Lecture 3 PDF - Thermodynamics Lecture Notes

10/15/2024 Z = f (x , y)  Z  dZ    Z  dx     dy 2Z 2Z...

10/15/2024 Z = f (x , y)  Z  dZ    Z  dx     dy 2Z 2Z   x y  y x  yx xy ( ) ( ) ( ) =− cyclic rule First Law of Thermodynamics E = q - w E = q - PdV Compression in a single step. Compression in more steps. Expansion in a single step. Expansion in more steps. V 2 V2 W   P ex dv   P gas dv qv= E qP= H V1 V1   E  H  Cv   C p       T v  T p 1  10/15/2024 Special case of work Free Expansion Work done by free expansion = 0 In a free expansion, gas is allowed to expand into a vacuum. This happens quickly, so there is no heat transferred. No work is done because the gas does not displace anything. There is no change in internal energy, so the temperature is constant Measurement of (E/V)T (The Joule experiment) Bulb A was filled with air and the bulb B was evacuated. The two bulbs were connected by a stop cock. The surrounding water bath was stirred and its temperature was recorded. The stop cock was then opened A B and the temperature was again recorded. No temperature change was observed 2  10/15/2024 In this process no work is done by the gas, as it expands into vacuum w = 0. and no heat exchanged with the surrounding water Joule detected no temperature change i.e. dq = 0. the first law of thermodynamics along with the results of Joule's experiment gives dE = dq – dw = 0 E = f (T, V) E E E dE= ( )V dT +( )T dV 0= CV dT +( )T dV T V V dT = 0 E  E  dE= CV dT +( )T dV E dV  0   0 V 0= ( )T dV  V T V Thus the energy is independent of volume. This means that the energy of the gas is a function of temperature only. This behavior is Joule's law which may be expressed as E = f (T). However, this is valid only for ideal gases in which internal pressure or force of attraction between the molecules of the gas is zero. 3  10/15/2024 Relation between CP and CV H = E + PV dH = dE + PdV +VdP 1 E  f (T , V )  E   E  dE    dT   dV 2  T V  V T Substitute Therefore dH equals, from 2 to 1  E   E  dH    dT    dV  PdV VdP  T  V  V T  E   E  dH    dT    dV  PdV  VdP  T  V  V  T Divide the last equation by dT at constant pressure we get  H   E   E   V   V   P  =        + + P  +V    T P  T V  V T  T P  T P  T P  E   H   P    =Cv ,   =Cp ,   =0   T V  T  P  T  P  E   V   V  CP =CV +     + P   +0  V T  T P  T P  E    V   E    V  C p = C v +   + P  C p  C v    + P      V T   T  P    V T   T  P 4  10/15/2024 Since (dV/dT)P > 0, the heat capacity at constant pressure CP is always larger than the heat capacity at constant volume.(N.B liquid water behaves abnormally between 0 oC and 4oC; the rate of volume change with temperature (dV/dT)P is negative and therefore in this range CP < CV ) For an ideal gas, PV = RT for one mole  E   V     R/ P    0 Joule E xp eriment  T P  V T  E    V  C  C v    +P  R C p  C v  0 + P    p    V T   T  P P CP – CV = R Isothermal reversible expansion of an ideal gas Consider that n moles of an ideal gas are enclosed in a cylinder fitted with an ideal piston. Let V1 = volume of the gas at constant T. P is the external pressure acting over the gas which is approximately equal to the pressure of the gas. The work done by the gas for the small volume change dV would be dw = pdv V2 w   pdV For an ideal gas PV = nRT nRT V1 P V 5  10/15/2024 V2 nRT V2 dV w  dV  nRT  V1 V V1 V V2  nRT[ln V]V 1 V2 V2 w  nRT ln w  2.303 nRT log V1 V1 P1V1 = P2V2 For ideal gas P P1 w  nRT ln 1 w  2.303 nRT log P2 P2 Internal energy change (E) For iso-thermal, process, the initial and final temperatures are equal or dT=0. So the internal energy of the ideal gas remains constant during expansion because E depends on temperature only. Thus, E = E2 – E1 = 0 (Joule's law) Mathematical form of the first law is q = E + w V2 V2 w  q  2.303 nRT log w  q  nRT ln V1 V1 6  10/15/2024 Enthalpy change (H) H  H2 - H1  H  E  PV  (E2 PV 2 2) (E1 P 1V1)  (E2  E1)  (PV 2 2  P1 V1 )  E  (PV 2 2  PV 1 1) Since E = 0 and P1V1 = P2V2 for an ideal gas. H  0  0  0 Example Calculate the minimum work that must be expanded in the compression of 1 kg of ethylene from 10-1 m3 to 10-2 m3 at a constant temperature of 3000 K, assuming the gas to be ideal. Solution The work done in the reversible isothermal compression of an ideal gas is given by V2 w  nRT ln V1 1000 n  35.65moles 28.05 10 2 w  35.65 * 8.314 * 300 * ln 1  2.047x105 J  204.7 kJ 10 7  10/15/2024 Example Calculate the maximum work obtainable by the isothermal expansion ,of 2 moles of nitrogen, assumed ideal, initially at 25oC, from 10 liters to 20 liters. When a gas is expanded reversibly, the work performed is: V2 w  nRT ln V1 V  w  nRT ln  2  max  V1   20  w  2*2*(273  25) ln    826.23 Cal max  10  Actual Expansion (Irreversible Case) Energy change E For Ideal gas and T is constant, ΔE  E 2  E 1  0 Heat change q q = E + w The pressure is sudden change from P1 to P2 nRT nRT q = w = P2 V = P2(V2-V1)  P2 (  ) P2 P1 P2 w = q = nRT(1- ) P1 8  10/15/2024 Enthalpy change H H= E + PV H = E + (PV) = E + (RT) E = 0; RT is constant, (RT) = 0 H=0+0=0 Example One mole of an ideal gas at 300 K and 10 atm expand to 1 atm both reversibly and irreversibly under isothermal conditions. Calculate w. q, E and H for (a) Isothermal-reversible, (b) isothermal-irreversible For reversible isothermal E= H = 0 P1 w  q  nRT ln P2 10 w = q= 8.314*300ln =5744 J 1 For irreversible isothermal E= H = 0  P   1 w = q=nRT 1- 2  = 8.314*300* 1-  =2245J  P1   10  9  10/15/2024 Example (Homework) Two moles of an ideal gas are held by a piston under 10 atm pressure and at 273.15 K. The pressure is suddenly released to 0.4 atm and the gas allowed to expand isothermally. Calculate w, q, E and H for the process. Since the pressure is released suddenly the expansion must be irreversible. Given data: n = 2 moles; P1 = 10 atm; P2 = 0.4 atm; T1 = T2 = 273.15 K  P  w  nRT 1  2   P1   0.4   2 x 8.314 x 273.15 1   10  the ideal gas expands = 2 x 8.314 x 273.15 x 0.96 isothermally dT = 0 = 4360 J E = 0 ; H = 0 First law gives E = q – w = 0 q = w = 4360 J 10  10/15/2024 Adiabatic Expansion In adiabatic process there is no exchange of heat between the system and the surroundings i.e. q = 0. E = q – w q=0 E = - w -E = w The work is done by the gas on the expense of the internal energy of the gas. Since internal energy is a function of temperature, so decreasing of internal energy results in the lowering of temperature. The final temperature is less than the initial temperature in case of adiabatic expansion of a gas, whereas there is an increase of temperature for adiabatic compression. The thermodynamic quantities can be calculated as discussed below: Adiabatic Reversible Expansion for an ideal gas w can be calculated if E is known and vise versa. E is calculated from equation E Cv      T v 2 T2 1 dE  T C V dT 1 E = E2 – E1 = CV (T2 – T1) for one mole E = nCV (T2 – T1) = -w for n moles 11  10/15/2024 E = nCV (T2 – T1) = -w w= nCV (T1 – T2) If T2, T1 and Cv are known, E and w can be calculated Enthalpy change H  H     C P   T P H2 T2  dH  C P dT H1 T1 ΔH  C P (T 2  T1 ) ΔH  nCP (T2  T1 ) 12