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Review of the Gas Laws

Review of the Gas Laws. PV = nRT. PV = nRT. Boyle’s Law (isothermal & fixed amount) Charles’s Law (isobaric & fixed amount) Avogadro’s Law (isothermal & isobaric) ????? Law (isochoric & fixed amount) ????? Law (isothermal & isochoric) ????? Law (isobaric & isochoric). Boyle’s Law.

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Review of the Gas Laws

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  1. Review of the Gas Laws PV = nRT

  2. PV = nRT • Boyle’s Law (isothermal & fixed amount) • Charles’s Law (isobaric & fixed amount) • Avogadro’s Law (isothermal & isobaric) • ????? Law (isochoric & fixed amount) • ????? Law (isothermal & isochoric) • ????? Law (isobaric & isochoric)

  3. Boyle’s Law • Pressure and volume are inversely proportional. • As pressure increases, volume decreases. • If pressure increases by 2x, volume cuts in half.

  4. Charles’s Law • Temperature and volume are directly proportional. • As temperature increases, volume also increases. • If temperature increases by 2x, volume also doubles. • Temperature must be measured in Kelvin.

  5. Avogadro’s Law • Moles of gas and volume are directly proportional. • As the number of moles increases, the volume also increases. • If the number of moles increases by 2x, the volume also doubles.

  6. ????? Lawisothermal & isochoric • Moles of gas and pressure are directly proportional. • As the moles of gas increase, the pressure also increases. • If the number of moles of gas increases by 2x, the pressure also doubles.

  7. ????? Lawisobaric & isochoric • Moles of gas and temperature are inversely proportional. • As the number of moles of gas increase, the temperature decreases.

  8. Units of Pressure • 1 atm = 760 torr = 760 mmHg • 1 atm = 101.325 kPa • 1 bar = 105 Pa = 100 kPa • 1 Pa =

  9. How does 1 atm = 101.325 kPa? Let the area of the base of a cylinder = 1 m2 Volume = area x height = 1 m2 x 0.76 m = 0.76 m3 Convert volume to cubic centimeters. Use the density of mercury and the acceleration due to gravity to calculate the weight of mercury in the column.

  10. ContinueHow does 1 atm = 101.325 kPa? Pressure is force (or weight) per unit area. Divide the weight of mercury by the area it is resting on.

  11. Let the area of the cylinder = 1cm2

  12. Barometric Formula As elevation increases, the height of the atmosphere decreases and its pressure decreases. Check units.

  13. Continue Derivation of Barometric Formula Write in differential form. density Rewrite PV = nRT as Therefore,

  14. Continue Derivation of Barometric Formula Substitute the expression for density into the differential eqn. Divide both sides of the above equation by P and integrate.

  15. Continue Derivation of Barometric Formula Integration of the left side and moving the constants outside the integral on the right side of the differential equation gives,

  16. Continue Derivation of Barometric Formula Evaluating the integral between the limits of Pi at zero height and Pf at height h, gives

  17. Sample Problem Using the Barometric Formula

  18. Dalton’s Law of Partial Pressures Pressure is additive. Write each pressure as,

  19. ContinueDalton’s Law of Partial Pressures Multiply through by V (the combined volume of the gases) and divide by R T. Moles are indeed additive.

  20. Mole Fraction & Partial Pressure Therefore,

  21. ContinueMole Fraction & Partial Pressure Show that

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