Chapter 11: The Behavior of Gases

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Presentation transcript:

Chapter 11: The Behavior of Gases Section 2: The Gas Laws

Robert Boyle- Boyle’s Law Pressure and Volume Boyle’s Law states that the pressure and volume of a gas at constant temperature are inversely proportional P1V1 = P2V2   If pressure of gas ↑ - volume of gas ↓ If pressure of gas ↓ - volume of gas ↑ If volume of gas ↑ - pressure of gas ↓ If volume of gas ↓ - pressure of gas ↑ According to the kinetic theory, if the temperature of a gas is constant and the gas is compressed, its pressure must rise

http://education- portal http://education- portal.com/academy/lesson/boyles- charles-gay-lussacs-laws-pressure- volume-and-temperature- relationships.html#lesson

Inverse Proportion

Boyle’s Law Practice problems 1) 60.0 mL of methane gas at 700 mm Hg ? mL of methane gas at 760 mm Hg   Use P1V1 = P2V2 (700 mm Hg) 60 mL = (760 mm Hg) ? mL ? mL = 55.3 mL

Boyles Practice 2) 1000 L at 1.0 atm ? L at 5.0 atm Use P1V1 = P2V2   Use P1V1 = P2V2 1.0 atm (1000 L) = 5.0 atm (? L ) ? L = 200 L

Jacques Charles -Charles’s Law: Temperature and Volume Charles’s Law states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature V1 = V2 T1 T2 If volume increase, temperature increases (vice-versa) If volume decreases, temperature decreases (vice-versa)

http://education- portal http://education- portal.com/academy/lesson/charles-law- gas-pressure-and-temperature- relationship.html#lesson

Charles Law direct proportion

Charles’s Law Practice Problems 1) 3.0 L of He at 310 K ? L of He at 340 K use V1 = V2 3.0 L = ? L T1 T2 310 K 340 K   ? L = (340K) 3.0L = 3.29 L 310 K

Charles Law Problems Cont 2) 4.0 L of methane at 30.0°C ? L of methane at 0°C   First convert °C to K: (Tk = Tc + 273) 0°C = 273 K 30.0°C = 30 + 273 = 303K use V1 = V2 4.0 L = ? L T1 T2 303K 273 K ? L = (273 K) 4.0L = 3.6 L 303 K

COMBINED GAS LAW Combination of Boyle’s Law and Charles’s Law You can solve for one variable (P, V or T), regardless of the order in which the other two changed.   P1V1 = P2V2 T1 T2 The set conditions for the combined gas law problems are 0.00 °C and 1 atm- called STANDARD TEMPERATURE AND PRESSURE or STP

1) 2.7 L N2 at 121 kPa and 288 K ? L N2 at 202 kPa and 303K Use P1V1 = P2V2 T1 T2 121 kPa (2.7 L) = 202kPa ( ? L ) 288 K 303 K 326.7 L ·kPa = 202kPa ( ? L) 288 K 303 K 1.13 L ·kPa = 0.66 kPa (? L) K K ? L = 1.70 L

LAW OF COMBINING GAS VOLUMES The observation that at the same temperature and pressure, volumes of gases combine or decompose in ratios of small whole numbers Gas Reactions show that volumes of gases always react in ratios of small whole numbers http://education- portal.com/academy/lesson/daltons-law-of- partial-pressures-calculating-partial-total- pressures.html#lesson

Amedeo Avogadro First to interpret the law of combining volumes in terms of interacting particles The volume of a gas at a given temperature and pressure depends on the number of gas particles AVOGRADO’S PRINCIPLE states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles http://education- portal.com/academy/lesson/molar-volume- using-avogadros-law-to-calculate-the- quantity-or-volume-of-a-gas.html#lesson