constants for methane are a = 0 . 2303 Pa- m6/mol2 and b = 4.306*10-5 m3/mol, and -3 -5 2 -8 3 -11 4 CP/R = 4.568 - 8.975*10 T + 3.631*10 T - 3.407*10 T + 1.091*10 T
Algorithm RT a P VbV 2 Calc V knowing P and T PV Z RT Calc Z
R a HRTZ (1) R V ClCalc H
53 B.Srinivas ChEngg GVP CE At the inlet P = 10 bar and T = 600 K
R V = 0.005, Z = 0.999 and H 1 = -48.926
At the outlet P = 2 bar and T = 400 K
R V = 0.017, Z = 0.998 and H 2 = -19.096 400 C HHRR RP dTH 21 600 R
= -19. 096 – 9289 – (-48. 926) = 9319
54 B.Srinivas ChEngg GVP CE 55 B.Srinivas ChEngg GVP CE 56 B.Srinivas ChEngg GVP CE 57 B.Srinivas ChEngg GVP CE 58 B.Srinivas ChEngg GVP CE 59 B.Srinivas ChEngg GVP CE For liquids which are poorly described a cubic EOS another approach is usually followed by defining isothermal compressibilty and volume expansivity.
For moderate changes in T and P these can be considered constant
and integrations can be carried out to determine changes in U, H and S.
60 B.Srinivas ChEngg GVP CE 61 B.Srinivas ChEngg GVP CE 62 B.Srinivas ChEngg GVP CE