8.3 Particle model and pressure — Practise

Work through these on your own. Keep the units with your numbers. Some questions ask for an explanation rather than a number, so use the particle model in your answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A fixed mass of gas at 100 kPa fills 0.50 m^3. It is compressed at constant temperature to 0.25 m^3. Calculate the new pressure.
A fixed mass of gas at 100 kPa fills 0.90 m^3. It expands at constant temperature to 150 kPa. Calculate the new volume.
A gas exerts a pressure of 120 kPa on a plate of area 0.025 m^2. Calculate the force on the plate.
Explain, using the particle model, why a gas exerts a pressure on the walls of its container.
A gas is sealed in a rigid container and then warmed. The volume does not change. Explain what happens to the pressure of the gas.
A force of 45 N acts at right angles on an area of 0.015 m^2. Calculate the pressure.
Say which relationship you will use, then calculate the pressure when 0.60 m^3 of gas at 100 kPa is compressed to 0.15 m^3 at a constant temperature.
H A student pushes the handle of a bicycle pump. Explain how doing work on the trapped air can raise its temperature.
H A force of 20 N pushes a piston through 0.25 m. Calculate the work done on the gas. Assume all of the work increases the internal energy of the gas.
A fixed mass of gas at 100 kPa fills 2.0 m^3 at a constant temperature. The gas expands until its pressure is 50 kPa. Calculate the new volume.

Answers and concise solutions

1. New pressure of the gas

Answer: $p_1V_1 = p_2V_2$ , so $p_2 = \dfrac{100 \times 0.50}{0.25} = 200\,\text{kPa}$ .

2. New volume of the gas

Answer: $V_2 = \dfrac{p_1V_1}{p_2} = \dfrac{100 \times 0.90}{150} = 0.60\,\text{m}^3$ .

3. Force on the plate

Answer: $F = pA = 120\,000 \times 0.025 = 3000\,\text{N}$ .

4. Why a gas exerts a pressure

The particles of the gas are in constant random motion and move quickly.
When a particle hits a wall it changes direction, so the wall feels a force. There are very many collisions each second.
The total force on each unit of area is the pressure of the gas.

5. Pressure when the gas is warmed

The particles gain kinetic energy and move faster.
They hit the walls more often, and each collision is harder.
The pressure of the gas therefore increases.

6. Pressure on the area

Answer: $p = \dfrac{F}{A} = \dfrac{45}{0.015} = 3000\,\text{Pa}$ .

7. New pressure after compression

Answer: Use $p_1V_1 = p_2V_2$ : $p_2 = \dfrac{100 \times 0.60}{0.15} = 400\,\text{kPa}$ .

8. Work done on a gas

Pushing the handle does work on the trapped air, because a force moves through a distance.
The work done is transferred to the internal energy of the air.
The particles gain kinetic energy, so the temperature of the air rises.

9. Work done on the gas

Answer: $W = Fs = 20 \times 0.25 = 5.0\,\text{J}$ . This becomes 5.0 J of internal energy in the gas.

10. New volume of the gas

Answer: $V_2 = \dfrac{p_1V_1}{p_2} = \dfrac{100 \times 2.0}{50} = 4.0\,\text{m}^3$ . The pressure halves, so the volume doubles.