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Homework+2+Solutions+_posted_ - BME 221 Homework#2...

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BME 221 Homework #2 Solutions (48 points) Due: Beginning of Class, Friday, October 2, 2009 1. Use the Maxwell distribution of speeds to estimate the fraction of N 2 molecules at 500K that have speeds in the range 290 to 300 m/s. Draw the Maxwell distribution curve of N 2 at this temperature and indicate on the graph the area under the curve corresponding to molecules in this speed range. (5 pts) 2 3/ 2 / 2 2 4 2 B mc k T B dN m c e dc N k T π π - = , where dN/N is the fraction of N 2 molecules, c = 290 m/s, m = mass of N 2 molecule = 4.65X10 -26 kg, k B = 1.38 X10 -23 J/K, T = 500K, and dc=10m/s. Then dN/N = 0.00884 0 500 1,000 1,500 molecular speed (m/s) number of molecules Speed Distribution of N2 Molecules at 500K fraction of molecules between 290-300m/s Note: The peak of the graph should be at c mp = (2RT/M) 1/2 = 545 m/s 2. a) Compartment A contains pure methane gas while Compartment B contains pure trichlorofluoromethane (CFCl 3 ) gas, both at the same temperature and pressure. The compartments are connected to Compartment C, which is initially a vacuum. Simultaneously, pin holes are opened such that the two gases can independently escape by effusion into Compartment C. After 10 min, what is the percent composition of the gas mixture in Compartment C? (3 pts)
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MW of CFCl 3 = 137.37g/mol MW of CH 4 = 16.04g/mol 3 4 3 4 137.37 2.93 16.04 CFCl CH CFCl CH MW r r MW = = = % CH 4 = 4 4 3 2.93/(2.93 1) CH CH CFCl r r r = + + = 74.5% % CFCl 3 = 25.5% b) Trichlorofluoromethane is also known by its commercial name Freon-11. Why is its use in the U.S. banned? (1 pt) CFCl 3 is a chlorofluorocarbon that was used as a refrigerant in air conditioners and refrigerators. When released into the atmosphere, CFCl 3 in the presence of ultraviolet light converts O 3 to O 2 , destroying the ozone layer that protects life from much of the sun’s UV light. In response to this environmental concern, chlorofluorocarbons were phased out in the U.S. beginning in 1978. 3. The binding enthalpy of a gas molecule to the surface of a protein at 25°C was determined to be 1 kJ/mol. Assume that a single gas molecule remains bound to the protein if the binding enthalpy is at least 100 times the kinetic energy of one gas molecule. Do you think that the gas molecule stays bound to the protein? Show your reasoning. (3 pts) No. Using the relation ½m<v 2 >=3/2*k b T, we find that the kinetic energy for a single gas molecule at 298K is 3/2*(1.38*10 -23 J/K)*(298K)=6.17*10 -21 J/molecule. The binding enthalpy is given to be 1 kJ/mol, or 1.66*10 -21 J/molecule. Thus, since the binding enthalpy for a single gas molecule is less than the kinetic energy for the gas molecule, the gas molecule will not remain bound to the protein. 4. a) State the first and second law of thermodynamics. (2 pts) 1 st : The internal energy of an isolated system is constant. OR Energy can neither be created nor destroyed.
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