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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.

Verified Answer :

E=q+ w =q+P(delta V)
E= 49.6 +4 (1.25)
E= 54.6 KJ

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