Measuring Gas Pressure

If we want to measure the pressure of a gas (e.g. what is the pressure of the gas in the Chemistry lab gas taps? Or how low is the pressure when we put a bell in a depressurised bell jar? Then we can use a bourdon gauge or a manometer.

The pressure due to a sealed volume of gas will act on any surface exposed to the gas.

The difference in the gas pressure and the atmospheric pressure is the excess pressure.

Bourdon Gauge

A bourdon gauge contains a coiled tube which tries to straighten when the pressure inside it is increased. This moves a needle around a dial and the pressure can be read.

typical bourdon gauge (bicycle pump)

 

Example

bicycle tyre markings give recommended pressures 3.0-5.0 bar 45-70 psi 1. What would be the recommended pressure in pascals? 1 bar = 100,000 Pa Thus recommended pressure is 3.0 x 10⁵ – 5.0 x 10⁵ Pa Show Answer 2. Would you expect the pressure of a car tyre to be higher or lower? I would expect the pressure of a car tyre to be lower than this. A higher pressure in the cycle tyre produced a very firm tyre which doesn’t deform much and so gives a very small contact patch in touch with the ground. This in turn keeps friction low and makes it easier to cycle. A car needs a larger area in contact with the ground due to the higher forces (torques) at the wheels. Typically a car tyre pressure will be 2.0 – 2.4 bar Show Answer

 

Manometer

A manometer is a U-tube partially filled with a liquid.

With both ends of the manometer open to the atmosphere there is the same pressure exerted on both sides and thus the level of liquids on both sides will be the same.

 

A rubber tube is connected to the right-hand tube of the manometer  and connected to a gas of a pressure we are trying to measure.This results in the the right side of the liquid being pushed down whilst the left side gets pushed up. The levels are separated by a distance h.

We can deduce that P_gas is a greater pressure than P_atm as the column of mercury has been pushed down more on the right side than the left.

Pressure at point A is caused by the atmospheric pressure above the tube and the height h of liquid above point A.

P_A = P_atm + h

Pressure at point B is just the pressure of the gas we are measuring P_gas.

P_B = P_gas

Looking at the above diagram we must realise that points A and B will be at the same pressure as they are points at the same level in the same body of liquid. Thus,

P_B = P_A

P_gas = P_atm + h

Note the manometer does not tell us the absolute pressure. It tells us the difference in pressures between the two sides of the manometer. The difference in the gas pressure and the atmospheric pressure is sometimes referred to as the excess pressure.

 

Example: Excess Pressure
What is the pressure of the gas P1 if atmospheric pressure is 76 cmHg. P1 = Patm + h P1 = 76 + 30 P1 = 106 cm Hg Show Answer

 

Sometimes the manometer will look like this:
Here we can conclude that P_atm is less than P_gas and easily see that:

P_gas = P_atm – h

 

 

Example 9
A manometer is used to measure the pressure of a gas supply as shown in the diagram.  Determine the excess pressure of the gas supply.  Excess Pressure = 16.0 cm – 6.0 cm = 10.0 cm oil (above atmospheric pressure) Show Answer

Example 10

Two immiscible liquids A and B of different densities are in equilibrium in an open U-tube of uniform cross-section as shown below.  (a) On the diagram, clearly mark with a cross (without calculation) and label two points, P and Q, one in each side of the U-tube, where the pressure is equal to atmospheric pressure.  (b) If liquid B is water, calculate the density of liquid A. (Density of water = 1.0 g cm–3 .)  Consider point R & S with equal pressures. (They are both in the body of liquid (liquid B) and are at the same vertical height.) PR = PS Using hρg, (11.1 cm –6.0 cm) x ρA g + Patm = (9.8 cm – 6.0 cm) x 1.0 g cm-3 x g + Patm ρA = 0.75 g cm-3 Show Answer

 

 

Example

The diagram below shows a manometer that is used to measure the pressure in a gas tank. (a) Given that atmospheric pressure is 76.0 cmHg, determine the pressure of the gas in the tank. Ptank = Patm + h Ptank = 76.0 + (64.1 – 13.4) Ptank = 76.0 + 50.7 Ptank = 126.7 cm Hg Show Answer (b) The pressure in the gas tank is increased such that the reading of the left-hand column of the manometer changes to 11.4 cm. State the new reading on the right-hand side of the column, and the new pressure in the gas pipe in cmHg. The mercury level on the left-hand side drops by (13.4 – 11.4) = 2.0 cm As mercury is incompressible and the volume of mercury is fixed, the right-hand side of the mercury level must  rise by 2.0 cm So the new reading on the right-hand side is now (64.1 + 2.0) 66.1 cm. Difference between the two side is now 66.1 – 11.4 = 54.7 cm The new pressure in the tank is given by: Pnew = Patm + h Pnew = 76.0 + 54.7 Pnew = 130.7 cm Hg Show Answer

 

Two Liquid Problem

If we put two immiscible liquids in a U-tube then we may have something like this.

Note that although both sides of the U-tube are at atmospheric pressure the levels of the two liquids are at different heights. This happens because the two liquids have differing densities.

The two points P₁ and P₂ are at the same level in the same liquid (both can be considered to be in liquid 1). They must therefore be at the same pressure.

The pressure on each tube can be considered to be resulting from the weight of the liquid above it (as well as the outside atmospheric pressure).

 

Pressure at P₁ given by:

P₁ = P_atm + h₁ρ₁g

Pressure at P₂ given by:

P₂ = P_atm + h₂ρ₂g

 

Thus

P₁ = P₂

P_atm + h₁ρ₁g = P_atm + h₂ρ₂g

h₁ρ₁ = h₂ρ₂

 

Note we are using pascals here as the unit (i.e. not using cm Hg) because there are two liquids – thus it is not possible to express everything in terms of the height of liquid.

Example

Oil and water are on either sides of a manometer tube as shown. Both ends of the manometer tube are open to the atmosphere. The density of water is 1.00 g/cm³ and the density of oil is 0.78 g/cm³. What if the height h of the oil column?     density of water = 1.00 g/cm³ = 1000 kg/m³ density of oil = 0.78 g/cm³ = 780 kg/m³   On the left-hand column the pressure under 12.0 cm of water is given by: Pleft = Patm + hwater ρwater g Pleft = Patm + 0.12 × 1000 × g On the right-hand column the pressure under h cm of water is given by: Pright = Patm + hwater ρwater g Pright = Patm + h × 780 × g   As the pressure at these points is the same: Pleft = Pright Patm + 0.12 × 1000 × g = Patm + h × 780 × g 0.12 × 1000 = h × 780 h = 0.154 m So the height of oil column needed to balance the 12.0 cm of water is 15.4 cm.   Show Answer

 

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