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Ideal Gas Law and Avogadro’s Hypothesis

Ideal Gas Law and Avogadro’s Hypothesis. Avogadro’s Hypothesis. Equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules Mathematically, n 1 n 2 V 1 V 2. =.

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Ideal Gas Law and Avogadro’s Hypothesis

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  1. Ideal Gas Law and Avogadro’s Hypothesis

  2. Avogadro’s Hypothesis • Equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules Mathematically, n1 n2 V1 V2 =

  3. One mole of gas occupies the same volume as one mole of another gas at the same temperature and pressure • Molar Volume is measured in L/mol • The molar volume of an ideal gas at STP is 22.4 L/mol

  4. Using Avogadro’s law to determine the volume of gases Example: What is the volume of 3.0 mol of nitrous oxide at STP? • Use: V2 = 67 L n1 n2 V1 V2 = 1mol 3.0 mol 22.4 V2 =

  5. Ideal Gas Law Do real gases always behave like ideal gases? • The particles of a real gas have a significant volume • Molecules do attract each other • Molecules do not necessarily move in straight lines • Collisions are not completely elastic

  6. Ideal Gas Law PV = nRT where R (universal gas constant) = 8.314 kPa L Mol K

  7. Guidelines for Using the Ideal Gas Law • Always convert the temperature to K • Always convert mass to moles. • Always convert volumes to Litres • Convert the pressures to kPa.

  8. Example 1: Use the ideal gas law to calculate the molar volume of a gas at SATP. The conditions for SATP are 298 K and 100 kPa What is given: P = 100 kPa n = 1.00 mol R = 8.314 kPaL/molK T = 298 K

  9. Step 2: Apply PV = nRT V= nRT P = 1.00 mol(8.314)(298) 100 kPa = 24.8 L The molar volume is 24.8 L/mol, which is slightly greater than 22.4 L/mol. This is expected since the pressure has decreased and the temperature has increased.

  10. Example 2: Dentists sometimes use laughing gas (N20) to keep patients relaxed during dental procedures. A cylinder of laughing gas has a diameter of 23.0 cm and a height of 140 cm. The pressure is 108 kPa and the temperature is 294 K. How many grams of laughing gas are in the cylinder?

  11. What is given? P = 108 kPa V = πr2h, where r = 11.5 cm and h=140 cm R = 8.314 kPaL/molK T = 294 K

  12. Step 1: Find the volume of the cylinder V = πr2h = π (11.52)(140) = 5.73 x 104 cm3 = 57.3 L

  13. Step 2: Use the gas law to find the number of moles PV = nRT (108)(57.3) = n (8.314)(294) n= 2.53 mol

  14. Step 3: Calculate the mass of the gas Molar mass of N20 = 2 (14.01)+16.00 = 44.02 g/mol 2.53 mol of N20 = 2.53 (44.02 g/mol) = 111 g

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