Calculating Pressure Difference in a Manometer with Oil/Mercury

[mikefitz]

A manometer using oil (density 0.9 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 0.75 cm of Hg. (a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (b) What would your answer be if the manometer used mercury instead?

I'm using 1 Pa for initial pressure and 1.75 Pa for final pressure

1 = .0009kg/cm^3(981 cm/s^2)(d) =>1.1326cm
1.75 = .0009kg/cm^3(981 cm/s^2)(d) =>1.9821cm

1.9821-1.1326 = .8495cm - this is wrong? any ideas why?

 

[andrevdh]

The two pressures need to balance each other - the pressures of the stated mercury column (rise in the pressure in the tank) and the pressure of the raised oil column. But the oil column will only rise to only half the balancing height due to the fact that on the tank side it lowers by half the height and on the open side it rises by the other half.

 

[kyle8921]

I would approach this problem by first identifying the variables and equations involved. The pressure difference in a manometer can be calculated using the equation P = ρgh, where P is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height difference in the fluid levels.

Using this equation, we can calculate the pressure difference in the manometer using oil as the fluid. The initial pressure in the air tank is 1 Pa, and the final pressure is 1.75 Pa, resulting in a pressure difference of 0.75 Pa. Plugging in the density of oil (0.9 g/cm3) and the acceleration due to gravity (981 cm/s2), we can solve for the height difference in the fluid levels.

(a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere?

Using the equation P = ρgh, we can rearrange it to solve for h, which represents the height difference in the fluid levels. Plugging in the values, we get:

0.75 Pa = 0.9 g/cm3 * 981 cm/s2 * h
h = 0.75 / (0.9 * 981) = 0.00085 cm

Therefore, the fluid level in the side of the manometer open to the atmosphere will rise by 0.00085 cm.

(b) What would your answer be if the manometer used mercury instead?

If the manometer used mercury instead, the density of mercury (13.6 g/cm3) would need to be used in the calculation. Using the same equation, we get:

0.75 Pa = 13.6 g/cm3 * 981 cm/s2 * h
h = 0.75 / (13.6 * 981) = 0.000053 cm

Therefore, the fluid level in the side of the manometer open to the atmosphere would only rise by 0.000053 cm if mercury was used instead of oil. This is significantly smaller than the rise with oil due to the higher density of mercury.