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butalik [34]
3 years ago
8

A reaction that absorbs 49.6 kj from the surroundings in a vessel that undergoes a volume decrease of 1.25 l at a constant 4.00

atm pressure has an energy change, δe = ________ kj.
Physics
2 answers:
Verdich [7]3 years ago
6 0

Answer : The energy change is, 49.09 kJ

Explanation :

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

\Delta E=q+w

or,

\Delta E=q+(-P\Delta V)

\Delta E=q-P\Delta V

where,

\Delta E = energy change

q = heat absorb = 49.6 kJ = 49600 J     (1 kJ = 1000 J)

P = pressure = 4.00 atm

\Delta V = change in volume = 1.25 L

Now put all the given values in the above formula, we get:

\Delta E=q-P\Delta V

\Delta E=(49600J)-(4.00atm\times 1.25L)

\Delta E=(49600J)-(5L.atm)

As we know that, (1 L.atm = 101.325 J)

\Delta E=(49600J)-(5\times 101.325J)

\Delta E=(49600J)-(506.625J)

\Delta E=49093.375J=49.09kJ

Therefore, the energy change is, 49.09 kJ

Licemer1 [7]3 years ago
4 0
E=q+ w =q+P(delta V)
E= 49.6 +4 (1.25)
E= 54.6 KJ
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