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 : 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