High School

An open-ended U-shaped manometer is attached to a container of gas that is exerting a pressure of 104.5 kPa. When the valve is opened:

i. State the direction of the mercury displacement in the open arm of the U-shaped manometer (upwards/downwards), with reasoning.

ii. Calculate the difference in the height of the mercury levels in the two arms once the mercury stops moving.

[Atmospheric pressure is 99.8 kPa, and [tex]\rho_{\text{mercury}} = 13600 \, \text{kg/m}^3[/tex]]

Answer :

An open-ended U-shaped manometer is a device used to measure the pressure of a gas. In this case, the manometer is attached to a container of gas that is exerting a pressure of 104.5 kPa.

i. When the valve is opened, the mercury in the open arm of the U-shaped manometer will displace upwards. This is because the pressure inside the container is greater than the atmospheric pressure.


ii. To calculate the difference in the height of the mercury levels in the two arms once the mercury stops moving, we need to consider the pressure difference between the gas in the container and the atmospheric pressure.

The pressure difference can be calculated using the equation: Pressure difference = Pressure inside the container - Atmospheric pressure

Pressure difference = 104.5 kPa - 99.8 kPa = 4.7 kPa

Since the density of mercury is given as 13600 kg/m^3, we can use the equation: Pressure difference = Density of mercury * g * Height difference

Height difference = Pressure difference / (Density of mercury * g)

Height difference = 4.7 kPa / (13600 kg/m^3 * 9.8 m/s^2)

Height difference = 0.00408 m or 4.08 mm

Therefore, the difference in the height of the mercury levels in the two arms of the manometer will be 4.08 mm.

In summary, when the valve is opened, the mercury in the open arm of the U-shaped manometer will displace upwards due to the higher pressure inside the container.

To know more about atmospheric pressure visit:

https://brainly.com/question/31634228

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