A brief note on Temperature, Gases and Thermal Expansion . Read it before a test. Try each question at the end before you open its answer.
The big ideas Heat flows from hot to cold; temperature decides the direction. Bodies in contact reach thermal equilibrium. Scales : T F = 9 5 T C + 32 T_F = \tfrac95 T_C + 32 T F = 5 9 T C + 32 ; T = T C + 273.15 T = T_C + 273.15 T = T C + 273.15 . A change of 1 K 1\ \text{K} 1 K equals one of 1 ∘ C 1\ ^\circ\text{C} 1 ∘ C . Ideal gas : p V = n R T = N k B T pV = nRT = Nk_BT p V = n R T = N k B T ; Boyle and Charles are special cases. Use kelvin. Absolute zero , 0 K = − 273.15 ∘ C 0\ \text{K} = -273.15\ ^\circ\text{C} 0 K = − 273.15 ∘ C , where every gas's pressure line meets zero. Expansion : Δ L = α L Δ T \Delta L = \alpha L\,\Delta T Δ L = α L Δ T ; area 2 α 2\alpha 2 α , volume γ = 3 α \gamma = 3\alpha γ = 3 α . Water is densest at 4 ∘ C 4\ ^\circ\text{C} 4 ∘ C . The facts at a glance Law Formula Boyle (steady T T T ) p 1 V 1 = p 2 V 2 p_1V_1 = p_2V_2 p 1 V 1 = p 2 V 2 Charles (steady p p p ) V 1 T 1 = V 2 T 2 \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} T 1 V 1 = T 2 V 2 Steady V V V p 1 T 1 = p 2 T 2 \dfrac{p_1}{T_1} = \dfrac{p_2}{T_2} T 1 p 1 = T 2 p 2 Ideal gas p V = n R T pV = nRT p V = n R T Linear expansion Δ L = α L Δ T \Delta L = \alpha L\,\Delta T Δ L = α L Δ T Volume expansion Δ V = γ V Δ T \Delta V = \gamma V\,\Delta T Δ V = γ V Δ T
A worked example A weather balloon holds 50 50 50 moles of helium. Find its volume on the ground ( 1.0 × 10 5 Pa 1.0 \times 10^{5}\ \text{Pa} 1.0 × 1 0 5 Pa , 288 K 288\ \text{K} 288 K ) and 10 km 10\ \text{km} 10 km up ( 2.6 × 10 4 Pa 2.6 \times 10^{4}\ \text{Pa} 2.6 × 1 0 4 Pa , 223 K 223\ \text{K} 223 K ).
Ground. V = 50 mol × 8.31 J mol − 1 K − 1 × 288 K 1.0 × 10 5 Pa = 1.20 m 3 V = 50\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times \dfrac{288\ \text{K}}{1.0 \times 10^{5}\ \text{Pa}} = 1.20\ \text{m}^3 V = 50 mol × 8.31 J mol − 1 K − 1 × 1.0 × 1 0 5 Pa 288 K = 1.20 m 3 . High up. V = 50 mol × 8.31 J mol − 1 K − 1 × 223 K 2.6 × 10 4 Pa = 3.56 m 3 V = 50\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times \dfrac{223\ \text{K}}{2.6 \times 10^{4}\ \text{Pa}} = 3.56\ \text{m}^3 V = 50 mol × 8.31 J mol − 1 K − 1 × 2.6 × 1 0 4 Pa 223 K = 3.56 m 3 . So it swells to about three times its size, though the air there is colder. Practice with new numbers
A program to try The program works out the balloon at three heights.
Read it, then run it.
Add 15 km 15\ \text{km} 15 km up: 1.2 × 10 4 Pa 1.2 \times 10^{4}\ \text{Pa} 1.2 × 1 0 4 Pa and 217 K 217\ \text{K} 217 K . How big is it now?
Mistakes to avoid Celsius in the gas laws. Always use kelvin. Gauge pressure in the gas laws. Use absolute pressure. Litres in p V = n R T pV = nRT p V = n R T . Change to cubic metres. 3 α 3\alpha 3 α for a liquid's γ \gamma γ . A liquid has only a volume coefficient. A hole shrinking when a plate is heated. It grows with the plate. Check yourself What is − 40 ∘ C -40\ ^\circ\text{C} − 40 ∘ C in Fahrenheit? Stuck? Show a hint Show the answerA gas at a steady volume goes from 27 ∘ C 27\ ^\circ\text{C} 27 ∘ C to 327 ∘ C 327\ ^\circ\text{C} 327 ∘ C . By what factor does its pressure grow? Stuck? Show a hint Show the answerHow many molecules are in 1.0 mol 1.0\ \text{mol} 1.0 mol of any gas? Stuck? Show a hint Show the answerA steel beam 12 m 12\ \text{m} 12 m long warms by 30 K 30\ \text{K} 30 K . How much does it grow? Stuck? Show a hint Show the answerWhy is a gap left at the joints of a concrete road? Stuck? Show a hint Show the answer