HELP!!!!~~~THXXX大家

Be aware that the gas in part A and part B is of the same type of gas. You have missed out the point that "...both parts A and B are heated to the same temperature...".

The reason is simply because the sum of volume of space occupied by A and B together is constant (the water column is incompressible). At the same temperature, the gas, if seen altogheter, should be of the sam pressure, hence resulted in the same volume occupied by the gas in parts A and B.

If using the kinetic theory, pressure P is given by
P = (1/3).d.c^2
where d is the gas density, and c is the rms speed.

At the same temperature, the gas should have the same density d and rms speed c, and hence the same pressure. That said, the pressures of gas in part A and part B are equal at the new temperature. The water column thus doesn't move.

Your arguments:

(i) 因為 V大左兩倍 P會變少2倍...另一邊 V一樣 P無變

This is entirely not correct. Why pressure P is reduced to half when volume V is doubled. Remember that the mass of gas in part B is double to that in part A.

(ii) GAS A應該係: PV/T...GAS B: (P/2)乘(V乘2)/T

Again, this is not correct
PV/T = constant
but the "constant" for part A is NOT equal to the "constant" for part B
, simply because the mass of gas in part B is not the same as that in part A.
The full ideal gas equation is: PV = nRT
For gas in part A: PV/T = nR, where n is the number of moles of gas in part A.
For gas in part B: P(2V)/T = (2n)R, where 2n is the number of moles of gas in part B.
You could see that the "constant" for part B gas is, in fact, double to the "constant" for part A gas.

Hence, the gas volume ratio is obtained by dividing the second equation by the first,
[P(2V)/T]/[PV/T] = (2n)R/nR
i.e. (2V)/V = 2
or (volume of gas in B)/(volume of gas in A) = 2:1
the ratio is independent of temperature.