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Boyle's Law states that the product of the pressure and volume of an ideal gas under constant temperature is constant. That is: $P_i V_i = P_f V_f$.

But what happens when temperature is not constant? I suppose squeezing affects the temperature of the gas (neglecting the heat transfer between the balloon and hand)? How do we calculate the work done in this case?

Note that if we assume the temperature constant, then the work must be zero as $P . V$ yields energy and it is always constant, which does not make sense to me.

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(a) "But what happens when temperature is not constant? I suppose squeezing affects the temperature of the gas (neglecting the heat transfer between the balloon and hand)? How do we calculate the work done in this case?"

This is the so-called adiabatic change: no heat transfer to or from surroundings. The work that you do in squeezing the balloon to make its volume decrease goes to raising the internal energy of the gas inside it, and so raising its temperature. For n moles of an ideal monatomic gas the work done is given by $$W=\tfrac32nR\Delta T$$ in which $\Delta T$ is the temperature rise of the gas. This can be related to what happens to the pressure and volume of the gas using the ideal gas equation and a little gentle calculus. It is standard textbook stuff.

(b) "Note that if we assume the temperature constant, then the work must be zero as P x V yields energy" No. $pV$ has the dimensions of energy and the product $pV$ does feature in various thermodynamic energy formulae, but your conclusion about work simply doesn't follow.

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  • $\begingroup$ Thank you Philip. For a) Would you expand your answer by providing the standard textbook stuff? Because PV=nRT does not seem to yield the temperature rise to me. For b) assuming constant temperature means we remove heat from the system while squeezing the balloon which might keep the work at 0. Don't you think? $\endgroup$
    – Xfce4
    Commented Dec 14, 2020 at 16:44
  • $\begingroup$ (a) "Would you expand your answer by providing the standard textbook stuff?" I don't mean to be rude, but don't you have a textbook? And have you checked websites? (b) If you're squeezing the balloon and making its volume smaller, then you are doing work on the gas. The work can't be zero. In an isothermal compression of an ideal gas, heat must leave the gas at the same rate that work is being done on it. This follows from the First Law of Thermodynamics and the fact that for an ideal gas the internal energy depends only on temperature. $\endgroup$
    – Philip Wood
    Commented Dec 14, 2020 at 22:28
  • $\begingroup$ Hey Philip. I already checked the pages and formulas about ideal gases before writing here. The thing is non-constant-tempereture case is not an easy "standard textbook stuff". Here is the page from Wikipedia: en.wikipedia.org/wiki/Ideal_gas_law. Can you point me two formulas that will solve a situation with two variables (pressure and temperature) having unknown final values. That is, we know all the initial values and the final volume but we don't we know the final pressure and the final temperature. According to you, what are the formulas yielding the equilibrium point? $\endgroup$
    – Xfce4
    Commented Dec 20, 2020 at 19:01
  • $\begingroup$ Your original question asked about reversible adiabatic changes (no heat transfer). Are these still what you're interested in? $\endgroup$
    – Philip Wood
    Commented Dec 20, 2020 at 19:42
  • $\begingroup$ To Philip. I take your last comment as a hint. So it looks like I need to use the Isentropic process formulas where entropy is constant. Since we know $V2/V1$ ratio the related formulas seem to be: $P2 = P1(V2/V1)^{-γ}$ and $T2 = T1(V2/V1)^{(1-γ)}$, where γ is typically 1.4 for diatomic gases and 1.6 for mono atomic gases. What are your comments? $\endgroup$
    – Xfce4
    Commented Dec 20, 2020 at 22:19

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