7.2 Ideal gases and molecular kinetic theory — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Keep the temperature in kelvin, and keep the volume in cubic metres when you use the ideal gas equation.

Warm-up: remember

A molecule of mass $4.8 \times 10^{-26}\,\text{kg}$ moves at 500 m s⁻¹. Find its kinetic energy.
The same molecule strikes a wall at 500 m s⁻¹ and rebounds at 500 m s⁻¹. Find the change in its momentum.
Write 200 cm³ in cubic metres.

Answers

$E_k = \tfrac{1}{2}mv^2 = 0.5 \times 4.8 \times 10^{-26} \times 500^2 = 6.0 \times 10^{-21}\,\text{J}$ .
The velocity reverses, so $\Delta p = 2mv = 2 \times 4.8 \times 10^{-26} \times 500 = 4.8 \times 10^{-23}\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .
$200\,\text{cm}^3 = 200 \times 10^{-6}\,\text{m}^3 = 2.0 \times 10^{-4}\,\text{m}^3$ .

If question 1 goes wrong, remember that kinetic energy is $\tfrac{1}{2}mv^2$ . If question 2 goes wrong, the momentum changes direction, so the change is twice the momentum. If question 3 goes wrong, one metre is 100 cm, so one cubic metre is $10^{6}\,\text{cm}^{3}$ .

Practise with help

1. A gas occupies $4.0 \times 10^{-3}\,\text{m}^{3}$ at a pressure of $1.0 \times 10^{5}\,\text{Pa}$ . It is compressed at constant temperature to $1.0 \times 10^{-3}\,\text{m}^{3}$ . Find the new pressure.

Hint

The temperature is constant, so this is Boyle's law.

Next step

Use $p_1V_1 = p_2V_2$ and make $p_2$ the subject: $p_2 = \dfrac{p_1V_1}{V_2}$ . The volume falls to one quarter, so predict a pressure four times larger before you substitute.

Full working

\begin{aligned} p_2 &= \frac{p_1V_1}{V_2} \\ &= \frac{1.0 \times 10^{5} \times 4.0 \times 10^{-3}}{1.0 \times 10^{-3}} \\ &= 4.0 \times 10^{5}\,\text{Pa} \end{aligned}

The new pressure is $4.0 \times 10^{5}\,\text{Pa}$ .

2. A fixed mass of gas at 290 K occupies $2.0 \times 10^{-3}\,\text{m}^{3}$ . At constant pressure it is warmed to 348 K. Find its new volume.

Hint

The pressure is constant, so this is Charles's law. The temperature must be in kelvin, and it already is.

Next step

Use $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ and make $V_2$ the subject: $V_2 = \dfrac{V_1T_2}{T_1}$ . The temperature rises by a factor of $\dfrac{348}{290}$ , so the volume rises by the same factor.

Full working

\begin{aligned} V_2 &= \frac{V_1T_2}{T_1} \\ &= \frac{2.0 \times 10^{-3} \times 348}{290} \\ &= 2.4 \times 10^{-3}\,\text{m}^{3} \end{aligned}

The new volume is $2.4 \times 10^{-3}\,\text{m}^{3}$ .

3. Find the number of moles in $0.020\,\text{m}^{3}$ of gas at $1.0 \times 10^{5}\,\text{Pa}$ and 300 K. Use $R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$ .

Hint

Use the ideal gas equation $pV = nRT$ . The pressure, volume and temperature are already in SI units.

Next step

Make $n$ the subject: $n = \dfrac{pV}{RT}$ . Work out the top line and the bottom line separately before you divide.

Full working

\begin{aligned} n &= \frac{pV}{RT} \\ &= \frac{1.0 \times 10^{5} \times 0.020}{8.31 \times 300} \\ &= 0.802\,\text{mol} \end{aligned}

The gas contains 0.802 mol.

4. Find the number of molecules in the gas in the previous problem. Use $n = 0.802\,\text{mol}$ and $N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}$ .

Hint

One mole contains the Avogadro constant, so multiply the number of moles by $N_A$ .

Next step

Use $N = nN_A = 0.802 \times 6.02 \times 10^{23}$ .

Full working

N = nN_A = 0.802 \times 6.02 \times 10^{23} = 4.83 \times 10^{23}

The gas contains $4.83 \times 10^{23}$ molecules.

5. Find the average random kinetic energy of a gas molecule at 300 K. Use $k = 1.38 \times 10^{-23}\,\text{J}\,\text{K}^{-1}$ .

Hint

The average molecular kinetic energy is $\dfrac{3}{2}kT$ . The temperature is already in kelvin.

Next step

Substitute $k$ and $T$ into $\dfrac{3}{2}kT$ .

Full working

\tfrac{1}{2}m(c_{\text{rms}})^2 = \tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\,\text{J}

The average molecular kinetic energy is $6.21 \times 10^{-21}\,\text{J}$ .

6. Find the rms speed of nitrogen molecules at 300 K. The molar mass of nitrogen is $0.028\,\text{kg}\,\text{mol}^{-1}$ and $R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$ .

Hint

Use $c_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$ . Keep the molar mass in kilograms per mole.

Next step

Work out the top line: $3 \times 8.31 \times 300 = 7479$ . Then divide by 0.028 and take the square root.

Full working

\begin{aligned} c_{\text{rms}} &= \sqrt{\frac{3RT}{M}} \\ &= \sqrt{\frac{3 \times 8.31 \times 300}{0.028}} \\ &= 517\,\text{m}\,\text{s}^{-1} \end{aligned}

The rms speed is 517 m s⁻¹.

7. Explain, in terms of the molecules, how a gas exerts a pressure on the walls of its container.

Hint

Think about what happens to a molecule's momentum when it hits a wall and bounces back, and about how many molecules hit the wall each second.

Next step

Say that the molecules move at random and hit the wall. Then link the change in momentum to the force on the wall.

Full working

The molecules move quickly and at random. They collide with the walls of the container. When a molecule hits a wall it rebounds, so its momentum changes. The change in momentum means the wall feels a force from that molecule. There are very many collisions every second, so the wall feels a steady force. Pressure is force divided by area, so the gas exerts a steady pressure on the walls.