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.