1.17 Adiabatic Compression or Expansion (4) S.J.T.0. Phase Transformation and Applications Helium,ideal gas Valve First gas,？ Insulated 50-liter,25 C,20 atm Quench chamber,?10 atm (L,),-(H。)6im。+8Q+8w=dU=0 System -boundary om m。 H,=H。 T=T, H(T,P)=H(T) Quench chamber.?10 atm,Close system Define the system as the quantity of gas remaining in the tank when the pressure reaches 10 atm Adiabatic expansion T3=T(0.758)=298×0.758=226K 226K/-47C SJTU Thermodynamics of Materials Fall 2012 ©X.J.Jin Lecture 3 Second law IPhase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2012 © X. J. Jin Lecture 3 Second law I 1.17 Adiabatic Compression or Expansion (4) Helium, ideal gas Insulated 50-liter, 25 °C, 20 atm Quench chamber, ?, 10 atm (H ) m − (H ) m + Q + W = dU = 0 i δ i o δ o δ δ Valve H i = H o ( ) 5 2 0.4 1 2 1 2 ) (0.5) 2010 = ( ) = ( = R R CR P PP TT T2 = T1(0.758) = 298×0.758 = 226 K 226K / -47 °C System boundary δmi δmo First gas, ? Ti = To Quench chamber, ?, 10 atm， Close system H(T, P) = H(T) Define the system as the quantity of gas remaining in the tank when the pressure reaches 10 atm Adiabatic expansion