8.3 Particle model and pressure — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Section 8.2 and Section 2.4.

Calculate the energy needed to warm $0.50\,\text{kg}$ of water by $40\,^\circ\text{C}$ . The specific heat capacity of water is $4200\,\text{J/kg}\,^\circ\text{C}$ .
A force of $60\,\text{N}$ acts on an area of $0.020\,\text{m}^2$ . Calculate the pressure.

Answers

$\Delta E = mc\Delta\theta = 0.50 \times 4200 \times 40 = 84\,000\,\text{J}$ .
$p = \dfrac{F}{A} = \dfrac{60}{0.020} = 3000\,\text{Pa}$ .

If question 1 went wrong, look again at Section 8.2: the temperature changes, so use specific heat capacity. If question 2 went wrong, remember that pressure is the force on each unit of area.

Practise with help

1. A fixed mass of gas at $100\,\text{kPa}$ fills $1.2\,\text{m}^3$ . The gas is compressed at constant temperature until its volume is $0.40\,\text{m}^3$ . Calculate the new pressure.

Hint

The mass of gas and the temperature do not change, so you can use $p_1V_1 = p_2V_2$ .

Next step

Make $p_2$ the subject: $p_2 = \dfrac{p_1V_1}{V_2}$ . Put in the values and divide.

Full working

\begin{aligned} p_1V_1 &= p_2V_2 \\ p_2 &= \frac{p_1V_1}{V_2} \\ &= \frac{100 \times 1.2}{0.40} \\ &= 300\,\text{kPa} \end{aligned}

2. A fixed mass of gas at $100\,\text{kPa}$ fills $0.30\,\text{m}^3$ . It expands at constant temperature until its pressure is $120\,\text{kPa}$ . Calculate the new volume.

Hint

This time the unknown is the volume. Rearrange $p_1V_1 = p_2V_2$ for $V_2$ .

Next step

Write $V_2 = \dfrac{p_1V_1}{p_2}$ , then divide $100 \times 0.30$ by $120$ .

Full working

\begin{aligned} p_1V_1 &= p_2V_2 \\ V_2 &= \frac{p_1V_1}{p_2} \\ &= \frac{100 \times 0.30}{120} \\ &= 0.25\,\text{m}^3 \end{aligned}

3. A gas exerts a pressure of $250\,\text{kPa}$ on a flat plate of area $0.040\,\text{m}^2$ . Calculate the force on the plate.

Hint

Use $p = \dfrac{F}{A}$ and make the force the subject. Change the pressure into pascals first.

Next step

Write $F = pA$ with $p = 250\,000\,\text{Pa}$ and $A = 0.040\,\text{m}^2$ .

Full working

\begin{aligned} F &= pA \\ &= 250\,000 \times 0.040 \\ &= 10\,000\,\text{N} \\ &= 1.0 \times 10^4\,\text{N} \end{aligned}

4. Explain, using the particle model, why the pressure of a gas rises when the gas is compressed at a constant temperature.

Hint

The particles now have less space. Think about how often each particle hits the walls.

Next step

Say that the particles are closer together, then link that to the number of collisions with the walls each second.

Full working

When the gas is compressed, the same number of particles is squeezed into a smaller volume. The particles are closer to the walls and to each other, so each particle hits the walls more often. The average kinetic energy of the particles does not change, because the temperature is constant. The walls receive more collisions each second, so the pressure rises.

This answer is a model answer for you to compare with your own wording.

5. A sealed metal container holding a gas is warmed. The volume of the container does not change. Explain what happens to the speed of the particles and to the pressure of the gas.

Hint

Think about the average kinetic energy of the particles when the gas gets hotter.

Next step

Say what happens to the particle speed, then say what that does to the number and the strength of the collisions with the walls.

Full working

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

This answer is a model answer for you to compare with your own wording.

6. Say which relationship you will use, then calculate the pressure when $0.80\,\text{m}^3$ of gas at $100\,\text{kPa}$ is compressed to $0.20\,\text{m}^3$ at a constant temperature.

Hint

The gas is the same fixed mass, and the temperature is constant. That tells you which relationship applies.

Next step

Use $p_1V_1 = p_2V_2$ with $p_1 = 100\,\text{kPa}$ , $V_1 = 0.80\,\text{m}^3$ and $V_2 = 0.20\,\text{m}^3$ .

Full working

\begin{aligned} p_1V_1 &= p_2V_2 \\ p_2 &= \frac{p_1V_1}{V_2} \\ &= \frac{100 \times 0.80}{0.20} \\ &= 400\,\text{kPa} \end{aligned}

The volume is divided by four, so the pressure is multiplied by four.

7. A student pushes the handle of a bicycle pump quickly. Explain why the air inside the pump becomes warmer.

Hint

Pushing the handle moves a force through a distance. That is work.

Next step

Say that work is done on the gas, then say what happens to the internal energy of the gas and to the particles.

Full working

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

This answer is a model answer for you to compare with your own wording.