Gases Chapter 5 Measurements on Gases Properties of

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Gases Chapter 5

Gases Chapter 5

Measurements on Gases • Properties of gases – – Gases uniformly fill any container

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

Measurements on Gases • Measuring pressure – The barometer – measures atmospheric pressure •

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

Measurements on Gases – The manometer – measures confined gas pressure

Measurements on Gases – The manometer – measures confined gas pressure

Measurements on Gases – Units • mm Hg (torr) – • Kilopascal (k. Pa)

Measurements on Gases – Units • mm Hg (torr) – • Kilopascal (k. Pa) – • 101. 325 k. Pa = standard pressure Atmospheres – • 760 torr = standard pressure 1 atmosphere (atm) = standard pressure STP = 1 atm = 760 torr = 760 mm. Hg = 101. 325 k. Pa

Measurements on Gases • Example: Convert 0. 985 atm to torr and to k.

Measurements on Gases • Example: Convert 0. 985 atm to torr and to k. Pa. 0. 985 atm 760 torr 1 atm 0. 985 atm 101. 325 k. Pa 1 atm = 749 torr = 98. 3 k. Pa

The Gas Laws of Boyle, Charles and Avogadro • Boyle’s Law (Robert Boyle, 16271691)

The Gas Laws of Boyle, Charles and Avogadro • Boyle’s Law (Robert Boyle, 16271691) – The product of pressure times volume is a constant, provided the temperature remains the same • PV = k

The Gas Laws of Boyle, Charles and Avogadro • • • P is inversely

The Gas Laws of Boyle, Charles and Avogadro • • • 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

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

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 = ? (1. 3 atm)(27 L) = (3. 9 atm)V 2 = 9. 0 L

The Gas Laws of Boyle, Charles and Avogadro • Charles’ Law (Jacques Charles, 1746

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 = b. T V=b T or V 1 = V 2 T 1 T 2 V 1 = b = V 2 T 1 T 2

The Gas Laws of Boyle, Charles and Avogadro • Temperature must be measured in

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

The Gas Laws of Boyle, Charles and Avogadro • Example: A gas at 30°C

The Gas Laws of Boyle, Charles and Avogadro • Example: A gas at 30°C and 1. 00 atm has a volume of 0. 842 L. What volume will the gas occupy at 60°C and 1. 00 atm? • V 1 = 0. 842 L • T 1 = 30°C (+273 = 303 K) • V 2 = ? • T 2 = 60°C (+273 = 333 K)

0. 842 L 303 K = V 2 = 0. 925 L V 2

0. 842 L 303 K = V 2 = 0. 925 L V 2 333 K

The Gas Laws of Boyle, Charles and Avogadro • Avogadro’s Law (Amedeo Avogadro, 1811)

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 1 = a = V 2 n 1 n 2 V/n = a or V 1 = V 2 n 1 n 2

The Gas Laws of Boyle, Charles and Avogadro • Example: A 5. 20 L

The Gas Laws of Boyle, Charles and Avogadro • Example: A 5. 20 L sample at 18°C and 2. 00 atm contains 0. 436 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. 20 L • n 1 = 0. 436 mol • V 2 = ? • n 2 = 1. 27 mol + 0. 436 mol = 1. 706 mol

The Gas Laws of Boyle, Charles and Avogadro 5. 20 L = V 2

The Gas Laws of Boyle, Charles and Avogadro 5. 20 L = V 2 0. 436 mol 1. 706 mol x = 20. 3 L

The Ideal Gas Law • Derivation from existing laws V=k V=b. T V=an P

The Ideal Gas Law • Derivation from existing laws V=k V=b. T V=an P V = kba(Tn) P

The Ideal Gas Law Constants k, b, a are combined into the universal gas

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

The Ideal Gas Law R = PV using standard numbers will give you R

The Ideal Gas Law R = PV using standard numbers will give you R n. T STP • P = 1 atm • V = 22. 4 L • n = 1 mol • T = 273 K • Therefore if we solve for R, R = 0. 0821 L • atm/mol • K

The Ideal Gas Law • Example: A sample containing 0. 614 moles of a

The Ideal Gas Law • Example: A sample containing 0. 614 moles of a gas at 12°C occupies a volume of 12. 9 L. What pressure does the gas exert? P=? V = 12. 9 L n = 0. 614 mol R = 0. 0821 L • atm/mol • K T = 12 + 273 = 285 K

The Ideal Gas Law (P)(12. 9 L) = (0. 614 mol)(0. 0821 L •

The Ideal Gas Law (P)(12. 9 L) = (0. 614 mol)(0. 0821 L • atm/mol • K)(285 K) 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)

The Ideal Gas Law Solving for new volumes, temperature, or pressure (n remaining constant) • Combined Law P 1 V 1 = n. R = P 2 V 2 T 1 T 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 0. 848

The Ideal Gas Law • Example: A sample of methane gas at 0. 848 atm and 4. 0°C occupies a volume of 7. 0 L. 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 = 0. 848 atm P 2 = 1. 52 atm • V 1 = 7. 0 L V 2 = ? • T 1 = 4+273 = 277 K T 2 = 11+273 = 284 K

The Ideal Gas Law (0. 848 atm)(7. 0 L) 277 K V 2 =

The Ideal Gas Law (0. 848 atm)(7. 0 L) 277 K V 2 = 4. 0 L = (1. 52 atm)(V 2) 284 K

 • Continue next time…

• Continue next time…