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A sample of an ideal gas goes through the process shown in Figure . From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy E int,B −E int.A

Respuesta :

The difference in internal energy E int,B −E int.A = =4.29×10 4 J for the ideal gas.

Because the fueloline is going via a cyde, the general extrade in inner electricity have to be zero. ThereforeΔE int =ΔE int,AB +ΔE int,BC +ΔE int,CD +ΔE int,DA =zero

→ΔE int,AB =−ΔE int,BC −ΔE int,CD −ΔE int,DA

Recognize that ΔE int =zero for the isothermal method CD and replacement from the primary regulation for the alternative inner electricity changes

  • ΔE int,AB =−(Q BC +W BC )−(Q DA +W DA )
  • =−(Q BC −P B ΔV BC )−(Q DA −P D ΔV DA )
  • =−(Q B C+Q DA )+(P B ΔV BC +P D ΔV DA )
  • =−(345kJ−371kJ)+[(3.00atm)(0.310m 3 )+(1.00atm)(−1.00m 3 )]×( 1atm1.013×10 5 Pa )
  • =4.29×10 4 J

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