7.2 Ideal gases and molecular kinetic theory — Practise

Work through these on your own. Convert every temperature to kelvin and every volume to cubic metres before you use a gas equation. Write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

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

Write these temperatures in kelvin: (a) -23 C; (b) 27 C.
A gas at a pressure of 1.2e5 Pa occupies 3.0e-3 m3. It is compressed at constant temperature to 1.5e-3 m3. Find the new pressure.
A fixed mass of gas occupies 250 cm3 at 290 K. It is heated at constant pressure to 348 K. Find its new volume.
A sealed rigid container holds gas at 1.0e5 Pa and 300 K. The gas is heated to 360 K. Find the new pressure.
Find the number of moles in 0.050 m3 of gas at 1.0e5 Pa and 290 K. R = 8.31 J/K/mol.
Find the number of molecules in the gas in the previous question. Use n = 2.07 mol and N_A = 6.02e23 /mol.
Find the average random kinetic energy of a molecule of a gas at 27 C. k = 1.38e-23 J/K.
Find the rms speed of oxygen molecules at 300 K. The molar mass of oxygen is 0.032 kg/mol and R = 8.31 J/K/mol.
A gas in a container of volume 1.5e-3 m3 has 8.0e22 molecules, each of mass 3.2e-26 kg. The rms speed is 600 m/s. Use pV = (1/3)Nm(c_rms)^2 to find the pressure.
A student measures the pressure p of a fixed mass of gas for several values of 1/V at constant temperature. The points are (1/V in 1e3 m-3, p in 1e5 Pa): (1.0, 1.02), (2.0, 1.96), (3.0, 3.05), (4.0, 3.98). Find the gradient of the graph and say what the gradient represents.
A weather balloon rises, where the temperature is constant but the pressure is lower. Explain, in terms of the molecules, why the balloon expands.
Explain, in terms of the molecules, why heating a gas at constant volume increases its pressure.

Answers and concise solutions

1. The temperatures in kelvin

Answer: (a) $T = -23 + 273 = 250\,\text{K}$ . (b) $T = 27 + 273 = 300\,\text{K}$ .

2. The new pressure

Answer: $p_2 = \dfrac{p_1V_1}{V_2} = \dfrac{1.2 \times 10^{5} \times 3.0 \times 10^{-3}}{1.5 \times 10^{-3}} = 2.4 \times 10^{5}\,\text{Pa}$ .

3. The new volume

Answer: $V_2 = \dfrac{V_1T_2}{T_1} = \dfrac{250 \times 348}{290} = 300\,\text{cm}^{3}$ .

4. The new pressure

Answer: At constant volume $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$ , so $p_2 = \dfrac{1.0 \times 10^{5} \times 360}{300} = 1.2 \times 10^{5}\,\text{Pa}$ .

5. The number of moles

Answer: $n = \dfrac{pV}{RT} = \dfrac{1.0 \times 10^{5} \times 0.050}{8.31 \times 290} = 2.07\,\text{mol}$ .

6. The number of molecules

Answer: $N = nN_A = 2.07 \times 6.02 \times 10^{23} = 1.25 \times 10^{24}$ .

7. The average kinetic energy

Answer: The temperature is $27 + 273 = 300\,\text{K}$ . Then $\tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\,\text{J}$ .

8. The rms speed

Answer: $c_{\text{rms}} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3 \times 8.31 \times 300}{0.032}} = 483\,\text{m}\,\text{s}^{-1}$ .

9. The pressure

Answer: $p = \dfrac{Nm(c_{\text{rms}})^2}{3V} = \dfrac{8.0 \times 10^{22} \times 3.2 \times 10^{-26} \times 600^{2}}{3 \times 1.5 \times 10^{-3}} = 2.0 \times 10^{5}\,\text{Pa}$ .

10. The gradient and its meaning

Answer: The gradient is about 1.0. The graph is a straight line through the origin, so p is proportional to 1/V, which is Boyle's law. The gradient is the constant pV.

11. A model answer

Outside the balloon there are fewer molecules per unit volume, so they hit the outside surface less often.
The molecules inside still hit the inside surface at the same rate, so the force per unit area inside is larger than outside.
The greater pressure inside pushes the surface outwards, so the balloon expands.
At constant temperature this is Boyle's law: as the pressure outside falls, the volume increases so that pV stays constant.

Answer: At lower pressure there are fewer collisions on the outside surface. The collisions inside are unchanged, so the pressure inside is greater and pushes the balloon outwards. At constant temperature, pV is constant, so V increases as p falls.

12. A model answer

Heating raises the average random kinetic energy, so the molecules move faster.
Each molecule hits the wall with more momentum, so the change in momentum is larger and the force from each collision is larger.
The faster molecules also hit the walls more often.
More forceful and more frequent collisions give a larger force per unit area, so the pressure rises.

Answer: Heating makes the molecules move faster. Each collision with the wall changes momentum by more, and the collisions are more frequent, so the force on each unit of area increases. The pressure rises.