Lecture 9 2025 PDF

Gas Processes Most thermodynamic processes, the heat is transferred (from or to) system. During processes, the pressure, volume, and temperature change. The study of thermodynamic process calculating the work done, the change in internal energy, and the amount of heat exchanged. Its syst...

Gas Processes Most thermodynamic processes, the heat is transferred (from or to) system. During processes, the pressure, volume, and temperature change. The study of thermodynamic process calculating the work done, the change in internal energy, and the amount of heat exchanged. Its system is Gas. Some processes in thermodynamics 1- The isochoric process or the isometric process This process is characterized by constant system volume. This means (work = zero). Because the change in volume = zero. So isochoric processes does not produce any work. Then the first law of the thermodynamic will be in the form (Q=U). Because the amount of heat absorbed by the thermodynamic system does not result in any kind of work, this means that the amount of heat absorbed by the thermodynamic system in isochoric processes tend to increase the internal energy of system only. When the internal energy of the system is equal to the amount of heat absorbed, the work is equal to zero. (Q=U) This process is represented by the diagram as pressure- volume is a straight line parallel to the pressure axis. P1V1 P2V2   V1  V2 T1 T2 P1 P2   T1 T2 W   PdV  dV  0 W  0  Q  dU  PdV  dV  0  Q  dU  SV dT  SV T2  T1   A container contains air, under pressure of 20 K Pa and temperature 30°C. Calculate the amount of heat required to increase the pressure of gas to 40 K Pa knowing that its specific heat under a constant volume is 0.727 J/kg. K.  Q  SV T2  T1  P1 P2 P2  T1   T2  T1 T2 P1 4  10  30  273 4  T2   606K 2  10 4  Q  0.727  606  303  220.28 J / kg  2- The isobaric process It process produce work through the amount of heat absorbed by system. Part of its absorbed heat is converted to work, the other part of its transformed to increase internal energy of system. Then the first law of thermodynamic will be the form (Q=U+W). This process is represented by the diagram pressure- volume is a straight line parallel to the volume axis.  P1V1 P2V2    P1  P2 T1 T2 V1 V2   T1 T2 V2 W   PdV V1 W  P V2  V1   Q  dH  VdP  Q  dH  S P dT  S P T2  T1  The final temperature in this process T  T f  V f  i   Vi  A gas kept in container at pressure 1.5 bars and volume 4 m3. What is the work done by its gas in the following cases? a- It is compressed at constant pressure to quarter its initial volume. b- It is expanded at constant pressure to twice its initial volume. a- It is compressed at constant pressure to quarter its initial volume.  W  P V f  Vi  W  1.5  105  1  4 W  450000J The work is -ve sign (compression process); this work is on the gas b- It is expanded at constant pressure to twice its initial volume.  W  P V f  Vi  W  1.5  105  8  4 W  600000J The work is +ve sign (expansion process), this work is by the gas.  3- The isothermal process It process is characterized by constant temperature of system. (T=0). When constant temperature of system then constant its internal energy of system. (U=0) It process is considered one of the best thermodynamic processes because its process the amount of heat absorbed converted into work by 100%. Then the first law of thermodynamic for its process will be the form (Q=W).  P1V1 P2V2    T1  T2 T1 T2  PV  const.  P1V1  P2V2 Vf Pi Wi f  n  R  T  ln  n  R  T  ln Vi Pf  PiVi  n  R  T Vf Pi  Q  Wi f  PiVi  ln  PiVi  ln Vi Pf The final pressure in this process Q  U  W  Vi  U  const.  U  0 Pf  Pi   V   W  Q  f  0.1kg from Hydrogen expanded at 270C to dual its initial volume. Find the work done Vf Wi f  nRT ln Vi m Vf Wi f  RT ln M Vi 100 2V Wi f   8.31 300 ln 2 V 100 Wi f   8.31 300 ln 2 2 Wi f  864000 J  4- Adiabatic process (repressed) (Suddenly) It process take place when the amount of heat of system is constant. Then the amount of heat remain constant, the temperature of system must change, and leads to change in the internal energy of system. Therefore it process is constant the amount of heat or process that change the internal energy of system. The work in its process depend on the change in the internal energy of the system. It require Isolated system to complete isolation from the surrounding environment. Wi f  1   1 PiVi  Pf V f    Vi  PiVi  Pf V f  Pf  Pi     V   f   A- The process of adiabatic expansion In this procedure, the internal energy of system decreases, and then decrease the final temperature of its system, this explains to us why hot water kept in an insulated thermos cools after a long period of time B- Adiabatic compression process In this procedure, the internal energy of system increases, and then increase the final temperature of its system so that the amount of heat for the system remains constant. In adiabatic expansion (Tf  Ti ) but in adiabatic compression (Tf  Ti )   PV  const.  Vi   Pf  Pi      PiVi  Pf V f V   f  Wi f  1   1 PiVi  Pf V f  1  1 1  Vi   Pi   T f  Ti   , T f  Ti   V  P   f   f  An ideal gas with a temperature of 75 C and a pressure of 50 bars suddenly expands to 8 times its original volume. Calculate final temperature, final pressure, work done. (= 1.4) Ti  75  273  348K , Pi  50 105 N / m 2 , VF  8Vi  1 1.4 1  Vi   Vi   T f  Ti    348   1510 K V   f   8Vi    Vi   Pi.Vi  Pf.V f  Pf  Pi     V   f  1.4 V   Pf  50 105  i   2.7 105 N / m 2  2.7bar  8Vi  W 1   1   PiVi  Pf V f , Pf V f  RT f RT f V f   4.6 10 3 m 3 Pf Vf V f  8Vi Vi   5.8 10  4 m 3 8 W  4145J Equation of state of isothermal processes  Equation of state of adiabatic processes   PiVi  Pf V f PiVi  Pf V f Slope of the isothermal curve  Slope of the adiabatic curve P  P slope  slope  V V We note that the adiabatic curve is steeper than the isothermal curve Because its slope is multiplied by the coefficient  this coefficient is always greater than one. =SP /SV  Isothermal process 1-Equation of state (PV=const.  Pi Vi = Pf Vf). 2- The work done W if = nRT ln (Vf / Vi W if = nRT ln (Pi / Pf W if = Pi Vi  ln (Vf / Vi 3 – The amount of heat absorbed (W = Q Q= W if ). 4 –The change in internal energy (dU = 0). 5 – The work done depend on the amount of heat absorbed 2liters of N2 at 300C and 20 atm is compressed isothermally to quarter its original volume. Calculate: 1-The work is done on the gas during this process. 2- Final pressure. 3-The quantity of heat is given up by the gas in this process. (Take n=1mole, R=8.31J/mole. 0C). 1  Vf   Vi  W  n  R  T  ln    1 8.31 303  ln  4   Vi   Vi      1 W  1 8.31 303  ln    3490.5 J 4  Pi  Vi  Pf  V f Pi  Vi 20 10 5  4  Vi  Pf    80atm Vf Vi  Q  W  3490.5 J Adiabatic process (suddenly) (change in internal energy) (Isolated process) (At constant heat ) 1-An adiabatic expansion The internal energy will be decreased. Then the final temperature will be decreased. 2 - An adiabatic compression The internal energy will be increased. Then the final temperature will be increased. 1-Equation of state    P  V  conat.  Pi  Vi  Pf  V f 2- The work done Wi f  1   P  V  Pf  V f   1 i i   Pi  Vi  m  R  Ti , Pf  V f  m  R  T f Wi f  m R  T T   1 i f   An ideal gas at 75 0C and 50 bar is expanded with change in its internal energy to 8 times its initial volume. Calculate the final temperature, the final pressure and the work done. Write comment on these results. Ti  75  273  348K , Pi  50 105 N / m 2 , VF  8Vi  1 1.4 1  Vi  V   T f  Ti   T f  348 i   1510 K V   f   8Vi   Pi.Vi   Pf.V f   Vi  1.4    Pf  Pi    50 105  Vi   2.7 105 N / m 2  2.7bar V   8V   f   i W  1   1  PiVi  Pf V f  RT f  Pf V f  RT f V f   4.6 10 3 m 3 Pf Vf V f  8Vi Vi   5.8  10  4 m 3 W  4145J 8