1 / 8

The Gas Laws

The Gas Laws. The density of a gas decreases as its temperature increases. Ideal Gas Law. PV = nRT. Brings together gas properties. Can be derived from experiment and theory . Ideal Gas Law.

janae
Download Presentation

The Gas Laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Gas Laws The density of a gas decreases as its temperature increases.

  2. Ideal Gas Law PV = nRT Brings together gas properties. Can be derived from experiment and theory.

  3. Ideal Gas Law At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: • 1. Temperature (expressed in K) • 2. Volume (expressed in liters) • 3. Amount (expressed in moles) • 4. Pressure (given in atmospheres) These variables are not independent — if the values of any three of these quantities are known, the fourth can be calculated.

  4. Ideal Gas Equation Universal Gas Constant Volume P V = 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

  5. PV = nRT Standard Temperature and Pressure (STP) T = 0 oC or 273 K P = 1 atm = 101.3 kPa = 760 mm Hg P = pressure V = volume T = temperature (Kelvin) n = number of moles R = gas constant 1 mol = 22.4 L @ STP Solve for constant (R) PV nT Recall: 1 atm = 101.3 kPa Substitute values: (1 atm) (22.4 L) (1 mole)(273 K) (101.3 kPa) R = 0.0821 atm L mol K = R = 8.31 kPa L mol K ( 1 atm) R = 0.0821 atm L / mol K or R = 8.31 kPa L / mol K

  6. 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?

  7. 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 gram; recall iodine is diatomic (I2) x mol I2 = 500 g I2(1mol 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 = 0.9737 atm R = 0.0821 atm . L / mol . K

  8. 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

More Related