Question Video: Liquid Column Manometer Uneven Unknown Gas Comparison | Nagwa Question Video: Liquid Column Manometer Uneven Unknown Gas Comparison | Nagwa

Question Video: Liquid Column Manometer Uneven Unknown Gas Comparison Physics • Second Year of Secondary School

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. Which of the following correctly relates the pressure of the gas and the pressure of the atmosphere, π_(gas) and π_(atm)? [A] π_(gas) = π_(atm) [B] π_(gas) > π_(atm) [C] π_(gas) < π_(atm)

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Video Transcript

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. Which of the following correctly relates the pressure of the gas and the pressure of the atmosphere, π sub gas and π sub atm? (A) The pressure of the gas equals the pressure of the atmosphere. (B) The pressure of the gas is greater than the pressure of the atmosphere. (C) The pressure of the gas is less than the pressure of the atmosphere.

The manometer described in our problem statement is this U-shaped tube here. Fluid is put into a manometer like this, and the pressure on either side is indicated by the height of the fluid. Any difference in the height of the fluid on either side of the U-shaped tube indicates a difference in pressure on either side. Here, we can see that whatever the liquid in the manometer is, the height of that liquid on the left-hand side is greater than the height on the right. We can understand this by thinking about whatβs going on at the interface of this fluid on either side.

Since the left half of the manometer is open to the atmosphere, that means at the air and liquid interface on the left, thereβs atmospheric pressure pushing down like this. Since this whole system is in equilibrium, we know that the liquid in our manometer on this side is also pushing up with that same pressure. And something similar is true on the right-hand side. Thereβs some amount of pressure created by the gas reservoir pushing on the surface of our liquid. And then, again, because our liquid is in equilibrium, it pushes back on the gas with an equal and opposite pressure.

The question is, which is greater, the pressure of the atmosphere here or the pressure of the gas here? To figure this out, we can go back to the fact that on either side of our U-shaped tube, the height of the liquid in the tube is indeed different. Specifically, the overall height of the liquid on the left-hand side of our manometer is greater. We could think of this difference as the outcome of a competition, we could call it, between this pressure due to the atmosphere and this pressure due to the gas. The gas presses on the liquid so that not only does it match the atmospheric pressure here pushing down, but itβs also able to support the pressure created by this difference in height between the liquid on either side of the manometer.

Recall that the pressure created by a fluid of density π and a height β equals π times π times β. So then, whatever this height is here, that amount of liquid creates some amount of pressure. That pressure is transmitted downward through the liquid and then all throughout the rest of the column up to the right-hand side. The pressure of the gas in the reservoir must equal the sum of this pressure due to the height of our fluid in the manometer plus atmospheric pressure over here.

If we call the pressure due to this height of our liquid π sub liquid, then we can write that the pressure of the gas in the reservoir equals atmospheric pressure plus the pressure of the liquid due to its additional height on the left-hand side of the manometer. Since π sub liquid is greater than zero, that tells us that the pressure of the gas is greater than the pressure of the atmosphere. And that tells us our answer. The pressure of the gas is greater than the pressure of the atmosphere, which is indicated by answer choice (B).

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