Initial state: v1= sh; p 1 = P0；
Final state: v2 = s (h-△ h); P2 = P0+ρGH；
Isothermal change of gas. According to Boyle's law, there is P 1V 1=P2V2.
That is p0sh = (p0+rhogh) s (h-△ h).
Get △h = rhoghp0+rhogh.
(2) If the temperature rises to at least t'℃ and the gas changes with the same volume, we can get P0273+t = P0+ρ GH 273+t' from Charlie's Law.
t′=(P0+rhogh)(273+t)P0？ 273(℃)
Answer: (1) If you press the piston hard, the piston will move down slowly, and the temperature will remain the same during the whole process. In order to make water flow out from the top of the thin tube, the moving distance of the piston should be at least ρ ghhp0+ρ gh;
(2) If the piston is kept in the initial position, the temperature will rise slowly, at least to (P0+ρgh)(273+t)P0, so that water can flow out from the top of the tubule. 273 degrees Celsius.