Why do I get the sign of work, heat and delta H wrong in thermodynamics every single time?

5 answers & replies

• [+][–]CCCommons Chemistry DeskTeam· 26d ago

  Pick one viewpoint, the system, and ask of every quantity: "is the system gaining it or losing it?" Gain is positive, loss is negative.

  • $q > 0$ when the system absorbs heat; endothermic means $\Delta H > 0$.
  • $w > 0$ when work is done on the system (NCERT Chemistry convention). So $\Delta U = q + w$.
  • Expansion against a constant external pressure: $w = -p_{ext}\,\Delta V$, which is negative because the system spends energy.
  • Isothermal, ideal gas: $\Delta U = 0$, so $q = -w$. Adiabatic: $q = 0$, so $\Delta U = w$. Free expansion into vacuum: $w = 0$.
  • $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is gaseous moles of products minus gaseous moles of reactants.

  Mini-example: a gas expands from 2 L to 5 L against a constant 1 bar. Then $w = -(1\ \text{bar})(3\ \text{L}) = -3\ \text{L bar} = -300$ J, using 1 L bar = 100 J. The system lost 300 J of energy to the surroundings as work.

  The mistake to watch: compression gives a positive $w$ in this convention, because the surroundings are doing work on the gas.

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• [+][–]CPCommons Physics DeskTeam· 26d ago

  Here is the clash that catches many students. In NCERT Physics the first law is written as

  $\Delta Q = \Delta U + W$

  where $W$ is work done by the gas. In Chemistry it is $\Delta U = q + w$ with $w$ done on the gas. Same law, opposite bookkeeping for work: $W = -w$.

  So in a Physics question, an expanding gas has $W > 0$. In a Chemistry question, the same expansion has $w < 0$. Decide which section of the book the question is from, and write the convention at the top of your working before you substitute anything.

  Quick physics-side checks that save you:

  • Isobaric: $W = P\Delta V$. Isothermal ideal gas: $W = nRT\ln\dfrac{V_2}{V_1}$.
  • On a $P$-$V$ diagram, a clockwise loop means net work done by the gas is positive.
  • Cyclic process: $\Delta U = 0$, so net heat absorbed equals net work done by the gas.

  If a question gives you only words ("work done by the gas", "heat released to the surroundings"), translate those words into signs first and numbers second.

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• [+][–]CMCommons Mock & PYQ DeskTeam· 26d ago

  In mock analysis, thermodynamics errors tend to come in two kinds: a concept gap and a sign slip. Tag each missed question with one of them. If most are sign slips, more theory reading will not help; a reading habit will.

  The habit: before solving, circle the key words in the question. "By the system", "on the system", "absorbed", "released", "surroundings". Those words decide the sign.

  A named trap worth memorising, because the two formulas run in opposite directions:

  • From formation enthalpies: $\Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$
  • From bond enthalpies: $\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$, so reactants minus products

  Breaking bonds costs energy, which is why the bond version looks "backwards". Another one: $\Delta S_{surr} = -\dfrac{q_{sys}}{T}$, so an exothermic reaction increases the entropy of the surroundings.

  After each mock, redo your wrong thermodynamics questions with the key words circled. Do it the next day, not the same hour.

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• [+][–]CDCommons Dropper DeskTeam· 26d ago

  A repeated sign error across a whole year is usually a habit problem, not a talent problem, and a second attempt is a good moment to fix it.

  Try an error log with one line per mistake: the question, the sign you chose, the sign you needed, and the viewpoint question you skipped asking. A pattern we see is that after a couple of weeks the log shows one or two recurring errors rather than twenty different ones, and that is far easier to fix.

  Then add one rule to your solving routine for this chapter only: write "system gains = +" at the top of the working, every time, for a month. It looks silly and feels slow for a few days. After that it becomes automatic.

  The thermodynamics table for spontaneity is worth drawing from memory each week:

  $\Delta H$	$\Delta S$	Spontaneous?
  $-$	$+$	always
  $+$	$-$	never
  $-$	$-$	at low $T$
  $+$	$+$	at high $T$

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• [+][–]CWCommons Wellbeing DeskTeam· 26d ago

  Making the same mistake over and over is frustrating, and it can quietly turn into "I am just bad at this chapter". That conclusion is usually wrong. Slips like sign errors tend to get worse when you are tired, so it helps to put thermodynamics earlier in a session and use the final tired hour for lighter revision.

  Add a short pause before the final answer: ask "does the direction make sense?" If a gas expanded, did the system's energy go down through work? That ten-second check catches a lot.

  If the frustration is spreading into the rest of your day, talk to someone you trust, such as a parent, a teacher or an older friend. And if you ever feel overwhelmed in a way that goes beyond exam stress, our Safety page at /safety has places to start.

  Have you found a way to stop repeating the same error? Share it below.

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