Questions: A manometer using oil (density 0.900 g / cm^3 ) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 7.98 mmHg. Density of mercury is 13.6 g / cm^3. By how much does the fluid level rise in the side of the manometer that is open to the atmosphere if the manometer used mercury instead? cm

Solution Steps

We need to determine the rise in fluid level in a manometer when the fluid is mercury, given that the pressure increase in the tank is 7.98 mmHg. The densities of oil and mercury are provided.

First, convert the pressure increase from mmHg to Pascals (Pa). The conversion factor is: \[ 1 \text{ mmHg} = 133.322 \text{ Pa} \]

So, the pressure increase is: \[ 7.98 \text{ mmHg} \times 133.322 \text{ Pa/mmHg} = 1063.9 \text{ Pa} \]

Using the pressure formula \( P = \rho g h \), where:

• \( P \) is the pressure,
• \( \rho \) is the density,
• \( g \) is the acceleration due to gravity (9.81 m/s²),
• \( h \) is the height change.

Rearrange to solve for \( h \): \[ h = \frac{P}{\rho g} \]

For mercury (\( \rho = 13.6 \text{ g/cm}^3 = 13600 \text{ kg/m}^3 \)): \[ h = \frac{1063.9 \text{ Pa}}{13600 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2} \]

\[ h = \frac{1063.9}{133416} \text{ m} \] \[ h = 0.00798 \text{ m} \] Convert meters to centimeters: \[ h = 0.00798 \text{ m} \times 100 \text{ cm/m} = 0.798 \text{ cm} \]

Final Answer