Part A : This is the first law of thermodynamics question.

case 1 - the system expands and the surroundings get hotter
case 2 - the system expands and the surroundgins get colder
case 3 - the system contracts and the surroundings get hotter
case 4 - the system contracts and the surroundgins get colder

For each case, is the change in internal energy a)Positive b) Negative c) Not enough data

Part B: A gaseous reaction occurs at a constant pressure of 30.0 and releases 57.4 of heat. Before the reaction, the volume of the system was 6.40. After the reaction, the volume of the system was 2.30 .
Calculate the total internal energy change in kilojoules.

I know that that the change in internal energy = q+w

thank you

Part A:

To determine the change in internal energy for each case, we need to consider the signs of heat (q) and work (w) exchanged between the system and the surroundings.

1. The system expands and the surroundings get hotter:
In this case, heat is flowing into the system (q > 0) and work is being done by the system on the surroundings (w < 0). Therefore, the change in internal energy (ΔU) can be positive, negative, or zero, depending on the magnitudes of q and w. The answer is c) Not enough data.

2. The system expands and the surroundings get colder:
In this case, heat is flowing out of the system (q < 0) and work is being done by the system on the surroundings (w < 0). Therefore, the change in internal energy (ΔU) can be positive, negative, or zero, depending on the magnitudes of q and w. The answer is c) Not enough data.

3. The system contracts and the surroundings get hotter:
In this case, heat is flowing into the system (q > 0) and work is being done on the system by the surroundings (w > 0). Since both q and w are positive, the change in internal energy (ΔU) would be positive (ΔU > 0). The answer is a) Positive.

4. The system contracts and the surroundings get colder:
In this case, heat is flowing out of the system (q < 0) and work is being done on the system by the surroundings (w > 0). Since q is negative and w is positive, the change in internal energy (ΔU) can be positive, negative, or zero, depending on the magnitudes of q and w. The answer is c) Not enough data.

Part B:

You are correct. The change in internal energy (ΔU) can be calculated using the equation: ΔU = q + w, where q is the heat and w is the work.

Given:
Pressure (P) = 30.0 atm
Heat (q) = -57.4 kJ (negative because heat is released)
Initial volume (V1) = 6.40 L
Final volume (V2) = 2.30 L

First, let's calculate the work done (w) using the equation:
w = -PΔV
where ΔV = V2 - V1

ΔV = 2.30 L - 6.40 L = -4.10 L
w = -30.0 atm * (-4.10 L) = 123.0 L·atm

Now, we can calculate the change in internal energy (ΔU) by substituting the values into the equation:
ΔU = q + w = -57.4 kJ + 123.0 L·atm

Since kilojoules and liters·atm have different units, we need to convert one of them to match the other. We can use the conversion factor: 1 L·atm = 101.3 J
123.0 L·atm * (101.3 J / 1 L·atm) = 12,478 J

Finally, we can convert the result to kilojoules by dividing by 1000:
ΔU = (12,478 J) / 1000 = 12.478 kJ

Therefore, the total internal energy change is 12.478 kJ.