Question

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.

Answer 1

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

Answer 2

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