A balloon popped in a room This question is Question 23 from CAP exam 2017.
An inflated balloon, filled with a gas of temperature T, is held in a room with the same temperature.  If we make a small hole on the side of the balloon and let the gas escape, which statement is true about the temperature T’ of the gas right after leaving the balloon?
The correct answer according to the solutions is "T′> T". However, I cannot imagine why this would be the case.
By all reasoning, this is an adiabatic expansion. That means $PV^\gamma$ is conserved. Since $\gamma$ is bigger than $1$, this means that $PV$ should decrease when $P$ decreases, and thus temperature should decrease.
As another reasoning, the gas inside is doing work by expanding outwards, and thus the energy would decrease.
 A: You are correct in saying that the gas remaining at any time within the balloon has experienced an adiabatic reversible expansion, so its temperature is cooler than initially.  But, the problem does not appear to be asking you to compare the temperature of the gas exiting the hole at any time with the temperature at time zero.  It seems to be asking you to compare the temperature of the gas exiting the hole at time t with the temperature of the gas remaining within the balloon at time t.  
The gas passing through the hole at time t is experiencing a change in pressure at constant enthalpy per unit mass.  This is the so-called Joule Thomson effect, and the change in the gas temperature is determined by the Joule-Thomson coefficient:  the partial derivative of temperature with respect to pressure at constant enthalpy.  For an ideal gas, this is zero, but, for most real gases at room temperature and approximately atmospheric pressure, the Joule-Thomson coefficient is positive.  In my judgment, this is what the problem statement was driving at.  In any event, it was a pretty ambiguous (crappy) statement, since, for a handful of gases under these conditions (e.g., He and H2), the JT coefficient is negative. 
