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Gases - OCR GatewayGas pressure and volume

Gases and liquids exert pressure on objects and the walls of their containers due to collisions. The greater the force and frequency of these collisions, the greater the pressure.

Part of Physics (Single Science)Matter

Gas pressure and volume

In a where the stays the same the of a gas increases as the decreases

Remember that the of the collisions do not change unless the temperature changes. If the temperature of a gas stays the same, the pressure of the gas increases as the volume of its container decreases. This is because the same number of collides with the walls of the container more frequently as there is less space. However, the particles still collide with the same amount of force.

Boyle's law

J-shaped glass tube, filled with mercury, and a trapped air bubble at the smaller end.

The Irish scientist Robert Boyle investigated the relationship between pressure and volume in the 17th century. He trapped a bubble of air at the sealed end of a J-shaped tube using liquid mercury. He then added more mercury and observed what happened to the volume of the air bubble.

Boyle found that the higher the column of mercury in the left hand side of the tube:

  • the greater the force exerted on the trapped air, the smaller the bubble became as its pressure increased
  • volume is to pressure

For example, if the volume of a container is halved, the pressure is doubled. This can also be observed in a balloon of air. If the balloon is squeezed, it gets smaller.

Using Boyle's law

For a given of gas at a constant temperature:

pressure 脳 volume = constant

This is when:

  • pressure is measured in (Pa)
  • volume is measured in metres cubed (m3)

Example

The pressure in a syringe of air is 3.00 脳 105 Pa. The volume of the gas is 5.00 脳 10-6 m3. Calculate the pressure when the volume increases to 8.00 脳 10-6 m3 at a constant temperature.

pressure 脳 volume = constant

pressure 脳 volume = 3.00 脳 105 脳 5.00 脳 10-6

= 1.50

Rearrange the equation:

\(pressure = \frac{constant}{volume}\)

\(pressure = 1.50/(8.00 \times 10^{-6})\)

\(187,500~Pa~(1.88 \times 10^5~Pa~to~3~significant~figures)\)