In: Chemistry
A reaction released 1256 J of heat and expands from 1.3L to 14.7L at a constant pressure of 1.04 kPa. What is the total energy change for the system in kJ. Include the correct sign with respect to the system. (1 kPa * L = 1 J)
The amount of heat released , Q = -1256 J
Work done by the system , W = - PdV
= -1.04 kPa x ( 14.7 - 1.3) L
= - 1.04 x 13.4 KPa L
= - 13.9 J
From first law of thermodynamics Q = dE + W
Here dE = change in internal energy = ?
dE = Q - W
= -1256 -(-13.9) J
= -1242.1 J