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HW Sol#1

HW Sol#1 - CHEM 131 K Lindenberg Fall 2007 Solutions to...

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Unformatted text preview: CHEM 131 K. Lindenberg Fall 2007 Solutions to Homework No. 1 Posted Monday October 8 These solutions to Homework No. 1 were written partly by Dario and partly by Jim. Note: they chose how to divide up the task, so the solutions are not in the order in which they were assigned. However, the solutions to all the problems are there. Chemistry 131, Fall 2007 Solutions to Homework No. 1 Problem 1.1. (a) True. (Levine p. 3). (b) False. Different phases of a pure substance may coexist, in which case the macroscopic properties of the system are not constant throughout the system and the system is not homogeneous (Levine p. 5). (c) False. The SI units of P , Nm- 2 (Levine eq. 1.8), are not the reciprocal of the units of V , m 3 (Levine p. 11), therefore the product PV is not dimensionless and does have units. (d) False. The statement does not provide enough information for us to conclude that the statement is true in general. If the process was isothermal and involved the same number of moles, then the statement would be true. (e) False. The density is ρ = m/V = ( m/n ) / ( V/n ) = M/V m . At fixed temperature and pressure the molar volume is also fixed, because V m = RT/P . However, the molar mass de- pends on the gas, so M/V m cannot be the same for all ideal gases. (f) True. The number of moles per unit volume is the recipro- cal of the molar volume, which is fixed for an ideal gas when temperature and pressure are specified (See part e). Problem 1.2. Boyle’s law describes how the volume of an ideal gas varies with pressure at constant T and n , while Charles’ law de- scribes how the volume of the gas varies with temperature at constant P and n . To apply Boyle’s law we must allow the pressure to vary, but to apply Charle’s law we must keep the pressure constant. Therefore the two laws cannot be used simultaneously to describe a process in which P , V , and T vary at the same time. So it does not make sense to divide one equation by the other. Problem 1.3. Treat the gases in the bulbs as two separate thermodynamic systems with the same volume. Write the equation of state for each system: P 1 V 1 = n 1 RT 1 and P 2 V 2 = n 2 RT 2 . We are told that V 1 = V 2 = V . Also, once equilibrium is reached, the pressures will be equal, P 1 = P 2 = P , because the bulbs are connected by the narrow tube. Therefore the right-hand side of the first equation of state must equal the right-hand side of the second, n 1 T 1 = n 2 T 2 . The temperatures are given, T 1 = 175 K and T 2 = 250 K. The total number of moles is given, n tot = n 1 + n 2 = 1 . 00 mol. Hence, n 1 = n tot T 2 / ( T 1 + T 2 ) = 0 . 59 mol, and n 2 = 0 . 41 mol. Intuitively it should make sense that the hotter bulb contains less gas, if the bulbs are...
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HW Sol#1 - CHEM 131 K Lindenberg Fall 2007 Solutions to...

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