Oil-water manometer with gas cylinder

Data: tube $d=4$ mm, enlarged end $D=20$ mm (right limb, oil above the interface), left limb water open to atmosphere. $s_w=1.0$, $s_o=0.9$. The interface falls $x=25$ mm. The gas pressure $p$ acts on the oil surface.

Movement of the free surfaces

• Left water surface rises $x=25$ mm (water pushed into the left limb of equal bore).
• Oil surface falls: $y=x\left(\dfrac{d}{D}\right)^{2}=25\left(\dfrac{4}{20}\right)^{2}=1$ mm.

Before the gas is connected both sides are open to the atmosphere and balanced at the initial interface level, so $\gamma_wH_w=\gamma_oH_o$. After the change, balance pressures at the new interface level ($x$ below the old):

$$p+\gamma_o\,(H_o-y+x)=\gamma_w\,(H_w+x+x)$$

Subtracting the initial balance:

$$p=\gamma_w(2x)-\gamma_o(x-y)$$ $$p=9810(0.050)-0.9(9810)(0.025-0.001)$$ $$p=490.5-211.9=278.6\ \text{Pa}$$

Answer: gauge pressure in the cylinder $\approx278.6$ Pa (about 0.28 kPa, or 28 mm of water).