Question

Calculate the change in internal energy (δe) for a system that is giving off 23.0 kj of heat and is changing from 18.00 l to 15.00 l in volume at 1.50 atm pressure. (remember that 101.3 j = 1 l ⋅ atm)

Answer 1

The change in internal energy (δE) for a system giving off 23.0 kJ of heat and changing in volume at a constant pressure is -22445.5 J.

Step-by-step explanation:

To calculate the change in internal energy (δE) for a system that is giving off 23.0 kJ of heat (Q) and changing in volume at a constant pressure, we apply the first law of thermodynamics, which states that δE = Q - W, where W is the work done by the system. In this context, work (W) is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume. Given that heat is being released by the system, we treat Q as negative, and since the system is decreasing in volume, work is done on the system, and W is positive.

First, we need to convert 1.50 atm to joules by equating 1 L · atm to 101.3 J. The total work done is W = (1.50 atm)(15.00 L - 18.00 L) = (1.50 atm × -3.00 L)(101.3 J/L·atm) = -454.50 J.

Now, we convert the heat given off from kJ to J: Q = 23.0 kJ × 1000 J/kJ = -23000 J. Finally, we find the change in internal energy: δE = Q - W = -23000 J - (-454.5 J) = -22445.5 J.