A U-tube manometer contains a liquid with density 0.8 g/cm³. If the liquid levels in the open and gas sides are 9 cm above and 9 cm below a reference line, respectively, what is the gas pressure?

From the diagram:

• The height from the bottom of the U-tube (Z) to the dashed line is $19 \text{ cm}$.
• The liquid level on the gas supply side (left) is $9 \text{ cm}$ below the dashed line.
• The liquid level on the open side (right) is $9 \text{ cm}$ above the dashed line.

Therefore, the height of the liquid level on the gas side from Z is:

$$h_{left} = 19 \text{ cm} - 9 \text{ cm} = 10 \text{ cm}$$

The height of the liquid level on the open side from Z is:

$$h_{right} = 19 \text{ cm} + 9 \text{ cm} = 28 \text{ cm}$$

The difference in height ($\Delta h$) of the liquid column is:

$$\Delta h = h_{right} - h_{left} = 28 \text{ cm} - 10 \text{ cm} = 18 \text{ cm}$$

Convert this to meters:

$$\Delta h = 0.18 \text{ m}$$

Given density of the liquid ($\rho_{liquid}$) is $0.8 \text{ g/cm}^3$.

$$\rho_{liquid} = 0.8 \text{ g/cm}^3 \times \frac{1 \text{ kg}}{1000 \text{ g}} \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^3$$ $$\rho_{liquid} = 0.8 \times \frac{1}{1000} \times 1000000 \text{ kg/m}^3$$ $$\rho_{liquid} = 800 \text{ kg/m}^3$$

We use the formula $P = \rho g h$. We'll use $g = 10 \text{ m/s}^2$.

$$P_{liquid} = \rho_{liquid} \cdot g \cdot \Delta h$$ $$P_{liquid} = 800 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.18 \text{ m}$$ $$P_{liquid} = 1440 \text{ Pa}$$

Since the liquid level on the gas supply side is lower than on the open side, the gas pressure is greater than the atmospheric pressure.

$$P_{gas} = P_{atm} + P_{liquid}$$

Given $P_{atm} = 101400 \text{ Pa}$.

$$P_{gas} = 101400 \text{ Pa} + 1440 \text{ Pa}$$ $$P_{gas} = 102840 \text{ Pa}$$

The gas pressure is $\text{102840 Pa}$.