Page 13 - Spotlight A+ Physics Form 4.5

Form
4 Physics Chapter 4 Heat

Conclusion:
The smaller the volume of a gas, the higher the pressure of the gas. The hypothesis is accepted.
Discussion:
1. A syringe with a large volume is used to obtain a higher change of volume and pressure of the air
trapped in it.
2. The piston is pushed slowly into the syringe so that the temperature of the air in the syringe is kept
constant.

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4. The rate of collision between the molecules and
the wall of the container increases.

Robert Boyle (1627-1691) was a scientist who 5. The force per unit area on the surface of the
emphasized on the scientific method when wall increases. Thus, the gas pressure increases.
conducting investigations. From experimental data,
he made the conclusion that the volume of a gas is
inversely proportional to the pressure of the gas.
For Boyle's law, temperature is constant.
1. Boyle’s law states that pressure is inversely P × V Constant
proportional to the volume of a fixed mass of
gas at constant temperature.
T
1
P ∝
V
1
P = k 1 2 PV = constant
P V = P V
V
1 1
2 2
where k is a constant
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P = pressure of the gas (Pa)
3
V = volume of the gas (m )
CHAP Eduweb TV: Boyle's law
4 Therefore, PV = k http://bit.ly/2tb1y2I
Try question 1 in Formative Zone 4.4
2. Suppose a gas experiences a change of pressure
and volume from an initial state to a final state. Example 6
Initial state of the gas, P V = k
1 1
Final state of the gas, P V = k The air in a closed syringe has a volume of
2 2
60 cm and pressure 108 kPa. The piston of
3
Therefore, P V = P V the syringe was pushed to compress the air to
1 1 2 2
a volume of 48 cm . Find the pressure of the
3
compressed air.
Solution
Volume is P = 108 kPa, P = pressure of compressed air
1
2
decreased V = 60 cm V = 48 cm 3
3
1 2
The temperature of gas is constant,
Boyle’s law is used:
P V = P V
1 1 2 2
108 × 60 = P × 48
Figure 4.27 A fixed mass of gas compressed at 2
constant temperature P = 108 × 60
2 48
3. When the volume of the gas is decreased, the
number of molecules per unit volume increases. = 135 kPa
112 4.4.2