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CHM 1311 Lecture Notes - Lecture 18: Internal Energy, Kinetic Energy, Potential Energy

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24 Nov 2016

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University of Ottawa

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Document Summary

Energy: capacity to do work or to produce heat; e(cid:374)e(cid:396)gy (cid:272)a(cid:374) (cid:271)e t(cid:396)a(cid:374)sfe(cid:396)(cid:396)ed (cid:271)ut (cid:272)a(cid:374)"t (cid:271)e (cid:272)(cid:396)eated. Energy flowing into a system u > 0; energy flowing out of a system u < 0. First law of thermodynamics: the total energy of the universe if constant: u is energy of the system (textbook is e) Internal energy of the system (u or e): the internal energy of a system is the sum of the kinetic and potential energies in the system. Kinetic energy: molecular or atomic motion, movement of electrons. Potential energy: attractive or repulsive interaction btwn electrons, nuclei, atoms, and molecules. Heat: flow of thermal energy btwn substances that are at different temps: heat absorbed by system: + value; heat absorbed by surroundings (system loses heat): - value. Work: movement against a force; w = force x distance: work done on system: + value; work done by system: - value.

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Related Questions

1.) Which statement isincorrect?

(a) Energy is the capacity todo work or to transfer heat.

(b) Kinetic energy is theenergy of motion.

(c) Potential energy is the energy that a system possesses by virtue ofits position or composition.

(d) A process that absorbsenergy from its surroundings is called exothermic.

(e) The Law of Conservation of Energy is another statement of the FirstLaw of Thermodynamics.

2. Which of the following statementsabout the first law of thermodynamics and energy is false?

(a)Kinetic energy cannot be convertedto potential energy.

(b)Kinetic energy =1/2mv².

(c)All the energy in the universe isconserved.

(d)A system can never decrease itsenergy.

(e)Potential energy is the energy ofposition or composition.

3. Whichone of the following thermodynamic quantities isnot a state function?

(a)work (b) enthalpy (c)entropy (d) internalenergy (e) free energy

10. In this question, mean field theory will be used to derive the van der Waals equation. Consider a gas of interacting, indistinguishable particles. In principle, the potential energy from these interactions depends upon the positions of the particles so detailed structural information is needed to describe it. This is a difficult problem. However, we can take a much simpler, albeit approximate, approach by treating the interactions with a one-body energy term. The physical argument goes as follows. The interaction (potential) energies in principle will vary among microstates. However, we expect fluctuations to be small. Let's replace the exact potential U in the partition function with its average value (that is, the "mean field") eMF divided among the particles, and assume each particle has the same contribution regardless of its position. In other words, rather than considering the particles as interacting with each other (via a potential that depends upon particle positions) we treat them as interacting with the mean field (which is independent of particle positions and other particles). In this way, EMF becomes a one-body energy term that can be included in the usual non-interacting partition function as an additional energy contribution. That is, the interacting system is approximated by the non-interacting partition function with a one-body energy term approximating the particle interactions. With this mean-field approximation, the energy of our particles becomes E=E;deal + EMF , in which Eideal is the energy of a particle in an ideal gas. Since these energy contributions are independent, the particle partition function is q = qidealqMF . Normally, qideal would be the product of partition functions for translation, vibration, rota- tion, etc. but because the volume dependence will be important, let's write qidealc(T)V where c(T) includes all the constants and temperature dependencies arising from vibrations and rotations, and where the volume factor in the translational partition function has been explicitly accounted for. The remaining part of the translational partition function in included in c(T). In the present case though, we imagine the particles each occupy some volume b, so the "effective volume" of our system is V- Nb. In other words, take qideal cT)(V - Nb) We don't know the value of eMF but physical arguments suggest it must vary with the av- erage distance between particles, which in turn varies with density, N/V. So, let's suppose a linear dependence and write EMF =-aN/V with a some unknown constant. A minus sign is included here because it is anticipated EME will be negative (attractive interaction) so this will make a a positive number. This one energy is the same for each particle so 4 a) Derive the expression for the average pressure and show the system obeys the van der Waals equation of state. Note: as defined b is the volume per particle not per mole of particles as usually tabulated. Similarly, as defined a is per particle not per mole squared as usually tabulated. So, write Nb as nb, and aN2 as an2 if you want to use a and b as the usually tabulated molar values b) Show that the internal energy and heat capacity are given by an - - U ==ideal and Cv=CV,ideal, where Eideal and Cvideal are the internal energy and heat capacity for the ideal gas, respectively. Show that the first relation is consistent with (UaVTan2/V2, a relation derived from thermodynamics. Give a physical argument to explain why the van der Waals gas (an interacting system) could have the same Cv as the ideal gas (a non-interacting system)