Measuring Pressure — Question 1

Question 1

The higher liquid surface marks the lower gas pressure

A U-tube contains liquid of density $1200\,\mathrm{kg/m^3}$. The left arm is attached to a gas chamber and the right is open to atmosphere at $100\,\mathrm{kPa}$. The left liquid surface is $0.15\,\mathrm{m}$ above the right. Neglect gas density and capillarity; use $g=10\,\mathrm{m/s^2}$.

1. Find the chamber gauge and absolute pressures by tracing pressure through the liquid.

2. If chamber pressure is raised by $3.0\,\mathrm{kPa}$ while atmosphere stays fixed, which surface becomes higher and by how much?

Question 1 - Solution

The higher liquid surface marks the lower gas pressure

(a) Derivation and calculation

Descend from the higher left meniscus to the right-surface elevation:$$p_L+\rho g(.150)=p_{\mathrm{atm}}.$$$$\boxed{p_{g,L}=-(1200)(10)(.150)=-1800\,\mathrm{Pa}}.$$$$\boxed{p_L=100.0-1.8=98.2\,\mathrm{kPa}}.$$

(b) Result and interpretation

$$p_{g,L}^{\prime}=-1.8+3.0=+1.2\,\mathrm{kPa}.$$$$p_L^{\prime}-p_{\mathrm{atm}}=\rho g(z_R-z_L)\Rightarrow\boxed{z_R-z_L=1200/(1200\cdot10)=.100\,\mathrm m}.$$The right surface is now higher. The sign follows the ordered pressure path.

Physical check

The liquid surface is higher on the lower-pressure side in each equilibrium.