Thermodynamics: delta U = q + w ya q - w? bhai sign hi samajh nahi aata

11 answers & replies

• [+][–]Meera Patel11thSample· 20d ago

  Both are the same law, only the sign convention for work changes.

  Chemistry (IUPAC) convention: $w$ is work done on the system. So $\Delta U = q + w$. Work done on the system (compression) is positive, work done by the system (expansion) is negative. That's why $w = -p_{ex}\Delta V$: when the gas expands, $\Delta V > 0$ so $w < 0$.

  Physics convention: $W$ is work done by the system. So $Q = \Delta U + W$, i.e. $\Delta U = Q - W$. Here $W = p\Delta V$ is positive for expansion.

  Same physics, same numbers. $w_{chem} = -W_{phys}$. Just pick the convention your exam uses (NEET chemistry follows NCERT, so use $q + w$).

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  • [+][–]Yash Tiwari11thSample· 20d agoAuthor

    oh ok, toh dono me bas "kiske liye kaam" ka farak hai. thanks Meera. expansion me gas khud kaam kar rahi hai toh uski energy ghat rahi hai, isliye negative, right?

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    • [+][–]Meera Patel11thSample· 20d ago

      Right. The system spends energy, so $w$ is negative from the system's point of view.

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• [+][–]Kabir Joshi11thSample· 20d ago

  For your example: isothermal for an ideal gas means $\Delta U = 0$ (U depends only on temperature for an ideal gas). It doesn't mean q = 0. That would be adiabatic. In isothermal expansion the gas does work but its temperature stays constant, so it must be absorbing heat from the surroundings: $q = -w$, so $q > 0$ and $w < 0$.

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  • [+][–]Yash Tiwari11thSample· 20d agoAuthor

    ye wala point miss ho gaya tha, isothermal ≠ adiabatic. 😅 ok.

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• [+][–]Aman YadavDropperSample· 20d ago

  Your three cases with chemistry sign convention:

  1. Isothermal expansion of ideal gas: $\Delta U = 0$, $w < 0$, $q > 0$, $q = -w$. For reversible: $w = -2.303\,nRT\log\frac{V_2}{V_1}$.
  2. Adiabatic compression: $q = 0$, so $\Delta U = w$. Compression means $w > 0$, so $\Delta U > 0$ and temperature rises.
  3. Free expansion into vacuum: $p_{ex} = 0$ so $w = 0$. For an ideal gas with no heat exchange, $q = 0$ and $\Delta U = 0$, temperature unchanged.

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• [+][–]Harsh Gupta12thSample· 20d ago

  so in isothermal compression q is positive since we're putting energy in?

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  • [+][–]Rahul DubeyDropperSample· 20d ago

    Other way round, Harsh. In isothermal compression of an ideal gas, work is done on the gas ($w > 0$) but $\Delta U = 0$, so $q = -w < 0$. Heat flows out to the surroundings. That's how a bicycle pump gets warm and loses it to air, if you do it slowly.

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    • [+][–]Harsh Gupta12thSample· 20d ago

      ahh ok. so the energy you put in as work leaves as heat. makes sense. thanks Rahul

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• [+][–]Riya Singh12thSample· 19d ago

  I think of it as a bank balance. $\Delta U$ is the balance. $q$ and $w$ are deposits (positive) or withdrawals (negative) from the system's point of view. Whatever the convention, the system's own bank statement uses its own signs. Not the most scientific but it helped me stop panicking.

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  • [+][–]Yash Tiwari11thSample· 19d agoAuthor

    Update: dobara mock ke wo teeno sawal solve kiye, ab sahi aaye. Bank wali analogy ne bhi kaam kiya Riya 🙏

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