Answer:
the change in the internal energy of the system is 3,752.67 J
Step-by-step explanation:
Given;
initial volume of the gas, V₁ = 30.6 L
final volume of the gas, V₂ = 1.8 L
constant pressure of the gas, P = 1.8 atm
Energy released by the system, Q = 1.5 kJ = 1,500 J
Apply pressure-volume work equation, to determine the work done on the gas;
w = -PΔV
w = -P(V₂ - V₁)
w = - 1.8 atm(1.8 L - 30.6 L)
w = 51.84 L.atm
w = 51.84 L.atm x 101.325 J/L.atm
w = 5,252.67 J
The change in the internal energy of the system is calculated as;
ΔU = Q + w
Since the heat is given out, Q = - 1,500 J
ΔU = -1,500 J + 5,252.67 J
ΔU = 3,752.67 J
Therefore, the change in the internal energy of the system is 3,752.67 J