Graham’s Law and The Ideal Gas Law

1. Graham’s Law and The Ideal Gas Law

2. Ideal Gas Law The ideal gas law is the mathematical relationship among pressure, volume, temperature, and the number of moles of a gas. To use the ideal gas laws, gases must be assumed to be in an “ideal state.”

3. Ideal Gas Law The ideal gas law equation is: PV = nRT Where P = pressure (in atm, mmHg, or kPa) V = volume (in L) N = number of moles R = ideal gas constant T = temperature (in K)

4. Ideal Gas Law • R, the ideal gas constant, will differ depending on which units of pressure you use. • R = 0.0821 L x atm/mole x K • R = 62.4 L x mmHg/mol x K • R = 8.314 L x kPa/mol x K You will be given all R values, but you need to know when to use them!!

5. Example 1 What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0L container at 298 K? PV = nRT P x 10.0 = .500 x 0.0821 x 298 10.0 x P = 12.2329 10 10 P = 1.22 atm

6. Example 2 If 3.7 moles of propane are at a temperature of 28 ºC and are under 154.2 kPa of pressure, what volume does the sample occupy? PV = nRT 154.2 kPa x V = 3.7 x 8.314 x 301 154 x V = 9259.3 V = 60.1L 154 154

7. What pressure could be reached when ¼ lb of dry ice is placed in this 2 liter bottle? Temperature that night was 86 °F (30 °C) Facts:2 Liter bottle ¼ lb = 454 g ÷ 4 = 114g PV = nRT or P = nRT + 273 K V CO2= 12g/mol + 2*16g/mol = 44 g/mol 114 g = mol 1 mol 2.6 44 g

8. What pressure could be reached when ¼ lb of dry ice is placed in this 2 liter bottle? Temperature that night was 86 °F (30 °C) n R T 2.6 mol x 0.0821 atm*L x 303 K mol*K P = 2.0 L V P = 32.3 atmospheres 32.3 atm = psi 475 14.7 psi 1 atm

10. Graham’s Law • Remember that diffusion is the gradual mixing of two gases due to their random motion. • Effusion is the process where molecules confined to a container pass through a tiny opening of a container.

11. Graham’s Law Graham’s law of effusion states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

12. Graham’s Law So the greater the molar mass (i.e. the larger the gas molecule), the slower it will effuse. The temperature and pressure have to be the same to be able to compare effusion rates!

13. Graham’s Law The mathematical formula for graham’s law is: Rate of effusion of A = √ MB Rate of effusion of B = √ MA Where M is the molar mass of the gas.

14. Graham’s Law • The mathematical formula for Graham’s Law will let you compare the rate of effusion of two gases, and give you a numeric value for the comparison. • Examples: which gas will effuse faster, and how much faster (2 times, 4 times, ect).

15. Example 1: • Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Rate of effusion of H2 = √32.0 g/mol Rate of effusion of O2√2.0 g/mol

16. = √32/2 = 4.0 Hydrogen effuses 4 times faster than oxygen.

17. Example 2 • A nitrogen molecule travels at about 505 m/s at room temperature. Find the velocity of a helium atom at the same temperature. Rate of effusion of He= √28.0 g/mol Rate of effusion of N √4.0 g/mol

18. = √28/4 = 2.65 Helium effuses 2.65 times faster than nitrogen. If the velocity of nitrogen is 505 m/s, and helium effuses 2.65 times faster, then the velocity of helium is 505 x 2.65 = 1338 m/s