Chem%20161-2009%20Exam%20II%20%2B%20answers

Chem%20161-2009%20Exam%20II%20%2B%20answers - Chemistry 161...

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Chemistry 161 Exam II October 28, 2009 Chem 161-2009 Exam II Hill, Petrucci et al. Chapter 5 - Gases Kinetic molecular theory/Effusion and Diffusion/Real gases 1. A 2.00 L container of gas has a pressure of 1.00 atm at 300 K. The temperature of the gas is halved to 150K, and the measured pressure of the same 2.00 Liter sample is 0.420 atm. Which of the following is the best explanation for these observations? A. Pressure is proportional to temperature for a fixed volume of gas. B. The molecules of the gas occupy a significant portion of the volume. C. The molecules of the gas have negligible volume of their own. D . The molecules have significant attractive forces at 150 K. E. The gas is closer to an ideal gas at 150 K. This question could have been better written if the first sentence would have been, “A 2.00 L container of ideal gas has a pressure of 1.00 atm at 300 K.” My solution is based on the assumption that the gas under the initial conditions is an ideal gas. Conditions 1 Conditions 2 2.00L 2.00L 1.00 atm 0.420 atm 300K 150K Conditions 1: PV = nRT (ideal gas law) n = PV/RT (1.00 atm x 2.00L)/(0.08206Latmdeg -1 mol -1 ) x 300K = 0.08124 mol Consider condition 1 as the reference point. The issue is, does the gas still behave ideally when the temperature is lowered from 300K to 150K and the pressure is decreased from 1.00 atm to 0.420 atm? Condition 2: V = nRT/P = (0.08124 mol x 0.08206Latmdeg -1 mol -1 x 150K)/0.420 atm V = 2.38L if the gas is an ideal gas. But the observed volume is 2.00L. A different volume than is expected based on the ideal gas law demonstrates that the gas under these final conditions is “real”. A lower volume at a lower temperature is consistent with the molecules having significant attractive forces, which is option “D”. A. False. Whereas this statement is true, it isn’t the best explanation for the observation of a real gas being formed. B. False. This statement would be true for a positive deviation of the ideal gas law (that is, if the actual pressure was greater than the calculated pressure, i.e., PV/nRT > 1) , but since what is observed is a negative deviation from the ideal gas law (the actual pressure is less than the calculated pressure, making PV/nRT < 1), then this statement doesn’t apply to this case. C. False. Whereas this statement is true, it isn’t the best explanation for the observation of a real gas being formed. D. True. See discussion above. E. False. This would be true if the initial conditions were a real gas, and the final conditions were an ideal gas. But since we began with an ideal gas, and finished with a real gas then this statement is False. This problem also could have been solved using the combination gas law.
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P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 At constant moles and volume, P 1 /T 1 = P 2 /T 2 But 1.00 atm/300K 0.420 atm/150K. If the gas under the final conditions was ideal, then the pressure would have been 0.500 atm.
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Chem%20161-2009%20Exam%20II%20%2B%20answers - Chemistry 161...

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