A U-tube manometer measures the pressure of a gas supply. The water level on the gas supply side is 17 cm and on the open side is 15 cm. Atmospheric pressure is 74 cmHg. Calculate the gas pressure.

The water level on the gas supply side (left) is $17 \text{ cm}$ from the bottom.
The water level on the open side (right) is $15 \text{ cm}$ from the bottom.
The difference in height ($h$) is:

$$h = 17 \text{ cm} - 15 \text{ cm} = 2 \text{ cm}$$

Convert this height to meters:

$$h = 2 \text{ cm} = 0.02 \text{ m}$$

Given atmospheric pressure ($P_{atm}$) = $74 \text{ cmHg}$.
We use the formula $P = \rho g h$, where $\rho$ is the density of mercury, $g$ is the acceleration due to gravity, and $h$ is the height of the mercury column.
Density of mercury ($\rho_{Hg}$) = $13600 \text{ kg/m}^3$.
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$.
Height of mercury column ($h_{Hg}$) = $74 \text{ cm} = 0.74 \text{ m}$.

$$P_{atm} = \rho_{Hg} g h_{Hg}$$ $$P_{atm} = 13600 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 0.74 \text{ m}$$ $$P_{atm} = 98528 \text{ Pa}$$

The pressure due to the water column difference ($P_{water}$) is given by $P_{water} = \rho_{water} g h$.
Density of water ($\rho_{water}$) = $1000 \text{ kg/m}^3$.
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$.
Height difference ($h$) = $0.02 \text{ m}$.

$$P_{water} = 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 0.02 \text{ m}$$ $$P_{water} = 196 \text{ Pa}$$

Since the water level on the gas supply side (left) is higher than the water level on the open side (right), the gas pressure ($P_{gas}$) is less than the atmospheric pressure.

$$P_{gas} = P_{atm} - P_{water}$$ $$P_{gas} = 98528 \text{ Pa} - 196 \text{ Pa}$$ $$P_{gas} = 98332 \text{ Pa}$$

The gas pressure in SI units is:

$$\textbf{98332 Pa}$$