In previous lesson, we have discussed atmospheric pressure, including the proofs of existence, measurement of atmospheric pressure and applications of atmospheric pressure. In this lesson, we are going to continue with measurement of gas pressure by using Bourdon Gauge and manometer. In SPM, most questions about measurement of gas pressure are related to manometer. Therefore, our focus is on manometer. We will also discuss the example of U-Tube and trapped air in a capillary tube.

Let’s start with the manometer.

The U-shape tube filled with green liquid that connect to a gas tank filled with liquid methane and methane gas is a manometer. Manometer is a device used to measure gas pressure in a container.

The idea is pretty simple. Let’s say the pressure in the gas tank is P_{gas} and the pressure of the atmosphere is P_{atm}. The liquid inside the manometer has 2 ends. Let’s label it A and B. Surface A exposed to the methane gas in the gas tank. Well, I guess you still remember the pressure on the surface of a liquid is equal to the gas pressure of the gas it exposed. Therefore, the pressure at A is equal to the pressure of the methane gas. Which means the pressure at A is also equal to P_{gas}.

Surface B exposed to the atmosphere, therefore the pressure at B is equal to the atmospheric pressure, which is P_{atm}. Let’s say we have another point which at the same level as point A. Let’s call it point C.

Well, you should have learned how to find pressure in a liquid, right? So, what is the pressure at point C? Well, the pressure in a liquid is equal to the atmospheric pressure + the liquid pressure, right?

Okay, now is the final state. Pressure of a given liquid is the same at the same level. A and C are at the same level. Pressure at A is P_{gas}; pressure at C is atmospheric pressure + liquid pressure. Therefore, the gas pressure is equal the sun of atmospheric pressure and liquid pressure. This is the equation we used to determine pressure by using manometer: P_{gas} = P_{atm }+ P_{liquid}.

Let’s say we have a gas tank filled with liquid methane and methane gas. The manometer is filled with mercury. The density of mercury is 13600kgm^{-3}. The atmospheric pressure is equal to 76cmHg. Find the pressure of the methane gas in the gas tank in the unit of cmHg and Pascal.

Well, let’s say the pressure of the gas is P_{gas}. From previous slide, we learned that the gas pressure is equal the sum of atmospheric pressure and liquid pressure. The atmospheric pressure is 76cmHg. Well, what is the liquid pressure at point C? Point C is 12cm below the surface. Therefore, its pressure is 12cmHg. Plug in this number into the equation. The gas pressure is equal to 88cmHg.

Now, how do we convert the unit of pressure from cmHg to Pascal? We can convert cmHg to Pascal by using the equation, P = hρg. h = 88cm or 0.88m. Take note that h must be in meter. The density of mercury is 13,600kgm^{-3} and g is equal to 10 N/kg. Therefore, the gas pressure is equal to 119,680Pa. This is how we determine gas pressure in the unit of Pascal by using manometer.

Well, we have just finished our discussion on manometer. We are going to continue with the U-Tube.

This is a U-tube filled with 2 types of liquid. Usually, a U-tube is used to compare and measure density of liquids. In exam, you may be asked which liquid is denser, or what is the density of the second liquid if the density of the first liquid is given, or vice versa.

Let’s draw a line at the level that separates both the liquids. Let’s say the length here is h_{1}, and this is h_{2}. To answer the first question: The liquid with shorter length is denser than the liquid with longer length. In this case, liquid 2 is denser than liquid 1.

How about question 2? How do we find the density of the liquid? Well, let’s say the density of liquid 1 is ρ_{1} and the density of the liquid is ρ_{2}. The density of the liquid can be determined by using the equation h_{1}ρ_{1} = h_{2}ρ_{2}.

Well, we have just learned that questions related to the U-tube can be solved by using the equation h_{1}ρ_{1} = h_{2}ρ_{2}. In this slide, we are going to discuss how this equation is developed. Let’s label the pressure here P_{1} and the pressure here P_{2}.

When we discussed liquid pressure, we have learned that pressure in liquid is equal to the sum of atmospheric pressure and the liquid pressure. Atmospheric pressure is denoted by P_{atm}, while the liquid pressure can be calculated by using the equation hρg. Therefore, P_{1} = P_{atm} + h_{1}ρ_{1}g whereas P_{2} = P_{atm} + h_{2}ρ_{2}g. We have also learned that pressure of a given liquid is the same at the same level. Therefore, P_{1} = P_{2}, and hence P_{atm} + h_{1}ρ_{1}g = P_{atm} + h_{2}ρ_{2}g. Cancel the P_{atm} at both sides. Again, cancel the g. We get h_{1}ρ_{1} = h_{2}ρ_{2}. This is how the equation is developed.

The diagram shows a U-tube filled with water and a liquid P. Find the density of liquid P. (Density of water = 1000kg/m^{3})

Well, the density can be determined by using the equation h_{1}ρ_{1} = h_{2}ρ_{2}. h_{1} = 8cm; = 1000kg/m^{3}; h_{2} = 8cm; we need to find ρ_{2}. Plug in all these number into the equation. The density of liquid P is 400kgm^{-3}.

Well, we have just finished our discussion on U-Tube. We are going to continue with the “Air Trapped in a Capillary Tube”.

The diagram shows 3 capillary tubes and a J-tube with air trapped inside. In SPM, you may be asked to find the pressure of the trapped air.

For the capillary tube placed horizontally, the pressure of the trapped air will be equal to the atmospheric pressure.

For the capillary tube where the mercury column is above the trapped air, the pressure of the trapped air will be equal to the atmospheric pressure + the pressure cause by the mercury.

For the capillary tube with the mercury column below the trapped air, the pressure of the trapped air will be equal to the atmospheric pressure – the pressure caused by the mercury.

For the J-tube, the pressure of the trapped air will be equal to the atmospheric pressure + the pressure cause by the mercury column higher than the lower end of the trapped air.