Some more gas laws Combined Ideal Grahams Combined
Some more gas laws… Combined, Ideal & Graham’s!!
Combined gas law P 1 V 1 = P 2 V 2 T 1 n 1 T 2 n 2
Combined #1 • A balloon is filled with 3. 47 g of He that occupies 4. 25 L at 24°C and 762 mm. Hg. You put the balloon in a freezer at – 23. 3°C, and you let it sit for an hour. When you come back you find that the
Combined #1 (cont. ) atmospheric pressure inside the freezer has dropped to 755 mm. Hg, and you only have 3. 00 g of the He left in the balloon. What is the volume of the balloon after the hour spent in the freezer?
Combined #2 • How many m. L of oxygen gas at STP will be produced when 5. 25 g of potassium chlorate are heated and the temperature stabilizes at 27. 5 °C while the atmospheric pressure is 14. 6 psi? Assume none of the oxygen is lost.
Ideal gas law • Considers ideal conditions and applies them to the combined gas law • Ideal P? 1 atm • Ideal V? 22. 4 L • Ideal T? 273 K • Ideal n? 1 mol
Ideal gas law • So…when we plug those into the combined gas law… 1 atm*22. 4 L =PV 273 K*1 mol Tn
Ideal gas law • Or… 0. 0821 L-atm=PV K-mol Tn
Ideal gas law • Or…let 0. 0821 L-atm/K-mol be represented by R… PV R= Tn
Ideal #1 • How many m. L of oxygen gas will be produced when 5. 25 g of potassium chlorate are heated and the temperature stabilizes at 27. 5°C and the atmospheric pressure is 14. 6 psi? Assume none of the oxygen is lost.
Ideal #2 • If 20. 0 L of CO 2 are retrieved from the combustion of propane at 150°C and 764. 1 mm. Hg, how many grams of propane were combusted?
Graham’s law • Scottish chap who described effusion rate of gases relative to molar mass
Graham’s law • Effusion—gas flow from higher to lower concentration/pressure through a tiny orifice • Rate—something over time…in this case amount effused divided by time
Graham’s law • Since temperature is proportional to the kinetic energy of gas molecules, the kinetic energy of the two gas samples is also proportional. • Thus, you can write: • KE 1 = KE 2
Graham’s law • Since KE = (1/2)mv 2, you can write: • (1/2)m 1 v 12 = (1/2) m 2 v 22 • Or… m 1 v 12 = m 2 v 22 • Or… m 1 = v 2 m 2 v 1
Graham’s law • But…we tend to look at the rate of effusion rather than the volume of the gas in these problems • So… m 1 = rate 2 m 2 rate 1 • Rate will be V/t
Graham’s #1 • If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?
Graham’s #2 • Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape?
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