Unit 7 ~ Gases (Chapter 13) And you. 7-1 Introduction (Section 13.1) There are many variables needed to adequately describe the conditions of a gas 1)

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Presentation transcript:

Unit 7 ~ Gases (Chapter 13) And you

7-1 Introduction (Section 13.1) There are many variables needed to adequately describe the conditions of a gas 1) Pressure, 2) Volume, 3) moles (mass), 4) Temperature. Most students are familiar with volume and moles. However, pressure and temperature need additional attention.

Temperature Temperature is measured with a thermometer, and has units of: ºF, ºC, or K. We’ll use degrees Celsius and Kelvin. To convert between the two scales use: Kelvin = Celsius Example:30ºC = ? K = 303 K

Temperature and Volume Negative ºC temperatures are possible, but for Kelvin, all values are positive. What did we do to the straight line in the left graph to obtain the straight line in the right graph? We added 273 Can you guess the x-axis value for the left graph? The right graph’s x-axis has a special name: absolute zero (0 vol. at T = 0 kelvin) L On the atomic level, temperature is proportional to the average kinetic energy of the molecules. As T↑, the average KE↑ (and vice versa).

Pressure Pressure is defined as a force divided by the area of contact: P = F/A. For gases, each particle exerts a force over a small area when it collides with something. Therefore, pressure for a chemist depends on the number of collisions against the side of a container. Two common instruments to measure pressure are the barometer and the manometer.

Barometer Purpose: to measure atmospheric pressure. Construction: an inverted tube with graduations (“eudiometer” contains Hg, which is used because of its density (13.6x that of water). At sea level: the pressure of the gases in the atmosphere will push down on the Hg in the container, forcing some up into the tube to a height of 760 mm. This gives rise to the pressure unit “mm Hg” which also goes by the name of “torr”. Higher atmospheric pressure, higher mm Hg

Manometer

Manometers measure the pressure of a gas in a flask and typically come in two forms: Closed end and Open end. In figure (a) there is no pressure above the Hg in the U tube. The pressure in the flask is determined by measuring the height h 1 (at zero pressure, the levels would be equal). In figures (b) and (c), the right end is open to atmosphere which has to factored into the calculation as shown above.

Summary of various Pressure units: 1 atm = 760 mm Hg = 760 torr = 101,325 Pa = kPa = 14.7 psi Pa = Pascal, the metric unit of pressure. 1 Pa is small, about the pressure a dollar bill exerts when lying flat. Therefore, kilopascals are typically used.

Standard Temperature and Pressure (STP) To ensure that chemists can compare results, a “standard” temperature and pressure have been chosen: Standard temperature = 0 ºC (273 K) and 760 mm Hg (1 atm)

7-2 The Gas Laws (Sections 13.2, 13.3) PT (Pressure and Temperature) or Gay-Lussac’s Law Problem:The pressure in a hydrogen tank is 3000 psi. If the temperature is cooled from 20°C to 15°C, a) will the pressure increase or decrease? Note: In gas law calculations, temperatures MUST be in units of Kelvin ( K) !!!!!!!!!!!!!!!

What would the final pressure be? 3000 (288) / 293 = 2949 psi

PT (Pressure and Temperature) or Gay-Lussac’s Law y = mx T P

PV (Pressure and Volume) or Boyle’s Law animated.gif animated.gif

PV (Pressure and Volume) or Boyle’s Law V P 1/P

Problem:The pressure of a 1.5L sample of gas changes from 56 torr to 150 torr. a) Did the volume increase or decrease? 56 (1.5) = 150 (.56)

VT (Volume and Temperature) or Charles’s Law _Gay-Lussac%27s_Law_animated.gif _Gay-Lussac%27s_Law_animated.gif

V V -273 T ( o C) T (K)

A sample of gas at 15°C has a volume of 2.58 L. The temperature is then raised to 38°C (at constant pressure) what is the new volume? Note: temp. must be in Kelvin!!!!!! V 1 T 1 = V 2 T (288) = 2.39 (311)

PVT (Pressure, Volume, and Temperature) or Combined Gas Law P 1 P 2 and V 1 V 2 and P 1 V 1 = P 2 V 2 T 1 = T 2 Put it all together into one equation: P 1 V 1 = n = P 2 V 2 T 1 T 2 Where n is a constant

100 ml of N 2(g) at 27 o C and 1.13 atm. of Pressure changes to 500 ml of gas at 1.89 atm. of pressure. What is the Temp. of the gas? Temp. must be in Kelvin!!!!!!!!!!!!!!!!!!!!!!!! P 1 V 1 = n = P 2 V 2 T 1 T (100) = 1.89 (500)