A, b, if the top of the pipeline is divided into vacuum, atmospheric pressure P0 = L 1+L2+L3. At this time, PA = L 1 < Pb = L 1+L2. If two parts of AB gas are mixed, because the upper end of L 1 is vacuum, the mixing pressure PAB = L 1. Therefore, the total volume after mixing is greater than the sum of the original volumes, and the mercury column L3 does not move, so the mercury column L 1 will rise. Since the atmospheric pressure is constant, it can be seen from P0=L 1+L2+L3 that the total length of the three mercury columns will not change, that is, L 1+L4 = L 1+L2+L3. Therefore, A is correct.

C. if there is air c at the top of the pipe, there is PC < pa = PC+l1< Pb = PC+l1+L2. The pressure of mixed gas is greater than that of PA, which makes the pressure of C increase and the volume decrease, so L 1 rises. So, C is wrong.

D, the gas is compressed, the volume decreases, the pressure increases, the pressure difference between the mixed gas and the outside is less than L2+L3, and the supported mercury column is shortened accordingly, that is, the mercury flows out of the tubule, which also promotes the volume increase of B, so the total length of the mercury column decreases, that is, it is correct to have L 1+L4 < L 1+L2+L3.

So choose: AD