First Law: The change in the internal energy of a closed system is equal to the heat added to the system plus the work done on the system.
In symbols: \$ΔU = q + W\$
• \$ΔU\$ – increase in internal energy (J)
• \$q\$ – heat transferred to the system (J)
• \$W\$ – work done on the system (J)
| Quantity | Positive When | Negative When |
|---|---|---|
| \$q\$ | Heat added to the system 🔥 | Heat removed from the system ❄️ |
| \$W\$ | Work done on the system 💪 | Work done by the system (expansion) 🚀 |
Think of a thermos bottle as a closed system. When you pour hot coffee into it, heat (\$q\$) flows into the bottle. If you then squeeze the bottle, you do work (\$W\$) on the coffee, increasing its internal energy (\$ΔU\$). The bottle keeps the coffee warm because it limits heat loss to the surroundings.
Problem: A piston contains 1 mol of an ideal gas at 300 K. The gas is compressed quasi‑statically, doing 200 J of work on the gas, while 50 J of heat is transferred into the gas. What is the change in internal energy?
🔹 Remember the sign convention: Heat added → \$q>0\$; Work done on the system → \$W>0\$.
🔹 Check units: All terms in joules (J). If you see kJ, convert to J.
🔹 Use the equation in the correct form: \$ΔU = q + W\$ (not \$ΔU = q - W\$ unless you’re using the sign convention where work done by the system is positive).
🔹 Show your work: Write the equation, plug in the numbers, and state the final answer clearly.