First Law Of Thermodynamics

Question

A gas absorbs 120 J of heat and expands against a constant external pressure, performing 50 J of work on the surroundings. The change in internal energy (ΔU) of the gas is:

RankGuru · Accepted Answer

1. **State the First Law**: $\Delta \cup = q + w$. By IUPAC sign convention, heat absorbed by the system is positive and work done *by* the system is negative.
2. **Assign Values**: $q = +120$ J (heat absorbed); $w = -50$ J (expansion work done by the gas).
3. **Calculate**: $\Delta \cup = 120 + (-50) = +70$ J.
4. **Interpret**: The internal energy rises by 70 J — the system retained the portion of absorbed heat that was not used for expansion work.
5. **Why Not the Others?**
   - +170 J would result from incorrectly adding q and |w|.
   - −70 J reverses the sign of ΔU.
   - −170 J has no physical basis for this scenario.

First Law Of Thermodynamics

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