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Quark (2/13)
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I am a beginner in QE and I have a question. During I was reading about Output file, I saw a part where have it: "The calculation goes on step by step until convergence is reached. Then we show all the energies of all bands (eigenvalues of the Kohn-Shan orbitals) for each k-point, the Fermi energy, the total energy, the contributions of all the terms to the total energy, and the calculation time used by each subroutine of the program." Output file: ... iteration # 9 ecut= 100.00 Ry beta=0.70 Davidson diagonalization with overlap ethr = 1.00E-13, avg # of iterations = 1.0 negative rho (up, down): 1.151E-06 0.000E+00 total cpu time spent up to now is 28.4 secs End of self-consistent calculation k = 0.0000 0.0000 0.0000 ( 3909 PWs) bands (ev): -17.8569 -5.9595 -1.3957 -1.3957 4.4242 9.2594 9.9561 9.9561 k = 0.0000 0.0550 0.0000 ( 3909 PWs) bands (ev): -17.8034 -5.8955 -1.6155 -1.5115 4.4968 9.3266 9.9837 10.1886 ... -10.9411 -10.9411 -8.9101 1.6562 1.6562 12.4084 14.2298 14.2298 the Fermi energy is 1.6562 ev ! total energy = -22.80493136 Ry Harris-Foulkes estimate = -22.80493136 Ry estimated scf accuracy < 1.5E-15 Ry The total energy is the sum of the following terms: one-electron contribution = -44.02527202 Ry hartree contribution = 24.36506286 Ry xc contribution = -6.99363523 Ry ewald contribution = 3.85762303 Ry Dispersion Correction = -0.00870488 Ry smearing contrib. (-TS) = -0.00000512 Ry convergence has been achieved in 9 iterations Writing output data file grafeno.save init_run : 2.26s CPU 2.29s WALL ( 1 calls) electrons : 20.10s CPU 20.24s WALL ( 1 calls) ... Parallel routines fft_scatter : 4.35s CPU 4.44s WALL ( 23020 calls) PWSCF : 28.46s CPU 28.70s WALL This run was terminated on: 11:51: 0 5Sep2017 =------------------------------------------------------------------------------= JOB DONE. =------------------------------------------------------------------------------=
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So, is the true value for work +22, because work is done on the system? I used the 𐤃U = q+w. because I studied in a chemical class. However, I considered the contribution to work as negative. The book on the other hand, used a positive signal for work. The images is from the book.
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I have a question about the resolution of this exercise: I tried to solve this problem with the following steps: For the 1º law of thermodynamics, 𐤃U = q+w. As q = 𐤃H at constant pressure, I assumed that in this process q = - 15 kJ. As a result, 𐤃U = -15 - 22 (because was an expansion) and for this reason, 𐤃U = -37 kJ. However, the book has this resolution I understood why the book used this method, but I did not understand why my method is wrong.
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Scienc changed their profile photo
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Why in an irreversible expasion, the pressure is constant
Scienc posted a topic in Classical Physics
Why in an irreversible process, is the pressure constant? I understood why I should use Wirrev = -PΔV, but I did not understand why the pressure is constant in this process. -
Why does CH3CO2H have a greater nucleophilicity than CH3OH?
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An orbital can contain up to 2 electrons, and the electrons and orbitals are described by wave functions, are the 3 wave functions the same? for example, two electrons that are in the psi 2 0 0 orbital do they have the same wave function as that orbital?
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What does "air is saturated with water" or "air is saturated with something" and how can I relate this to vapor pressure?
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What is the physical meaning of enthalpy?
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The volume that 6 moles occupy 134.4l and 12 moles occupy 268.8 moles, since 1 mol = 22.4l, for this the gas needs to expand
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What is the difference between the work done by a gas and the workflow that the same gas performs to exist? why does the variation in the internal energy of the equation dH = dQ + nRT not consider the work done by the gas to exist (working flow) but consumes only the heat transfer? In the question below, the author considers dU = Q. He performs the calculation of the work done by the gas, separated from the internal energy (with an equation dH = dU + nRT ). If you know that in systems with constant volume dU = Q, but the gas volume varies (From 6 moles to 12 moles of gas), this does not imply an internal energy variation equal to dU = w + q (As stated in the 1st law of thermodynamics) and why consider dU = w + q and not dU = Q **? In practice or in what he did, the logic he did not use was. Question: “A constant volume calorimeter showed that the fuel loss in burning 1 mole of glucose is equal to -2559kJ in 298K, that is dU = -2555KJ”
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A variation of enthalpy at constant pressure is numerically equal to the variation of heat in a chemical reaction, but what is enthalpy in itself? what is its difference for internal energy, is it calculated as H = U + PV, or does that mean physically?
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Scienc started following 1. Thermodynamic work
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Guys, why in physics does the work done by the system have the POSITIVE sign, but in chemistry does the work done by the system have the NEGATIVE sign?
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Why does the equation τ = -n.R.T.ln (v_2 / v_1) only work for reversible processes?