# Category Archives: 06 Pressure

## Another P1V1 = P2V2 question

Solution: Option C

## When to use the concept PV = constant and P1V1 = P2V2 to solve?

When **temperature is constant** (for o-level), when a __fixed mass of gas (fixed number of air molecules)__ is __compressed__ in a closed system (e.g. piston), the __volume V decreases__ and __pressure P increases__, and vice versa.

But when you multiply pressure and volume, PV, it is always a constant.

**PV = constant**

Hence we can always equate the PV of the first scenario = to the PV of the second scenario, provided there is no addition or removal of air molecules from the system.

Hence, you have **P1V1 = P2V2**

The followings are 4 different questions which require this concept to solve. Do revise them.

**Solutions: Option D** (refer to the worked solutions below)

**Solutions: Option D**

**Solutions: A**

4)

**Solutions:**

5)

**Solution: Option C**

## Fluid pressure in a capillary tube

A uniform capillary tube closed at one end, contained air trapped by a thread of mercury 85 mm long. When the tube was held horizontal, the length of the air column was 50 mm. When it was held vertically with the closed end downwards, the length was 45 mm. Find the atmospheric pressure in Pa. (Density of mercury = 14 x 10^{3 }kg/m^{3 })

**Solutions:**

## Manometer – mercury levels difference changes with different density of liquid used

The pressure of a gas is measured using a manometer as shown in the diagram.

The mercury in the manometer is replaced with a liquid which is less dense. How does the value of h change?

**A** It becomes zero.

**B** It decreases, but not to zero.

**C** It stays the same.

**D** It increases.

**Solutions: Option D**

The pressure to be measured remains constant. Since P = pgh, where p is the density of the liquid used in the manometer. If a liquid of lower density is used, height h of the liquid (level difference) will be greater. The gravitational field strength g remains constant.