Thermodynamics Lecture Notes PDF

First law of thermodynamic E = q – P  V qv= E qp = H  E  C v    H   E    V  C    C p  C v    + P    T v...

First law of thermodynamic E = q – P  V qv= E qp = H  E  C v    H   E    V  C    C p  C v    + P    T v p   T p  V  T   T   Work done by free expansion = 0  E  CP – CV = R for ideal gad    0 Joule experiment  V T Isothermal reversible process V2 P1 E = 0 w  n R T ln w  n R T ln V1 P2 H  0  Work in reversible isothermal V2 P1 E= H = 0 q  w  n R T ln q  w  n R T ln V1 P2 Work in irreversible isothermal P2 w = q = nRT(1- ) E= H = 0 P1 in reversible adiabatic Decrease of internal energy results in the lowering of temperature in case of adiabatic q=0 expansion E = - w increase of temperature for adiabatic compression E = nCV (T2 – T1) = -w w= nCV (T1 – T2) H = nCp (T2 – T1) Relation between temperature and volume in reversible adiabatic process dE= dq - dw dE= dq – pdv  E  C v    dE = nCV dT   T v For adiabatic expansion dq = 0 then we get dE = – pdv nCv dT = -PdV for an ideal gas pv = nRT n RT C v dT dV nCv dT   dV   R T V V  T 2 V 2 C v dT dV     R T1 T V1 V Integration between suitable limits, we get C v T2 V1 ln  ln 2 R T1 V2 X Y 1/  T2  CV / R V1 X Y ln    ln  T1  V2 R/  T2   V1  C V /R  T2   V1        Taking antilogarithm, we have T1  V2   T1   V2   CP – CV = R CP -CV /CV CP /CV 1 T2  V1  T2  V1      T1  V2  T1  V2  1 T2  V1     T1  V2   = Cp / Cv > 1 R / Cv  1 T2 V2  T1V1 Relation between pressure and temperature in reversible adiabatic process R /C V T2  V1      V = RT / P T1  V2  R / CV R/C V R/C V T2  RT1 / P1  T2  T1   P2          T1  RT2 / P2  T1  T2   P1  R/CV R/CV  T2   T2   P2          T1   T1   P1  1 R / C V R / CV CV  R / CV R / CV  T2   P2   T2   P2            T1   P1   T1   P1  CP / CV R / CV R CV  T2   P2   T 2   P2  * CV C P           T1   P1   T1   P1   R / CP CP CV / CP T2  P2  T2  P2         T1  P1  T1  P1  1 1 T2  P2      T1  P1  Relation between pressure and volume in reversible adiabatic process R/CV R / CP T2  V1  T2  P2          T1  V2  T1  P1  R / CP R / CV  P2   V1        P1   V2  CP/CV  P2   V1        P1   V2      P2   V1       PV PV 1 1 2 2  P1   V2  Example A quantity of air at 25oC, is compressed adiabatically and reversibly from a volume of 10 liters to 1 liter Assuming ideal behaviour and taking CV (for air) = 5 cal. deg-1 mole-1 calculate the final temperature of air. R /C V T  V  2   1  T V  1  2  T 2  ( 1 0 ) 2 /5 298 1 T  749o K  476oC 2 omework xample quantity of air is allowed to expand adiabatically and reversibly from 200 m. to 20 atm assuming ideal behaviour calculate the final temperature if he initial temperature is 25oC and Cv = 5 cal deg-1 mole-1  1 Cal = 4.18 J , 1 J = 107 erg Example Calculate the work of compression in ergs when the pressure of 1 mole of an ideal gas is changed adiabatically and reversibly from 1 to 5 atm, initial temp 25oC (Cv = 5 cal mole-1 deg-1). w = n CV (T1 – T2) P2 R/C w  n C V T1 [1  ( ) P1 T2 P2 R/CP P2 R/CP ( ) T2  T1 ( ) 5 2/7 T1 P1 P1  5  298 [1  ( ) ] c 1 5 Cp = 5 + 2 = 7 cal deg-1 mole-1  5  298  4.18 107 [1  ( ) 2/7 1   3.64  1010 ergs  Adiabatic Irreversible Expansion In an irreversible expansion the pressure is constant and the work of expansion is given by w  P2dV= P2 (V2 V) 1 irr Energy change Since q = 0 for adiabatic changes, the first law equation gives E = -w= -P2 (V2 – V1) E = nCV (T2 – T1) Also E = nCV (T2 – T1) = - P2 (V2 – V1) = P2(V1 – V2)  E = nCV (T2 – T1) = P2(V1 – V2)  n R T1 nR T 2  nC V (T 2 - T 1 ) = P2     P1 P2  P2 C v (T2  T1 )  RT1  RT2 P1 C v  ( P2 / P1 ) R T 2  T1 Cp From the last equation we can calculate T2 and hence E Enthalpy change H= n CP (T2 – T1) Using the value of T2 from equation C v  (P2 / P1 )R T2  T1 Cp H can be evaluated. Homework Example One mole of an ideal gas at 300 K and 10 atm expand to 1 atm both reversibly and irreversibly under isothermal and adiabatic conditions. Calculate w. q, E and H for (a) Isothermal-reversible, (b) isothermal-irreversible (c) adiabatic-reversible and (d) adiabatic- irreversible expansion of the gas. (Given that CV = 3/2 R)  Comparison between isothermal and adiabatic process Plot P versus V for the isothermal and adiabatic reversible expansion for ideal gas from the same initial point to the final points The higher curve which has the lower negative slope represents the isothermal reversible expansion the lower curve which has the higher negative slope represents the adiabatic reversible expansion. To prove that, comparison between adiabatic an isothermal process. Calculation of the slope of each process as follows. lope for reversible isothermal expansion can be obtained from the general quation PV = RT. P = RT/ V dP RT RT 1 P - 2    dV V V V V For the adiabatic reversible expansion we use the equation, -γ PV = constant, P = constant V dP - γ -1 dP γ -γ -1 P   γ const V   γPV V   γPV -1   γ dV dV V  > 1,  The slope in case of adiabatic reversible expansion is lower than the isothermal expansion. The area under the P-V carve for the same volume expansion is greater in isothermal reversi expansion than adiabatic reversible expansion. Example One mole of benzene is converted reversibly into vapour at its boiling point 80.2oC by supplying heat. The vapour expands against the pressure of 1 atm. The heat of vaporization of benzene is 395 J/g. Calculate q,w,  and H for the process. The process is C6H6(liquid) C6H6 (vapour) Heat supplied for conversion of 1 mole of benzene into its vapour q = Heat of vaporization per gm. × mol. Wt. of benzene = 395 × 78 q = 30810 J mol-1. w= PV = P(Vv – VL) = PVv if Vv > VL then neglect VL = RT if vapour is ideal = 8.314 x 353.2; T = 273 + 80.2 = 353.2K = 2936.4 J mol-1. E= q – w First law = 30810.0 – 2936.4 = 27873.6 J H = qP = 30810 J mol-1.  Example Calculate the work done when 18.0 g of liquid water is vaporized at 100oC against a constant external pressure of 1.00 atm, given that: R= 0.0820 liter atm mole-1deg-1 = 1.99 cal mole-1deg-1 At 100oC density of H2O (l) = 1.00 g cm-3 and that of water vapor is 0.58824 g/liter.The molecular weight of water = 18.0 g mole-1. m D  V w  PdV  P(V  V ) m f i V  D VF = V (vapour) = 18/0.58824 = 30.5997 L Vi = V(L) = 18/1 = 18 cm3 = 0.018 L w  PdV  1*(30.6  0.018)  30.582L.atm  (30.582 / 0.082)*1.99  742.17Cal Example Calculate ΔE in joule for a system that performs 213 kJ of work on its surroundings and loses 79 kJ of heat E = q- w = -79-213 =-292 kJ the system absorbs heat from the surroundings, q is positive. f work is done by the system against the surroundings,(expansion) PdV is positive. Work done on the system (compressing a gas) is negative. Example Calculate the change in the internal energy of the system for a process i which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. E = q- w =140-85 =55 J