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### fnt15

Course: PHY 7A, Winter 2008
School: UC Davis
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Word Count: 922

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from FNTs DL 15 FNT 2) What is the physical basis for the difference in measured values of C p and Cv for a gas? The physical reason why Cp and Cv have different values is that Cv is measured when our volume is constant and unchanging, so no work is done on or by the system. Heat capacity is equal to the heat gained or lost by the system, divided by the change in temperature [ C = Q/(change in T) ]. When no work...

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from FNTs DL 15 FNT 2) What is the physical basis for the difference in measured values of C p and Cv for a gas? The physical reason why Cp and Cv have different values is that Cv is measured when our volume is constant and unchanging, so no work is done on or by the system. Heat capacity is equal to the heat gained or lost by the system, divided by the change in temperature [ C = Q/(change in T) ]. When no work is done, our change in internal energy (delta U) is equal to the heat. When we calculate Cp, our system remains at constant pressure but our volume can change. So whenever we add Q (and increase T), our volume changes (PV = nRT), which means work is done ( W = -P *(change in Volume)). In this case, our heat is equal to our change in internal energy minus whatever energy changes are due to work. Using PV = nRT and W = - P*(delta V) and p(delta V) = nR (delta T) , we can find the relationship between Cpand Cv to be: Cp = Cv + nR so Cp > Cv always 3) (I) Q =0 (II) change in internal energy =0 (III) W=0 a)An isochoric (constant volume) process: III b/c of Thermodynamic model [W = P(change in V)] b)An isobaric (constant pressure) process: None of these are necessarily true c)An isothermal (constant temperature) process: II, only if an ideal gas, b/c of Model of Thermal Energy [Eth=(#modes/atom)RT], and b/c of Thermodynamic model [(change in U) = (change in Eth) + (change in Eb)] d)A process where no heat is transferred to/from the system (adiabatic): I b/c of Thermodynamic model e)A process where no work is transferred to/from the system: III b/c of Thermodynamic model 4) a) B>A>C>D b) B=A=C=D because the four processes have the same initial and final states, so the changes in their state variables, including internal energy, will be the same. If work is done on the system, change in volume is negative (final-initial) so W>0 and energy is gained If work is done by the system, change in volume is positive (final - initial) so W<0 and energy is lost 5) (a) is true, this gas is monatomic, there are only three modes. We can figure out the number of modes with the given information using the model of Thermal Energy: (change in Eth) = (#modes/atom)R(change in T) 6) a) I - Temperature is an indicator for internal energy, if temperature increases then internal energy increases. (change in Eth) = (#modes/atom)R(change in T) (change in U) = (change in Eth) + (change in Eb) III b) - For a gas, work done is always indicated by a change in volume. Work = - (area under graph) = - P * (change in V) W = -1150 J if you care to calculate it reminder: when we say ideal gas, we mean we can apply PV = nRT because we are close enough to approximate ideal conditions: 1) our gas has no mass ( is very small compared to volume) 2) there are no interactions between molecules (interactions are too small to have a great affect) 7) a) W = -1000 J over one complete cycle. The amount of work done from state I to II is 2200 J (it is negative because the volume is increasing), the amount of work done from state II to III is 0, since there is no change in volume, and the amount of work done from III to I is 1200 J, since the volume is decreasing. So the total W = -2200 +1200 = -1000 J. b) the change in internal energy of one complete cycle is zero, since (change in U) = Uf - Ui and we begin and end at the same values. Remember that the change in a state function (U is a state function) doesn't depend on the process by which it happens, it only depends on the initial and final conditions. c) During the process from III to I, we calculated in part a) that W = 1200 J, now we are told that Q = -3000 J. So for the process between III and I, the change in U = Q + W = -3000J + 1200J = -1800J We know that (change in U) = (change in E th) + (change in Eb), but there is no change in bond energy in this case because we remain in the gas state. If we can figure out the number of modes for a gas, we should be able to figure out whether its monatomic, diatomic, etc. (change in U) = (1/2)(#modes/atom)R (change in T) We don't know the change in T, but we do know the pressure and the change in volume, and we know that this is an ideal gas, so P(change in V) = nR(change in T) making that substitution: (change in U) = (1/2) (#modes/atom)R(P(change in V)/nR) -1800J = (1/2)(#modes/atom)1200 J -----> we wind up with 3 modes per atom, so this is a monatomic ideal gas. 8) a)W = Area under the curve = -1150J, we know its negative because the change in volume is positive, this is energy our system is losing. b) (change in U) = (change in Eth) + (change in Eb), since this is a gas and no phase change is occuring, there is no change in bond energy, so (change in U) =(change in Eth) = (1/2)(#modes/atom)R(change in T) = 332 J c) (change in U) = Q + W so Q = (change in U) - W = 332 - (-1150) = 1482 J
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