1. **Problem statement:** Calculate the change in internal energy for a gas or system given heat absorbed or released and work done. 2. **Formula used:** The first law of thermodynamics states that the change in internal energy $\Delta U$ is given by: $$\Delta U = Q + W$$ where $Q$ is the heat added to the system and $W$ is the work done on the system. 3. **Important rules:** - Heat absorbed by the system is positive $Q > 0$. - Heat released by the system is negative $Q < 0$. - Work done on the system is positive $W > 0$. - Work done by the system on surroundings is negative $W < 0$. 4. **Calculations:** **(2) Gas absorbs 100 J heat and does 40 J work on surroundings:** - $Q = +100$ J (heat absorbed) - $W = -40$ J (work done by system) $$\Delta U = 100 + (-40) = 60 \text{ J}$$ **(3) System does 50 J work, 30 J heat flows into system:** - $Q = +30$ J - $W = -50$ J (work done by system) $$\Delta U = 30 + (-50) = -20 \text{ J}$$ **(4) Piston compresses gas doing 90 J work, 30 J heat enters system:** - $Q = +30$ J - $W = +90$ J (work done on system) $$\Delta U = 30 + 90 = 120 \text{ J}$$ **(5) System receives 150 J heat and does 50 J work on surroundings:** - $Q = +150$ J - $W = -50$ J $$\Delta U = 150 + (-50) = 100 \text{ J}$$ **Final answers:** - (2) $\Delta U = 60$ J - (3) $\Delta U = -20$ J - (4) $\Delta U = 120$ J - (5) $\Delta U = 100$ J These results show how internal energy changes depending on heat and work directions.