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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 45.0 atm and releases 68.0 kj of heat. before the reaction, the volume of the system was 6.40 l . after the reaction, the volume of the system was 2.80 l . calculate the total internal energy change, δe, in kilojoules.

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W0lf93
To calculate the total internal energy change in kilojoules we need to calculate U which is equal to q + w . We know that w = P delta V where P is the pressure and V is the volume then we have that w = 45 atm * ( 6.40L - 2.80L) = 180 atm * L . Then we know that 1 atm * L = 101.325 J then we have that w = 180 * 101.325 J = 18,238.5 J . Then we have that 1 kJ = 10^3 J . Then w = 18.2385 kJ . Then we have that U = 18.238 - 68.0 kJ = -49.7615 kj since heat was released. Thus we have that the total internal energy change was -49.7615.

The total internal energy change, in the system is 49.761 kJ.

Internal energy formula is,

[tex]\rm \bold{\Delta U = Q-W}[/tex]

Where,

Q = heat transferred to the system

W= Work done by the system

we know,

[tex]\rm \bold{P\times \Delta V = W}[/tex]

where

P- pressure

V - change in volume

By solving the equation we get W = 18.238 kJ

Put the value in

U = 68 - 18.238

U = 49.761 kJ

Hence we conclude that the total internal energy change, in the system is 49.761 kJ.

To know more about internal energy, refer to the link:

https://brainly.com/question/11278589?referrer=searchResults

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