Gases Chapter 5

Measurements on Gases Properties of gases –Gases uniformly fill any container –Gases are easily compressed –Gases mix completely with any other gas –Gases exert pressure on their surroundings Pressure = Force Area

Measuring pressure –The barometer – measures atmospheric pressure Inventor – Evangelista Torricelli (1643) Measurements on Gases

–The manometer – measures confined gas pressure Measurements on Gases

–Units mm Hg (torr) –760 torr = standard pressure Kilopascal (kPa) – kPa = standard pressure Atmospheres –1 atmosphere (atm) = standard pressure STP = 1 atm = 760 torr = 760 mmHg = kPa Measurements on Gases

Example: Convert atm to torr and to kPa. Measurements on Gases atm760 torr 1 atm = 749 torr atm kPa 1 atm = 98.3 kPa

The Gas Laws of Boyle, Charles and Avogadro Boyle’s Law (Robert Boyle, ) –The product of pressure times volume is a constant, provided the temperature remains the same PV = k

P is inversely related to V The graph of P versus V is hyperbolic Volume increases linearly as the pressure decreases –As you squeeze a zip lock bag filled with air (reducing the volume), the pressure increases making it difficult to keep squeezing The Gas Laws of Boyle, Charles and Avogadro

–At constant temperature, Boyle’s law can be used to find a new volume or a new pressure P 1 V 1 = k = P 2 V 2 or P 1 V 1 = P 2 V 2 –Boyle’s law works best at low pressures –Gases that obey Boyle’s law are called Ideal gases The Gas Laws of Boyle, Charles and Avogadro

Example: A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? P 1 = 1.3 atm V 1 = 27 L P 2 = 3.9 atm V 2 = ? The Gas Laws of Boyle, Charles and Avogadro (1.3atm)(27L) = (3.9atm)V 2 V 2 = 9.0 L

The Gas Laws of Boyle, Charles and Avogadro Charles’ Law (Jacques Charles, 1746 – 1823) –The volume of a gas increase linearly with temperature provided the pressure remains constant V = bT V = b V 1 = b = V 2 T T 1 T 2 or V 1 = V 2 T 1 T 2

The Gas Laws of Boyle, Charles and Avogadro Temperature must be measured in Kelvin ( K = °C + 273) –0 K is “absolute zero”

Example: A gas at 30°C and 1.00 atm has a volume of 0.842L. What volume will the gas occupy at 60°C and 1.00 atm? V 1 = 0.842L T 1 = 30°C (+273 = 303K) V 2 = ? T 2 = 60°C (+273 = 333K) The Gas Laws of Boyle, Charles and Avogadro

0.842L = V 2 303K 333K V 2 = 0.925L

The Gas Laws of Boyle, Charles and Avogadro Avogadro’s Law (Amedeo Avogadro, 1811) –For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles, n V = an V/n = a V 1 = a = V 2 or V 1 = V 2 n 1 n 2

The Gas Laws of Boyle, Charles and Avogadro Example: A 5.20L sample at 18°C and 2.00 atm contains moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? V 1 = 5.20L n 1 = mol V 2 = ? n 2 = 1.27 mol mol = mol

The Gas Laws of Boyle, Charles and Avogadro 5.20 L = V mol mol x = 20.3L

The Ideal Gas Law Derivation from existing laws V=k V=bT V=an P V = kba(Tn) P

The Ideal Gas Law Constants k, b, a are combined into the universal gas constant (R), V = nRT or PV = nRT P

The Ideal Gas Law R = PV using standard numbers will give you R nT STP P = 1 atm V = 22.4L n = 1 mol T = 273K Therefore if we solve for R, R = L atm/mol K

The Ideal Gas Law Example: A sample containing moles of a gas at 12°C occupies a volume of 12.9L. What pressure does the gas exert? P = ? V = 12.9L n = mol R = Latm/mol K T = = 285K

The Ideal Gas Law (P)(12.9L) = (0.614mol)( Latm/mol K)(285K) Hint: Rearrange and re-write (watch out for units in numerator and denominator) P =1.11 atm

The Ideal Gas Law Solving for new volumes, temperature, or pressure (n remaining constant) Combined Law P 1 V 1 = nR = P 2 V 2 or P 1 V 1 = P 2 V 2 T 1 T 2

The Ideal Gas Law Example: A sample of methane gas at atm and 4.0°C occupies a volume of 7.0L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11.0°C? P 1 = atmP 2 = 1.52 atm V 1 = 7.0LV 2 = ? T 1 = = 277KT 2 = = 284K

The Ideal Gas Law (0.848atm)(7.0L) = (1.52atm)(V 2 ) 277K284K V 2 = 4.0L

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