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February 5, 2008

February 5, 2008. Go over Charles’s Law and Avogadro’s Law Homework Introduce Combined Gas Law Introduce Ideal Gas Law Work Sample Problems HOMEWORK: Pg. 480 -- #21, 23ac, 24ac, 25, 28, 29. Equation of State of an Ideal Gas. Robert Boyle ( 1662 ) found that at fixed temperature

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February 5, 2008

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  1. February 5, 2008 • Go over Charles’s Law and Avogadro’s Law Homework • Introduce Combined Gas Law • Introduce Ideal Gas Law • Work Sample Problems • HOMEWORK: Pg. 480 -- #21, 23ac, 24ac, 25, 28, 29

  2. Equation of State of an Ideal Gas • Robert Boyle (1662) found that at fixed temperature • Pressure and volume of a gas is inversely proportional PV = constant Boyle’s Law • J. Charles found that at fixed pressure • Volume of gas is proportional to change in temperature He Volume CH4 H2O H2 -273.15 oC Temp All gases extrapolate to zero volume at a temperature corresponding to –273.15 oC (absolute zero).

  3. Kelvin Temperature Scale • Kelvin temperature (K) is given by K = oC + 273.15 where K is the temperature in Kelvins, oC is temperature in Celsius • Using the ABSOLUTE scale, it is now possible to write Charles’ Law as V / T = constant Charles’ Law • Combining Boyle’s law, Charles’ law, and another law called Gay-Lussac’s Law (relating pressure and temperature) we can mathematically prove that P V / T = constant Charles

  4. Combined Gas Law • This brings us to the combined gas law:

  5. Practice Problem • A 1.50 L sample of neon gas at 1.10 atm and 25 °C is heated to 45 °C. The neon gas is the subjected to a pressure of 1.50 atm. Determine the new volume of the neon gas. P1 = 1.10 atm V1 = 1.50 L T1 = 25 °C = 298 K P2 = 1.50 atm V1 = ???? L T1 = 45 °C = 318 K V2 = 1.17 L

  6. The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 cm3. What volume will it occupy at 0 oC and 93.3 kPa? (101.3 kPa) x (500 cm3) = (93.3 kPa) x (V2) 273 K 273 K P1 = 101.3 kPa T1 = 273 K V1 = 500 cm3 P2 = 93.3 kPa T2 = 0 oC + 273 = 273 K V2 = X cm3 V2 = 542.9 cm3

  7. Ideal vs. Real Gases No gas is ideal. Most gases behave ideally (almost) at pressures of approximately 1 atm or lower, when the temperature is approximately 0 °C or higher. When we do calculations, we will assume our gases are behaving as ideal gases

  8. Ideal Gas Equation Universal Gas Constant Volume PV = nRT Pressure Temperature No. of moles R = 0.0821 atm L / mol K R = 8.314 kPa L / mol K Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366

  9. nRT V = P (500 g)(0.0821 atm . L / mol . K)(300oC) = V 740 mm Hg Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine T = 300oC P = 740 mm Hg R = 0.0821 atm . L / mol . K Step 2) Equation: PV = nRT Step 3) Solve for variable Step 4) Substitute in numbers and solve V = What MISTAKES did we make in this problem?

  10. What mistakes did we make in this problem? What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine Convert mass to moles; recall iodine is diatomic (I2) 500 g I(1 mole I2/254 g I2) n = 1.9685 mol I2 T = 300oC Temperature must be converted to Kelvin T = 300oC + 273 T = 573 K P = 740 mm Hg Pressure needs to have same unit as R; therefore, convert pressure from mm Hg to atm. x atm = 740 mm Hg (1 atm / 760 mm Hg) P = 0.8 atm R = 0.0821 atm . L / mol . K

  11. nRT V = P (1.9685 mol)(0.0821 atm . L / mol . K)(573 K) = V 0.9737 atm Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine n = 1.9685 mol I2 T =573 K (300oC) P =0.9737 atm(740 mm Hg) R = 0.0821 atm . L / mol . K V = ? L Step 2) Equation: PV = nRT Step 3) Solve for variable Step 4) Substitute in numbers and solve V = 95.1 L I2

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