# How to calculate the work done by squeezing an air balloon (or pushing a piston)?

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.

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.
• 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? Commented Dec 20, 2020 at 22:19