Question

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 50.0 atm and releases 65.0 kJ of heat. Before the reaction, the volume of the system was 7.60 L . After the reaction, the volume of the system was 2.20 L . Calculate the total internal energy change, ΔE, in kilojoules.

Express your answer with the appropriate units.

Answer 1

The ideal gas system at initial volume (vi) of 7.60L undergoes reaction at a constant pressure of 50 atm The final volume of the system becomes (V.) 2.2L. - The system releases heat energy of 65 kJ. It is irreversible process as the syentem is undergoing reaction at constant pressure. Siti Work done (W) Pext. (VF - Vi) (IUPAC) . Here, Pext = 50 atm, Vf = 2.2L, Viz 7.6L. . ..W2 - 50 atm ( 2.2 -7.6) L. [lliatm = 101-325 3] = + 270 Li atmi = (270x 101.325)3 = 27357.75J . it ve sign signifies that work done on the system (during irreversible process) is occured. Again, the system releases 65 kJ heat energy. :: 9 = - 65 KJ = -65X103 J PIKJ=1031] ve sign for heat every release by the system. - from the first law of thermodynamies, AE= gtw (as per IUPAC) . AE= internal energy change of the system. AE= -(65x103) J + 27375.75 J. LE = -37642.25 J ..AE = -37.64 kJ

1 L.atm= 1 dm? (1.013225x 105 Nm2) 3 10m3 x 1.01325x105 Nimi? = 101.325 N.m = 101.325 J Idma 10 m Iatmal.01325x16 Nim? IN.MAIJ