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Physics GRE

mflynn210 Englisch 58 Karten

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  1. 1\(k\) Boltzmann constant\[ 1.381 \times 10^{-23} J/K \]
    \[= 8.617 \times 10^{-5} eV/K \]
  2. 2Avogadro's Number \(N_A\)\[ 6.022 \times 10^{23} \]
  3. 3Gas Constant \(R\)\[8.315 \frac{J}{mol \cdot K}\]
  4. 4Planck's constant \(h\)\[ 6.626 \times 10^{-34} J \cdot s\] \[=4.136 \times 10^{-15} eV \cdot s \]
  5. 5speed of light \(c\)\[2.998 \times 10^8 m/s \]
  6. 6Gravitation constant \(G\)\[ 6.673 \times 10^{-11} \frac{N \cdot m^2}{kg^2} \]
  7. 7charge of electron \(e\)\[ 1.602 \times 10^{-19} C \]
  8. 81 eV in Joules\[ 1.602 \times 10^{-19} J \]
  9. 91 atomic mass unit \(u\)\[ 1.661 \times 10^{-27} kg \]
  10. 10mass of electron \(m_e\)\[ 9.109 \times 10^{-31} kg \]
  11. 11mass of proton \(m_p\)\[ 1.673 \times 10^{-27} kg\]
  12. 12ideal gas lawPV = nRT
  13. 13Derive average translational kinetic energy of an ideal gasImagine the molecules bouncing off the side of a box. Take advantage of the fact that they must completely reverse their momentum i.e. \(\Delta p = -2mv_x \) in the amount of time that it takes to traverse the distance, which will be \(\Delta t = 2 dx/v_x \) plugging into a force equation with pressure we get \[ F_{normal} = -PA = -\frac{-2m v_x} { 2 dx /v_x }.\] Averaging over all molecules and using the ideal gas law we get something like \[PV = NkT = N m \bar{v_x^2}.\] So \(\bar{mv_x^2} = kT\), for all directions. Adding this up for all directions gives us \[\bar{K}_{trans} = \frac{1}{2}m\bar{(v_x^2+v_y^2 + v_z^2)} = \frac{3}{2} k T \]
  14. 14Translational kinetic energy of a room temperature air molecule\[ \frac{3}{2}kT = \frac{3}{2}(1.38 \times 10^{-23} J/K)(300 K) = 6.21 \times 10^{-21} J \] \[ = 0.03876404494 eV \]
  15. 15\(v_{rms}\) root mean squared velocity\[ \frac{1}{2} m v^2 = \frac{3}{2} k T \ \Rightarrow \ v_{rms} = \sqrt{\frac{3kT}{m}} \]
  16. 16Equipartition TheoremAt temperature \(T\), the average energy of any quadratic degree of freedom is \( \frac{1}{2} k T .\)
  17. 17Total thermal energy\[ U_{thermal} = N f \cdot \frac{1}{2}kT.\]
  18. 18Vibrational degrees of freedom in a diatomic molecule2, one for kinetic energy, one for potential
  19. 19Degrees of freedom of \(N_2\) and \(O_2\) at room temperature5, 3 translational and 2 rotational. The polar rotational axis does not count "because quantum mechanics" and vibrational modes are frozen out at room temperature
  20. 20degrees of freedom of a room temperature solid6 vibrational, 2 in each direction: kinetic and potential
  21. 21Why can't degrees of freedom be analyzed for a liquidAverage translational energy gets \(\frac{3}{2} kT \), but other intermolecular energies don't have nice quadratic functions, and so don't get exactly \(\frac{1}{2} k T\) from the partition function.
  22. 22How to test equi-partition theoremAdd some energy to a system, measure how the temperature changes, compare to \[U_{thermal} = N f \frac{1}{2} k T \]
  23. 23HeatSpontaneous flow of energy from one object to another, caused by a difference in temperature
  24. 24WorkThe transfer of energy to a system by means other than heat
  25. 25First Law of ThermodynamicsHeat is conserved via \[ \Delta U = Q + W, \] the change in energy is equal to the heat added plus work done. Just a restatement of conservation of energy.
  26. 26\[ [ J ] \]\[ kg \cdot m^2 / s^2 = N \cdot m\]
  27. 271 calorie in Joules\(1 \ \text{cal } =\ 4.186 \text{ Joules} \)
  28. 28Quasistatic approximationActions on a gas happen slow enough that the gas has time to react and re-equilibrate. Would violate if we punched gas with a piston moving close to the speed of sound in the gas.
  29. 29If a thermodynamic system has its volume increased at constant pressure, how much work is done on it, how much heat is added, what is the change in thermal energy?Work: \( W = - P \Delta V \). Thermal energy: \( \Delta U = \frac{f}{2} P \Delta V \). Heat Added: \( Q = \Delta U - W = \frac{f+2}{2} P \Delta V \)
  30. 30Derive adiabatic expansion relation.\(Q = 0\), so \(U = W\). This translates to the equation \(N f \frac{1}{2}k dT = - P dV.\) Using the ideal gas law and separating \(T\) and \(V\) we find that \[ \frac{f}{2} \int \frac{dT}{T} = - \int \frac{dV}{V}.\] We integrate to get \[ \frac{f}{2} \ln T = - \ln V + C .\] Putting the constants on one side and getting rid of \(T \) with the ideal gas law, we get \[ P V^{\frac{f + 2}{f}} = \text{Constant} \]
  31. 31Derive Dalton's Law of Partial PressuresFrom the ideal gas law \(PV=NkT\), if we have multiple gases, say a vapor and a gas, the ideal gas law becomes \[ P = \frac{N_{vapor}}{V}kT + \frac{N_{gas}}{V}kT = P_{vapor} + P_{gas}\]
  32. 32Heat Capacity\[ \frac{Q}{\Delta T} \]
  33. 33Specific heat capacity\[ \frac{Q}{m\Delta T} \]
  34. 34How does heat capacity depend on the circumstandces\[ C = \frac{Q}{\Delta T} = \frac{\Delta U - W}{\Delta T }. \] It depends on how much work you are doing on the material, and therefore could be anything.
  35. 35Heat Capacity at Constant Volume\[ C_V = \left ( \frac{\Delta U}{\Delta T} \right )_V = \left ( \frac{\partial U}{\partial T} \right )_V \]
    No work is being done on material since \(\Delta V= 0\) and \( W = -\int P \partial V \).
  36. 36Heat Capacity at Constant Pressure\[C_P = \left ( \frac{\Delta U - (-P\Delta V)}{\Delta T} \right )_P = \left ( \frac{\partial U}{\partial T} \right )_P + P \left ( \frac{\partial V}{\partial T} \right )_P .\] More relevant than at constant volume because most materials expand as they are heated, heat added must account for this negative work so \( C_P > C_V \).
  37. 37\(C_V\) for a material with \(f\) purely quadratic degrees of freedom\( \Delta U = Nf \frac{1}{2} kT \) so \[ C_V = \left ( \frac{\partial U}{\partial T} \right )_V = \frac{N f k}{2}. \] For a monoatomic gas \(C_V = \frac{3}{2} Nk \), for a solid \(C_V = 3 Nk\). However, these degrees of freedom tend to freeze out when the temperature gets lower, so not totally accurate (especially for solids).
  38. 38Specific heat a constant pressure for a material with \(f\) purely quadratic degrees of freedom\( \left ( \frac{\partial U}{\partial T} \right )_P = C_V \), and \( \left ( \frac{ \partial V} {\partial T} \right )_P = \frac{Nk}{P} \). So \[ C_P = C_V + Nk = Nk \left ( \frac{f + 2}{2} \right ) \]
  39. 39How can you get the adiabatic exponent \(\gamma = \frac{f+2}{f}\) from the specific heat and constant pressure \(C_P\) and the specific heat at constant volume \(C_V\)?\[ \frac{C_P}{C_V} = \frac{ Nk \left ( \frac{f + 2}{2} \right ) }{ Nkf/2} = \frac{f + 2}{f} = \gamma \]
  40. 40Latent HeatThe amount of heat needed to be added to conduct a phase transformation per mass of the substance \[ \frac{Q}{m} \]
  41. 41Latent heat for melting ice\( 333 J/g \)
  42. 42Latent heat for boiling water\(2260 J/g\)
  43. 43Define Enthalpy \( H \). Why is it useful?\[ H = U + PV .\] The amount of energy needed to create a substance and expand a space for it to fill at constant pressure \(P\). Useful for doing computations at constant pressure. Can be used to isolate non-compression work i.e. \[ \Delta U = Q + (-P \Delta V) + W_{other} \] so \( \Delta H = Q + W_{other} \). Also results in a simple formula for specific heat at constant pressure: \[ C_P = \left ( \frac{\partial H}{\partial T} \right )_P .\]
  44. 44Fourier Heat Conduction Law\[ \frac{Q}{\Delta T} = -k_t A \frac{ d T}{dx} .\]
  45. 455 orders of thermal conductivity\(k_t:\) air, 0.026; wood, 0.08; water, 0.6; glass, 0.8; iron 80; copper 400.
  46. 46Derive the mean free path in a dilute gasMean Free Path: When the volume of the path cylinder is equal to the average volume per molecule. The cylinder's radius is \(2 r \) because this is the distance between 2 particles of radius \(r\) where they will collide. \[ V_{path} = \pi (2r)^2 l\] \[ V_{particle} = \frac{V}{N} \] \[ l \approx \frac{1}{4 \pi r^2} \frac{V}{N} .\] This is not always exact because \(r\) is not always well defined (not spherical, etc.)
  47. 47Compute the average time between collisions for air at room temperature.For a nitrogen or oxygen molecule \(r = 1.5\) angstroms \( = 1.5 \times 10^{-10} m\). Volume per particle is \(V/N = kT/P = 4 \times 10^{-26}\ m^3\), so the average free path is \[ l = \frac{4 \times 10^{-26} m^3}{ 4 \pi (1.5 \times 10^{-10}\ m)^2} = 1.5 \times 10^{-7}\ m .\] From here we divide by the root mean squared speed \[ \Delta t = \frac{1.5 \times \ 10^{-7} \ m}{500 \ m/s} = 3 \times 10^{-10} \ s. \]
  48. 48Atmospheric pressure in standard units1 atm = \(1.013 \times 10^5\ N/m^2 \).
  49. 49Uranium Fuel cycleExploration/Mining (U/Th) =>
    Milling (U3O8) =>
    Processing (UF6) =>
    Enrichment (enriched UF6) =>
    Fuel Fabrication (UO2 pellets) =>
    Reactor =>
    Interim Storage =>
    Reprocessing (U233, Plutonium, U235)=>
    Waste disposal
  50. 506 Reference Reactors1. Boiling Water Reactor
    2. Pressure Tube graphite reactor
    3. Pressurized Water Reactor
    4. Pressurized Heavy-water reactor
    5. High Temperature gas-cooled reactor
    6. Liquid-metal fast-breeder reactor
  51. 51Fick's Law\[ J_x = -D \frac{dn}{dx} \]
  52. 52Derive Fick's second lawDraw a pipe section, the change in the number of particles in a section of length \(\Delta x\) is the difference between the number received from the fluxes on either side: \[ \frac{ \Delta N}{\Delta t} = F_- A - F_+ A. \] Using Fick's First Law we can convert this to a density difference \[ \frac{\Delta N}{A\Delta x \Delta t} = D \left ( \frac{\frac{dn_+}{dx} - \frac{dn_-}{dx}}{\Delta x} \right ). \] Converting this to differential form we get \[ \frac{d n}{dt} = D \frac{d^2 n}{dx^2} .\]
  53. 53Multiplicity of an Einstein solid with \( N \) molecules and \( q \) units of energy\[ { N + q - 1\choose q } \]
  54. 54Fundamental assumption of statistical mechanicsIn an isolated system in thermal equilibrium, all accessible microstates are equally probable
  55. 55Stirling's approximation (no log)\[ N! \approx N^N e^{-N} \sqrt{2 \pi N} \]
  56. 56Stirling's Approximation\[ \ln N! \approx N \ln N - N \]
  57. 57Derive Stirling's approximation for \(\ln \Omega\) for a Large Einstein Solid\[ \ln \Omega \approx \ln (q + N)! - \ln q! - \ln N! \] \[ \approx (q+N) \ln (q+N) - (q+N) - q \ln q + q - N\ln N + N \] \[ = (q +N) \ln (q+N) - q \ln q - N \ln N \] Using Taylors approximation for \( q \gg N \): \[ \ln(q+N) = \ln q \left ( 1 + \frac{N}{q} \right ) \approx \ln q + \frac{N}{q} \] so \[ \ln \Omega \approx N \ln \frac{q}{N} + N + \frac{N^2}{q} .\] The last term goes to zero if \( q\) is very large, raising both sides we get: \[ \Omega \approx \left ( \frac{eq}{N} \right )^N .\]
  58. 58Gaussian of multiplicity of two large einstein solids\[ \Omega = \Omega_{max} e^{-N(2x/q)^2} \] with \( \Omega_{max} = (e/N)^{2N}(q/2)^{2N} \) and the sharpness is \[ \Delta x = \frac{q}{2 \sqrt N} \]

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Karteikarten am Physics GRE, erstellt von mflynn210 am 09/07/2015.
Erstellt von
mflynn210 · Jul 2015
Aktualisiert
7 Mär 2018
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