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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 73.2 kj of heat. before the reaction, the volume of the system was 7.60 l . after the reaction, the volume of the system was 2.00 l . calculate the total internal energy change, δe, in kilojoules. express your answer with the appropriate units.

Respuesta :

Total internal energy change is equals to -44.83kJ

Q=-73.2kJ (negative sign indicates that heat was released by the system),

P= 50.0atm

ΔU= Q + W, FIRST LAW OF THERMODYNAMICS..........(1)

ΔV= Final volume - initial volume= 2.00 litre - 7.60litre= -5.60litre

work done by the system (w)= -PΔV

w= -(50.0×(-5.60)) atm×litre=  280atm litre

1 atm litre= 101.325J

w= 280 ×101.325 J= 28,371J

1kJ=1000J,

w=28.37KJ,

so putting in the values in equation (1)...

energy change(ΔU) = -73.2 kJ + 28.37 kJ

= - 44.83 kJ

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