1. Chapter 1. Part 2 Ideal Gases Section1.5, 1.6, 1.7, 1.8

2. Pressure Standard Atmosphere the pressure exerted by a column of mercury 760 mm high at 0 ºC SI Unit 1 atm = 101325 Pa = 760 Torr 1 Torr  1 mmHg 1 Pa = 1 Kg m -1 s -2 = 1 N m -2 1 Bar = 100 kPa =0.986923 atm Barometer 氣壓計

3. Manometer 壓力計 P =  g h P pressure  density g gravity h height

4. Example 1 例題 Compare the length of a column of mercury to that of a column of water required to produce a pressure of 1.000 bar. The densities of mercury and water at 0.00 ºC are 13.596 g cm -3 and 0.99987 g cm -3, respectively. Answer Formula P =   g  g h Hg =  water  g h water Unit Conversion 1 bar = 0.986923 atm; 1 atm = 760 mmHg 1 bar = 750.06 mm Hg 13.596 g /cm 3  g  750.06 mm = 0.99987 g/cm 3  g  h w h w = 10199 mm

5. Boyle’s Law At constant temperature, the pressure of a fixed amount of gas is inversely proportitional to its volume 1/P  V; P=constant / V PV = constant (n and T are fixed)

6. Gay-Lussac’s Law At constant pressure, the volume of a fixed amount of gas varies linearly with the temperature V  T V = constant  T PV = constant (n and P are fixed) Kelvin Temperature (K)

7. Avogadro’s Principle At a given temperature and pressure, equal volumes of gas contain the same number of molucules V  n Mole ( 莫耳 ) 任何物質與 0.012kg C-12 擁有相同的數量 1 mole = 6.022137  10 23 Molar volume V m = V / n The molar volume of a gas is almost the same for all gases at the same temperature and pressure

8. Ideal Gases Boyle’s Law : P  1/V Charle’s Law : V = A + B 

9. Equation of State for an Ideal Gas Pressure of the sample= f(amount, volume, temperature) PV = nRT P : pressure of the sample V : volume of the sample occupies n :amount of substances in the sample T : temperature of the sample Boyle’s Law : P  1/V Charle’s Law :V = A + B  Avogadro’s Principle :V  n http://www.phy.ntnu.edu.tw/demolab/java/idealGas/

10. Gas Constant R R=8.31451 J / K mol =8.31451 kPaL/K mol =1.98722 cal/ k mol =8.20578×10 -2 L atm /K mol =62.364 L Torr/K mol

11. 其他形式的理想氣體狀態方程式 namount of substance M molar mass of the substance  density of the gas (=m/M)

12. Example 2 例題 Calculate the average molar mass of air at sea level and 0 °C if the density of air is 1.29 kg m -3. Amswer